LuhnCalc and bpay MOD10 version 5 - php

I am using the following PHP code to calculate a CRN for BPay:
<?php
function LuhnCalc($number) {
$chars = array_reverse(str_split($number, 1));
$odd = array_intersect_key($chars, array_fill_keys(range(1, count($chars), 2), null));
$even = array_intersect_key($chars, array_fill_keys(range(0, count($chars), 2), null));
$even = array_map(function($n) { return ($n >= 5)?2 * $n - 9:2 * $n; }, $even);
$total = array_sum($odd) + array_sum($even);
return ((floor($total / 10) + 1) * 10 - $total) % 10;
}
print LuhnCalc($_GET['num']);
?>
However it seems that BPAY is version 5 of MOD 10, for which I can't find any documentation. It seems to not be the same as MOD10.
The following numbers where tested:
2005,1597,3651,0584,9675
bPAY
2005 = 20052
1597 = 15976
3651 = 36514
0584 = 05840
9675 = 96752
MY CODE
2005 = 20057
1597 = 15974
3651 = 36517
0584 = 05843
9675 = 96752
As you can see, none of them match the BPAY numbers.

This PHP function will generate BPay reference numbers based on the mod10 version 5 algorithm.
Who knows why BPay can't add this to their website. I only found an explanation by googling finding the algorithm being called "MOD10V05" instead of "Mod 10 version 5".
function generateBpayRef($number) {
$number = preg_replace("/\D/", "", $number);
// The seed number needs to be numeric
if(!is_numeric($number)) return false;
// Must be a positive number
if($number <= 0) return false;
// Get the length of the seed number
$length = strlen($number);
$total = 0;
// For each character in seed number, sum the character multiplied by its one based array position (instead of normal PHP zero based numbering)
for($i = 0; $i < $length; $i++) $total += $number{$i} * ($i + 1);
// The check digit is the result of the sum total from above mod 10
$checkdigit = fmod($total, 10);
// Return the original seed plus the check digit
return $number . $checkdigit;
}

Here's a way of implementing the "MOD10V5" algorithm (or "mod 10 version 5") using a t-sql user defined function in SQL server. It accepts a Customer ID up to 9 characters long, and return an 11 character CRN (Customer Reference Number).
I also prepended a version number onto the start of my CustomerID, you could do this too if you think you might end up changing it in the future.
CREATE Function [dbo].[CalculateBPayCRN]
(
#CustomerID nvarchar(9)
)
RETURNS varchar(11)
AS
BEGIN
DECLARE #NewCRN nvarchar(11)
DECLARE #Multiplier TINYINT
DECLARE #Sum int
DECLARE #SubTotal int
DECLARE #CheckDigit int
DECLARE #ReturnVal BIGINT
SELECT #Multiplier = 1
SELECT #SubTotal = 0
-- If it's less than 9 characters, pad it with 0's, then prepend a '1'
SELECT #NewCRN = '1' + right('000000000'+ rtrim(#CustomerID), 9)
-- loop through each digit in the #NewCRN, multiple it by the correct weighting and subtotal it:
WHILE #Multiplier <= LEN(#NewCRN)
BEGIN
SET #Sum = CAST(SUBSTRING(#NewCRN,#Multiplier,1) AS TINYINT) * #Multiplier
SET #SubTotal = #SubTotal + #Sum
SET #Multiplier = #Multiplier + 1
END
-- mod 10 the subtotal and the result is our check digit
SET #CheckDigit = #SubTotal % 10
SELECT #ReturnVal = #NewCRN + cast(#CheckDigit as varchar)
RETURN #ReturnVal
END
GO

Modula 10 V1 in PHP. Tested against my Windows dataflex routine and it is the same.
function generateBpayRef($number) {
//Mod 10 v1
$number = preg_replace("/\D/", "", $number);
// The seed number needs to be numeric
if(!is_numeric($number)) return false;
// Must be a positive number
if($number <= 0) return false;
$stringMemberNo = "$number";
$stringMemberNo = str_pad($stringMemberNo, 6, "0", STR_PAD_LEFT);
//echo " Padded Number is $stringMemberNo ";
$crn = $stringMemberNo;
for($i=0;$i<7;$i++){
$crnval = substr($crn,(5-$i),1);
$iPartVal = $iWeight * $crnval;
if($iPartVal>9){
//echo " Greater than 9: $iPartVal ";
$firstChar = substr($iPartVal,0,1);
$secondChar = substr($iPartVal,1,1);
$iPartVal=$firstChar+$secondChar;
//$iPartVal -= 9;
}
$iSum+=$iPartVal;
$iWeight++;
if ($iWeight>2){$iWeight=1;}
//echo " CRN: $crnval ] Weight: $iWeight ] Part: $iPartVal ] SUM: $iSum ";
}
$iSum %= 10;
if($iSum==0){
//echo " zero check is $iSum ";
//return $iSum;
}
else{
//return 10-$iSum;
$iSum=(10-$iSum);
}
//echo " Check is a $iSum ";
$BpayMemberNo = $stringMemberNo . $iSum ;
echo " New: $BpayMemberNo ";
return ($BpayMemberNo);
}

Here is a ruby class I whipped up quickly for Mod 10 v5
module Bpay
class CRN
attr_accessor :number, :crn
class << self
def calculate_for(number)
new(number).crn
end
end
def initialize(number)
#number = number
calculate
end
def calculate
raise ArgumentError, "The number '#{number}' is not valid" unless valid?
digits = number.to_s.scan(/\d/).map { |x| x.to_i }
raise ArgumentError, "The number '#{number}' must be at least 2 digits in length" if digits.size < 2
check_digit = digits.each_with_index.map { |d, i| d * (i + 1) }.inject(:+) % 10
#crn = "#{number}#{check_digit}"
end
def valid?
return false unless !!Integer(number.to_s) rescue false
return false if number.to_i <= 0
true
end
end
end

This is in C#, but this is what I have so far for BPay check digit generation:
private void btnBPayGenerate_Click(object sender, EventArgs e)
{
var originalChars = txtBPayNumber.Text.ToCharArray();
List<int> oddDigits = new List<int>();
List<int> evenDigits = new List<int>();
int oddTotal = 0, evenTotal = 0, total = 0, checkDigit ;
const int oddMultiplier = 3;
const int modulus = 10;
bool isOdd = true;
for (int x = 0; x < originalChars.Length; x++)
{
if(isOdd)
oddDigits.Add(Int32.Parse(originalChars[x].ToString()));
else
evenDigits.Add(Int32.Parse(originalChars[x].ToString()));
isOdd = !isOdd;
}
foreach (var digit in oddDigits)
oddTotal += digit;
foreach (var digit in evenDigits)
evenTotal += digit;
oddTotal = oddTotal * oddMultiplier;
total = oddTotal + evenTotal;
checkDigit = (modulus - (total % modulus));
lblBPayResult.Text = txtBPayNumber.Text + checkDigit.ToString();
}
I haven't completed testing this yet, I will post back once BPAY get back to me.
EDIT: try this: https://gist.github.com/1287893

I had to work out a version for javascript, this is what I came up with. It correctly generates the expected numbers in the original question.
var startingNumber = 2005;
var reference = startingNumber.toString();
var subTotal = 0;
for (var x = 0; x < reference.length; x++) {
subTotal += (x + 1) * reference.charAt(x);
}
var digit = subTotal % 10;
var bpayReference = reference + digit.toString();

Here is a function I created using vb.net to calculate a mod 10 version 5 check digit
Private Function CalcCheckDigit(ByRef psBaseNumber As String) As String
Dim lCheckDigit, iLoop As Integer
Dim dCalcNumber As Double
lCheckDigit = 0
dCalcNumber = 0
For iLoop = 0 To (psBaseNumber.Length - 1)
lCheckDigit = lCheckDigit + (psBaseNumber.Substring(iLoop, 1) * (iLoop + 1))
Next iLoop
lCheckDigit = lCheckDigit Mod 10
CalcCheckDigit = psBaseNumber & CStr(lCheckDigit)
End Function

Related

PHP String to Int for MySQL Unique ID

I would like to convert a 17 digit unique ID retrieved as a string from my MYSQL database to a number. I use the int to do so. However I get a zero at the end:
$num = 96435171263250434;
(int)$num --> 96435171263250430
I've checked I am running a 64 bit system. I get the following:
php -r 'echo PHP_INT_MAX;'
9223372036854775807
How do I fix this issue???
You cannot exceed PHP_INT_MAX
$num = "96435171263250434";
$x = (float) $num; // This should hold it but it's a float
$maxIntMult = 0;
$maxIntMult = intval($x / PHP_INT_MAX);
$remainder = $x - $maxIntMult * PHP_INT_MAX;
echo PHP_INT_MAX . " x " .$maxIntMult. " + " . $remainder; // function of two integer if you can't work with floats and you can make something of this
You can try to make use of the fact that ids don't have negative values, effectively doubling your range.
$num = PHP_INT_MAX + 50;
$x = (float) $num;
$intX = $num - PHP_INT_MAX;
echo $intX; // Shows 50 with the '0' being -PHP_INT_MAX
function getIdWithNonZeroOffset($stringId)
{
$x = (float) $stringId;
$intX = $x - PHP_INT_MAX;
return $intX;
}
function getStringFromNonZeroOfssetId($id)
{
return (string) ($id + PHP_INT_MAX);
}
echo getIdWithNonZeroOffset((string)(PHP_INT_MAX + 200)); // Gives 200 (store this in int column)
echo getStringFromNonZeroOfssetId(200); // Gives "2147483847" (my max int is "2147483647")

I need to find a smallest section of a string that contains all of the characters I'm searching for [duplicate]

I have recently come across an interesting question on strings. Suppose you are given following:
Input string1: "this is a test string"
Input string2: "tist"
Output string: "t stri"
So, given above, how can I approach towards finding smallest substring of string1 that contains all the characters from string 2?
To see more details including working code, check my blog post at:
http://www.leetcode.com/2010/11/finding-minimum-window-in-s-which.html
To help illustrate this approach, I use an example: string1 = "acbbaca" and string2 = "aba". Here, we also use the term "window", which means a contiguous block of characters from string1 (could be interchanged with the term substring).
i) string1 = "acbbaca" and string2 = "aba".
ii) The first minimum window is found.
Notice that we cannot advance begin
pointer as hasFound['a'] ==
needToFind['a'] == 2. Advancing would
mean breaking the constraint.
iii) The second window is found. begin
pointer still points to the first
element 'a'. hasFound['a'] (3) is
greater than needToFind['a'] (2). We
decrement hasFound['a'] by one and
advance begin pointer to the right.
iv) We skip 'c' since it is not found
in string2. Begin pointer now points to 'b'.
hasFound['b'] (2) is greater than
needToFind['b'] (1). We decrement
hasFound['b'] by one and advance begin
pointer to the right.
v) Begin pointer now points to the
next 'b'. hasFound['b'] (1) is equal
to needToFind['b'] (1). We stop
immediately and this is our newly
found minimum window.
The idea is mainly based on the help of two pointers (begin and end position of the window) and two tables (needToFind and hasFound) while traversing string1. needToFind stores the total count of a character in string2 and hasFound stores the total count of a character met so far. We also use a count variable to store the total characters in string2 that's met so far (not counting characters where hasFound[x] exceeds needToFind[x]). When count equals string2's length, we know a valid window is found.
Each time we advance the end pointer (pointing to an element x), we increment hasFound[x] by one. We also increment count by one if hasFound[x] is less than or equal to needToFind[x]. Why? When the constraint is met (that is, count equals to string2's size), we immediately advance begin pointer as far right as possible while maintaining the constraint.
How do we check if it is maintaining the constraint? Assume that begin points to an element x, we check if hasFound[x] is greater than needToFind[x]. If it is, we can decrement hasFound[x] by one and advancing begin pointer without breaking the constraint. On the other hand, if it is not, we stop immediately as advancing begin pointer breaks the window constraint.
Finally, we check if the minimum window length is less than the current minimum. Update the current minimum if a new minimum is found.
Essentially, the algorithm finds the first window that satisfies the constraint, then continue maintaining the constraint throughout.
You can do a histogram sweep in O(N+M) time and O(1) space where N is the number of characters in the first string and M is the number of characters in the second.
It works like this:
Make a histogram of the second string's characters (key operation is hist2[ s2[i] ]++).
Make a cumulative histogram of the first string's characters until that histogram contains every character that the second string's histogram contains (which I will call "the histogram condition").
Then move forwards on the first string, subtracting from the histogram, until it fails to meet the histogram condition. Mark that bit of the first string (before the final move) as your tentative substring.
Move the front of the substring forwards again until you meet the histogram condition again. Move the end forwards until it fails again. If this is a shorter substring than the first, mark that as your tentative substring.
Repeat until you've passed through the entire first string.
The marked substring is your answer.
Note that by varying the check you use on the histogram condition, you can choose either to have the same set of characters as the second string, or at least as many characters of each type. (Its just the difference between a[i]>0 && b[i]>0 and a[i]>=b[i].)
You can speed up the histogram checks if you keep a track of which condition is not satisfied when you're trying to satisfy it, and checking only the thing that you decrement when you're trying to break it. (On the initial buildup, you count how many items you've satisfied, and increment that count every time you add a new character that takes the condition from false to true.)
Here's an O(n) solution. The basic idea is simple: for each starting index, find the least ending index such that the substring contains all of the necessary letters. The trick is that the least ending index increases over the course of the function, so with a little data structure support, we consider each character at most twice.
In Python:
from collections import defaultdict
def smallest(s1, s2):
assert s2 != ''
d = defaultdict(int)
nneg = [0] # number of negative entries in d
def incr(c):
d[c] += 1
if d[c] == 0:
nneg[0] -= 1
def decr(c):
if d[c] == 0:
nneg[0] += 1
d[c] -= 1
for c in s2:
decr(c)
minlen = len(s1) + 1
j = 0
for i in xrange(len(s1)):
while nneg[0] > 0:
if j >= len(s1):
return minlen
incr(s1[j])
j += 1
minlen = min(minlen, j - i)
decr(s1[i])
return minlen
I received the same interview question. I am a C++ candidate but I was in a position to code relatively fast in JAVA.
Java [Courtesy : Sumod Mathilakath]
import java.io.*;
import java.util.*;
class UserMainCode
{
public String GetSubString(String input1,String input2){
// Write code here...
return find(input1, input2);
}
private static boolean containsPatternChar(int[] sCount, int[] pCount) {
for(int i=0;i<256;i++) {
if(pCount[i]>sCount[i])
return false;
}
return true;
}
public static String find(String s, String p) {
if (p.length() > s.length())
return null;
int[] pCount = new int[256];
int[] sCount = new int[256];
// Time: O(p.lenght)
for(int i=0;i<p.length();i++) {
pCount[(int)(p.charAt(i))]++;
sCount[(int)(s.charAt(i))]++;
}
int i = 0, j = p.length(), min = Integer.MAX_VALUE;
String res = null;
// Time: O(s.lenght)
while (j < s.length()) {
if (containsPatternChar(sCount, pCount)) {
if ((j - i) < min) {
min = j - i;
res = s.substring(i, j);
// This is the smallest possible substring.
if(min==p.length())
break;
// Reduce the window size.
sCount[(int)(s.charAt(i))]--;
i++;
}
} else {
sCount[(int)(s.charAt(j))]++;
// Increase the window size.
j++;
}
}
System.out.println(res);
return res;
}
}
C++ [Courtesy : sundeepblue]
#include <iostream>
#include <vector>
#include <string>
#include <climits>
using namespace std;
string find_minimum_window(string s, string t) {
if(s.empty() || t.empty()) return;
int ns = s.size(), nt = t.size();
vector<int> total(256, 0);
vector<int> sofar(256, 0);
for(int i=0; i<nt; i++)
total[t[i]]++;
int L = 0, R;
int minL = 0; //gist2
int count = 0;
int min_win_len = INT_MAX;
for(R=0; R<ns; R++) { // gist0, a big for loop
if(total[s[R]] == 0) continue;
else sofar[s[R]]++;
if(sofar[s[R]] <= total[s[R]]) // gist1, <= not <
count++;
if(count == nt) { // POS1
while(true) {
char c = s[L];
if(total[c] == 0) { L++; }
else if(sofar[c] > total[c]) {
sofar[c]--;
L++;
}
else break;
}
if(R - L + 1 < min_win_len) { // this judge should be inside POS1
min_win_len = R - L + 1;
minL = L;
}
}
}
string res;
if(count == nt) // gist3, cannot forget this.
res = s.substr(minL, min_win_len); // gist4, start from "minL" not "L"
return res;
}
int main() {
string s = "abdccdedca";
cout << find_minimum_window(s, "acd");
}
Erlang [Courtesy : wardbekker]
-module(leetcode).
-export([min_window/0]).
%% Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).
%% For example,
%% S = "ADOBECODEBANC"
%% T = "ABC"
%% Minimum window is "BANC".
%% Note:
%% If there is no such window in S that covers all characters in T, return the emtpy string "".
%% If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.
min_window() ->
"eca" = min_window("cabeca", "cae"),
"eca" = min_window("cfabeca", "cae"),
"aec" = min_window("cabefgecdaecf", "cae"),
"cwae" = min_window("cabwefgewcwaefcf", "cae"),
"BANC" = min_window("ADOBECODEBANC", "ABC"),
ok.
min_window(T, S) ->
min_window(T, S, []).
min_window([], _T, MinWindow) ->
MinWindow;
min_window([H | Rest], T, MinWindow) ->
NewMinWindow = case lists:member(H, T) of
true ->
MinWindowFound = fullfill_window(Rest, lists:delete(H, T), [H]),
case length(MinWindow) == 0 orelse (length(MinWindow) > length(MinWindowFound)
andalso length(MinWindowFound) > 0) of
true ->
MinWindowFound;
false ->
MinWindow
end;
false ->
MinWindow
end,
min_window(Rest, T, NewMinWindow).
fullfill_window(_, [], Acc) ->
%% window completed
Acc;
fullfill_window([], _T, _Acc) ->
%% no window found
"";
fullfill_window([H | Rest], T, Acc) ->
%% completing window
case lists:member(H, T) of
true ->
fullfill_window(Rest, lists:delete(H, T), Acc ++ [H]);
false ->
fullfill_window(Rest, T, Acc ++ [H])
end.
REF:
http://articles.leetcode.com/finding-minimum-window-in-s-which/#comment-511216
http://www.mif.vu.lt/~valdas/ALGORITMAI/LITERATURA/Cormen/Cormen.pdf
Please have a look at this as well:
//-----------------------------------------------------------------------
bool IsInSet(char ch, char* cSet)
{
char* cSetptr = cSet;
int index = 0;
while (*(cSet+ index) != '\0')
{
if(ch == *(cSet+ index))
{
return true;
}
++index;
}
return false;
}
void removeChar(char ch, char* cSet)
{
bool bShift = false;
int index = 0;
while (*(cSet + index) != '\0')
{
if( (ch == *(cSet + index)) || bShift)
{
*(cSet + index) = *(cSet + index + 1);
bShift = true;
}
++index;
}
}
typedef struct subStr
{
short iStart;
short iEnd;
short szStr;
}ss;
char* subStringSmallest(char* testStr, char* cSet)
{
char* subString = NULL;
int iSzSet = strlen(cSet) + 1;
int iSzString = strlen(testStr)+ 1;
char* cSetBackUp = new char[iSzSet];
memcpy((void*)cSetBackUp, (void*)cSet, iSzSet);
int iStartIndx = -1;
int iEndIndx = -1;
int iIndexStartNext = -1;
std::vector<ss> subStrVec;
int index = 0;
while( *(testStr+index) != '\0' )
{
if (IsInSet(*(testStr+index), cSetBackUp))
{
removeChar(*(testStr+index), cSetBackUp);
if(iStartIndx < 0)
{
iStartIndx = index;
}
else if( iIndexStartNext < 0)
iIndexStartNext = index;
else
;
if (strlen(cSetBackUp) == 0 )
{
iEndIndx = index;
if( iIndexStartNext == -1)
break;
else
{
index = iIndexStartNext;
ss stemp = {iStartIndx, iEndIndx, (iEndIndx-iStartIndx + 1)};
subStrVec.push_back(stemp);
iStartIndx = iEndIndx = iIndexStartNext = -1;
memcpy((void*)cSetBackUp, (void*)cSet, iSzSet);
continue;
}
}
}
else
{
if (IsInSet(*(testStr+index), cSet))
{
if(iIndexStartNext < 0)
iIndexStartNext = index;
}
}
++index;
}
int indexSmallest = 0;
for(int indexVec = 0; indexVec < subStrVec.size(); ++indexVec)
{
if(subStrVec[indexSmallest].szStr > subStrVec[indexVec].szStr)
indexSmallest = indexVec;
}
subString = new char[(subStrVec[indexSmallest].szStr) + 1];
memcpy((void*)subString, (void*)(testStr+ subStrVec[indexSmallest].iStart), subStrVec[indexSmallest].szStr);
memset((void*)(subString + subStrVec[indexSmallest].szStr), 0, 1);
delete[] cSetBackUp;
return subString;
}
//--------------------------------------------------------------------
Edit: apparently there's an O(n) algorithm (cf. algorithmist's answer). Obviously this have this will beat the [naive] baseline described below!
Too bad I gotta go... I'm a bit suspicious that we can get O(n). I'll check in tomorrow to see the winner ;-) Have fun!
Tentative algorithm:
The general idea is to sequentially try and use a character from str2 found in str1 as the start of a search (in either/both directions) of all the other letters of str2. By keeping a "length of best match so far" value, we can abort searches when they exceed this. Other heuristics can probably be used to further abort suboptimal (so far) solutions. The choice of the order of the starting letters in str1 matters much; it is suggested to start with the letter(s) of str1 which have the lowest count and to try with the other letters, of an increasing count, in subsequent attempts.
[loose pseudo-code]
- get count for each letter/character in str1 (number of As, Bs etc.)
- get count for each letter in str2
- minLen = length(str1) + 1 (the +1 indicates you're not sure all chars of
str2 are in str1)
- Starting with the letter from string2 which is found the least in string1,
look for other letters of Str2, in either direction of str1, until you've
found them all (or not, at which case response = impossible => done!).
set x = length(corresponding substring of str1).
- if (x < minLen),
set minlen = x,
also memorize the start/len of the str1 substring.
- continue trying with other letters of str1 (going the up the frequency
list in str1), but abort search as soon as length(substring of strl)
reaches or exceed minLen.
We can find a few other heuristics that would allow aborting a
particular search, based on [pre-calculated ?] distance between a given
letter in str1 and some (all?) of the letters in str2.
- the overall search terminates when minLen = length(str2) or when
we've used all letters of str1 (which match one letter of str2)
as a starting point for the search
Here is Java implementation
public static String shortestSubstrContainingAllChars(String input, String target) {
int needToFind[] = new int[256];
int hasFound[] = new int[256];
int totalCharCount = 0;
String result = null;
char[] targetCharArray = target.toCharArray();
for (int i = 0; i < targetCharArray.length; i++) {
needToFind[targetCharArray[i]]++;
}
char[] inputCharArray = input.toCharArray();
for (int begin = 0, end = 0; end < inputCharArray.length; end++) {
if (needToFind[inputCharArray[end]] == 0) {
continue;
}
hasFound[inputCharArray[end]]++;
if (hasFound[inputCharArray[end]] <= needToFind[inputCharArray[end]]) {
totalCharCount ++;
}
if (totalCharCount == target.length()) {
while (needToFind[inputCharArray[begin]] == 0
|| hasFound[inputCharArray[begin]] > needToFind[inputCharArray[begin]]) {
if (hasFound[inputCharArray[begin]] > needToFind[inputCharArray[begin]]) {
hasFound[inputCharArray[begin]]--;
}
begin++;
}
String substring = input.substring(begin, end + 1);
if (result == null || result.length() > substring.length()) {
result = substring;
}
}
}
return result;
}
Here is the Junit Test
#Test
public void shortestSubstringContainingAllCharsTest() {
String result = StringUtil.shortestSubstrContainingAllChars("acbbaca", "aba");
assertThat(result, equalTo("baca"));
result = StringUtil.shortestSubstrContainingAllChars("acbbADOBECODEBANCaca", "ABC");
assertThat(result, equalTo("BANC"));
result = StringUtil.shortestSubstrContainingAllChars("this is a test string", "tist");
assertThat(result, equalTo("t stri"));
}
//[ShortestSubstring.java][1]
public class ShortestSubstring {
public static void main(String[] args) {
String input1 = "My name is Fran";
String input2 = "rim";
System.out.println(getShortestSubstring(input1, input2));
}
private static String getShortestSubstring(String mainString, String toBeSearched) {
int mainStringLength = mainString.length();
int toBeSearchedLength = toBeSearched.length();
if (toBeSearchedLength > mainStringLength) {
throw new IllegalArgumentException("search string cannot be larger than main string");
}
for (int j = 0; j < mainStringLength; j++) {
for (int i = 0; i <= mainStringLength - toBeSearchedLength; i++) {
String substring = mainString.substring(i, i + toBeSearchedLength);
if (checkIfMatchFound(substring, toBeSearched)) {
return substring;
}
}
toBeSearchedLength++;
}
return null;
}
private static boolean checkIfMatchFound(String substring, String toBeSearched) {
char[] charArraySubstring = substring.toCharArray();
char[] charArrayToBeSearched = toBeSearched.toCharArray();
int count = 0;
for (int i = 0; i < charArraySubstring.length; i++) {
for (int j = 0; j < charArrayToBeSearched.length; j++) {
if (String.valueOf(charArraySubstring[i]).equalsIgnoreCase(String.valueOf(charArrayToBeSearched[j]))) {
count++;
}
}
}
return count == charArrayToBeSearched.length;
}
}
This is an approach using prime numbers to avoid one loop, and replace it with multiplications. Several other minor optimizations can be made.
Assign a unique prime number to any of the characters that you want to find, and 1 to the uninteresting characters.
Find the product of a matching string by multiplying the prime number with the number of occurrences it should have. Now this product can only be found if the same prime factors are used.
Search the string from the beginning, multiplying the respective prime number as you move into a running product.
If the number is greater than the correct sum, remove the first character and divide its prime number out of your running product.
If the number is less than the correct sum, include the next character and multiply it into your running product.
If the number is the same as the correct sum you have found a match, slide beginning and end to next character and continue searching for other matches.
Decide which of the matches is the shortest.
Gist
charcount = { 'a': 3, 'b' : 1 };
str = "kjhdfsbabasdadaaaaasdkaaajbajerhhayeom"
def find (c, s):
Ns = len (s)
C = list (c.keys ())
D = list (c.values ())
# prime numbers assigned to the first 25 chars
prmsi = [ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 , 97]
# primes used in the key, all other set to 1
prms = []
Cord = [ord(c) - ord('a') for c in C]
for e,p in enumerate(prmsi):
if e in Cord:
prms.append (p)
else:
prms.append (1)
# Product of match
T = 1
for c,d in zip(C,D):
p = prms[ord (c) - ord('a')]
T *= p**d
print ("T=", T)
t = 1 # product of current string
f = 0
i = 0
matches = []
mi = 0
mn = Ns
mm = 0
while i < Ns:
k = prms[ord(s[i]) - ord ('a')]
t *= k
print ("testing:", s[f:i+1])
if (t > T):
# included too many chars: move start
t /= prms[ord(s[f]) - ord('a')] # remove first char, usually division by 1
f += 1 # increment start position
t /= k # will be retested, could be replaced with bool
elif t == T:
# found match
print ("FOUND match:", s[f:i+1])
matches.append (s[f:i+1])
if (i - f) < mn:
mm = mi
mn = i - f
mi += 1
t /= prms[ord(s[f]) - ord('a')] # remove first matching char
# look for next match
i += 1
f += 1
else:
# no match yet, keep searching
i += 1
return (mm, matches)
print (find (charcount, str))
(note: this answer was originally posted to a duplicate question, the original answer is now deleted.)
C# Implementation:
public static Tuple<int, int> FindMinSubstringWindow(string input, string pattern)
{
Tuple<int, int> windowCoords = new Tuple<int, int>(0, input.Length - 1);
int[] patternHist = new int[256];
for (int i = 0; i < pattern.Length; i++)
{
patternHist[pattern[i]]++;
}
int[] inputHist = new int[256];
int minWindowLength = int.MaxValue;
int count = 0;
for (int begin = 0, end = 0; end < input.Length; end++)
{
// Skip what's not in pattern.
if (patternHist[input[end]] == 0)
{
continue;
}
inputHist[input[end]]++;
// Count letters that are in pattern.
if (inputHist[input[end]] <= patternHist[input[end]])
{
count++;
}
// Window found.
if (count == pattern.Length)
{
// Remove extra instances of letters from pattern
// or just letters that aren't part of the pattern
// from the beginning.
while (patternHist[input[begin]] == 0 ||
inputHist[input[begin]] > patternHist[input[begin]])
{
if (inputHist[input[begin]] > patternHist[input[begin]])
{
inputHist[input[begin]]--;
}
begin++;
}
// Current window found.
int windowLength = end - begin + 1;
if (windowLength < minWindowLength)
{
windowCoords = new Tuple<int, int>(begin, end);
minWindowLength = windowLength;
}
}
}
if (count == pattern.Length)
{
return windowCoords;
}
return null;
}
I've implemented it using Python3 at O(N) efficiency:
def get(s, alphabet="abc"):
seen = {}
for c in alphabet:
seen[c] = 0
seen[s[0]] = 1
start = 0
end = 0
shortest_s = 0
shortest_e = 99999
while end + 1 < len(s):
while seen[s[start]] > 1:
seen[s[start]] -= 1
start += 1
# Constant time check:
if sum(seen.values()) == len(alphabet) and all(v == 1 for v in seen.values()) and \
shortest_e - shortest_s > end - start:
shortest_s = start
shortest_e = end
end += 1
seen[s[end]] += 1
return s[shortest_s: shortest_e + 1]
print(get("abbcac")) # Expected to return "bca"
String s = "xyyzyzyx";
String s1 = "xyz";
String finalString ="";
Map<Character,Integer> hm = new HashMap<>();
if(s1!=null && s!=null && s.length()>s1.length()){
for(int i =0;i<s1.length();i++){
if(hm.get(s1.charAt(i))!=null){
int k = hm.get(s1.charAt(i))+1;
hm.put(s1.charAt(i), k);
}else
hm.put(s1.charAt(i), 1);
}
Map<Character,Integer> t = new HashMap<>();
int start =-1;
for(int j=0;j<s.length();j++){
if(hm.get(s.charAt(j))!=null){
if(t.get(s.charAt(j))!=null){
if(t.get(s.charAt(j))!=hm.get(s.charAt(j))){
int k = t.get(s.charAt(j))+1;
t.put(s.charAt(j), k);
}
}else{
t.put(s.charAt(j), 1);
if(start==-1){
if(j+s1.length()>s.length()){
break;
}
start = j;
}
}
if(hm.equals(t)){
t = new HashMap<>();
if(finalString.length()<s.substring(start,j+1).length());
{
finalString=s.substring(start,j+1);
}
j=start;
start=-1;
}
}
}
JavaScript solution in bruteforce way:
function shortestSubStringOfUniqueChars(s){
var uniqueArr = [];
for(let i=0; i<s.length; i++){
if(uniqueArr.indexOf(s.charAt(i)) <0){
uniqueArr.push(s.charAt(i));
}
}
let windoww = uniqueArr.length;
while(windoww < s.length){
for(let i=0; i<s.length - windoww; i++){
let match = true;
let tempArr = [];
for(let j=0; j<uniqueArr.length; j++){
if(uniqueArr.indexOf(s.charAt(i+j))<0){
match = false;
break;
}
}
let checkStr
if(match){
checkStr = s.substr(i, windoww);
for(let j=0; j<uniqueArr.length; j++){
if(uniqueArr.indexOf(checkStr.charAt(j))<0){
match = false;
break;
}
}
}
if(match){
return checkStr;
}
}
windoww = windoww + 1;
}
}
console.log(shortestSubStringOfUniqueChars("ABA"));
# Python implementation
s = input('Enter the string : ')
s1 = input('Enter the substring to search : ')
l = [] # List to record all the matching combinations
check = all([char in s for char in s1])
if check == True:
for i in range(len(s1),len(s)+1) :
for j in range(0,i+len(s1)+2):
if (i+j) < len(s)+1:
cnt = 0
b = all([char in s[j:i+j] for char in s1])
if (b == True) :
l.append(s[j:i+j])
print('The smallest substring containing',s1,'is',l[0])
else:
print('Please enter a valid substring')
Java code for the approach discussed above:
private static Map<Character, Integer> frequency;
private static Set<Character> charsCovered;
private static Map<Character, Integer> encountered;
/**
* To set the first match index as an intial start point
*/
private static boolean hasStarted = false;
private static int currentStartIndex = 0;
private static int finalStartIndex = 0;
private static int finalEndIndex = 0;
private static int minLen = Integer.MAX_VALUE;
private static int currentLen = 0;
/**
* Whether we have already found the match and now looking for other
* alternatives.
*/
private static boolean isFound = false;
private static char currentChar;
public static String findSmallestSubStringWithAllChars(String big, String small) {
if (null == big || null == small || big.isEmpty() || small.isEmpty()) {
return null;
}
frequency = new HashMap<Character, Integer>();
instantiateFrequencyMap(small);
charsCovered = new HashSet<Character>();
int charsToBeCovered = frequency.size();
encountered = new HashMap<Character, Integer>();
for (int i = 0; i < big.length(); i++) {
currentChar = big.charAt(i);
if (frequency.containsKey(currentChar) && !isFound) {
if (!hasStarted && !isFound) {
hasStarted = true;
currentStartIndex = i;
}
updateEncounteredMapAndCharsCoveredSet(currentChar);
if (charsCovered.size() == charsToBeCovered) {
currentLen = i - currentStartIndex;
isFound = true;
updateMinLength(i);
}
} else if (frequency.containsKey(currentChar) && isFound) {
updateEncounteredMapAndCharsCoveredSet(currentChar);
if (currentChar == big.charAt(currentStartIndex)) {
encountered.put(currentChar, encountered.get(currentChar) - 1);
currentStartIndex++;
while (currentStartIndex < i) {
if (encountered.containsKey(big.charAt(currentStartIndex))
&& encountered.get(big.charAt(currentStartIndex)) > frequency.get(big
.charAt(currentStartIndex))) {
encountered.put(big.charAt(currentStartIndex),
encountered.get(big.charAt(currentStartIndex)) - 1);
} else if (encountered.containsKey(big.charAt(currentStartIndex))) {
break;
}
currentStartIndex++;
}
}
currentLen = i - currentStartIndex;
updateMinLength(i);
}
}
System.out.println("start: " + finalStartIndex + " finalEnd : " + finalEndIndex);
return big.substring(finalStartIndex, finalEndIndex + 1);
}
private static void updateMinLength(int index) {
if (minLen > currentLen) {
minLen = currentLen;
finalStartIndex = currentStartIndex;
finalEndIndex = index;
}
}
private static void updateEncounteredMapAndCharsCoveredSet(Character currentChar) {
if (encountered.containsKey(currentChar)) {
encountered.put(currentChar, encountered.get(currentChar) + 1);
} else {
encountered.put(currentChar, 1);
}
if (encountered.get(currentChar) >= frequency.get(currentChar)) {
charsCovered.add(currentChar);
}
}
private static void instantiateFrequencyMap(String str) {
for (char c : str.toCharArray()) {
if (frequency.containsKey(c)) {
frequency.put(c, frequency.get(c) + 1);
} else {
frequency.put(c, 1);
}
}
}
public static void main(String[] args) {
String big = "this is a test string";
String small = "tist";
System.out.println("len: " + big.length());
System.out.println(findSmallestSubStringWithAllChars(big, small));
}
def minimum_window(s, t, min_length = 100000):
d = {}
for x in t:
if x in d:
d[x]+= 1
else:
d[x] = 1
tot = sum([y for x,y in d.iteritems()])
l = []
ind = 0
for i,x in enumerate(s):
if ind == 1:
l = l + [x]
if x in d:
tot-=1
if not l:
ind = 1
l = [x]
if tot == 0:
if len(l)<min_length:
min_length = len(l)
min_length = minimum_window(s[i+1:], t, min_length)
return min_length
l_s = "ADOBECODEBANC"
t_s = "ABC"
min_length = minimum_window(l_s, t_s)
if min_length == 100000:
print "Not found"
else:
print min_length

comparing two strings for percentage match

I am trying to compare a user submitted string to a string of a database record and see how close they are in terms of %
i have found this rather interesting code which looks like a good solution
Function Compare(ByVal str1 As String, ByVal str2 As String) As Double
Dim count As Integer = If(str1.Length > str2.Length, str1.Length, str2.Length)
Dim hits As Integer = 0
Dim i, j As Integer : i = 0 : j = 0
For i = 0 To str1.Length - 1
If str1.Chars(i) = " " Then i += 1 : j = str2.IndexOf(" "c, j) + 1 : hits += 1
While j < str2.Length AndAlso str2.Chars(j) <> " "c
If str1.Chars(i) = str2.Chars(j) Then
hits += 1
j += 1
Exit While
Else
j += 1
End If
End While
If Not (j < str2.Length AndAlso str2.Chars(j) <> " "c) Then
j -= 1
End If
Next
Return Math.Round((hits / count), 2)
End Function
firstly can anyone tell me what the language is used above, and can anyone help me convert it to php please?
i've tried to convert it but ran in to a bit of trouble early on
function compare($str1,$str2) as $double
{
$count = if(strlen($str1) > strlen($str2), strlen($str1) > strlen($str2));
$hits = 0;
$i - 0;
$j = 0;
for($i = 0; $i < strlen($str1); $i++)
{
if($str1[$i] == " ")
{
$i .= "1";
}
}
}
any help with this would be hugely appreciated
As an option, then, try something like this:
$teststr = "This is a test.";
$dbstr = "This was a test.";
$percent = (1 - levenshtein($teststr, $dbstr)/max( strlen($teststr),strlen($dbstr) ) ) * 100;
print "Percent match".$percent."\n";
Percent match: 92.857142857143
Far more info at: http://us3.php.net//manual/en/function.levenshtein.php

Finding next fibonacci number

I need to find a (the next) fibonacci number given a integer N. So let's say I have n = 13 and I need to output the next fibonacci number which is 21 but how do I do this? How can I find the previous number that summed up to form it?
I mean I could easily come up with a for/while loop that returns the fibonacci sequence but how can I find the next number by being given the previous one.
<?php
$n = 13;
while($n < 1000) {
$n = $x + $y;
echo($n."<br />");
$x = $y;
$y = $n;
}
?>
You can use Binet's Formula:
n -n
F(n) = phi - (-phi)
---------------
sqrt(5)
where phi is the golden ratio (( 1 + sqrt(5) ) / 2) ~= 1.61803...
This lets you determine exactly the n-th term of the sequence.
Using a loop you could store the values in an array that could stop immediately one key after finding the selected number in the previous keys value.
function getFib($n) {
$fib = array($n+1); // array to num + 1
$fib[0] = 0; $fib[1] = 1; // set initial array keys
$i;
for ($i=2;$i<=$n+1;$i++) {
$fib[$i] = $fib[$i-1]+$fib[$i-2];
if ($fib[$i] > $n) { // check if key > num
return $fib[$i];
}
}
if ($fib[$i-1] < $n) { // check if key < num
return $fib[$i-1] + $n;
}
if ($fib[$i] = $n-1) { // check if key = num
return $fib[$i-1] + $fib[$i-2];
}
if ($fib[$i-1] = 1) { // check if num = 1
return $n + $n;
}
}
$num = 13;
echo "next fibonacci number = " . getFib($num);
Please note that I haven't tested this out and the code could be optimized, so before downvoting consider this serves only as a concept to the question asked.
You can do it in 1 step:
phi = (1+sqrt(5))/2
next = round(current*phi)
(Where round is a function that returns the closest integer; basically equivalent to floor(x+0.5))
For example, if your current number is 13: 13 * phi = 21.034441853748632, which rounds to 21.

Convert septet to octet in php

I need to convert septets to octest like this c example:
private void septetToOctet()
{
int len = 1;
int j = 0;
int septetcount = septet.Count;
while (j < septet.Count - 1)
{
string tmp = septet[j]; // storing jth value
string tmp1 = septet[j + 1]; //storing j+1 th value
string mid = SWAP(tmp1);
tmp1 = mid;
tmp1 = tmp1.Substring(0, len);
string add = SWAP(tmp1);
tmp = add + tmp;// +"-";
tmp = tmp.Substring(0, 8);
//txtoctet.Text += tmp + " || ";
octet.Add(tmp);
len++;
if (len == 8)
{
len = 1;
j = j + 1;
}
j = j + 1;
}
}
..only problem is - i need to do it in php. Does anybody have an code example for this ?
Most of the code can be translated to PHP with little changes. For instance, the first two assignments would be just $len = 1 and $j = 0. There is no strong static typing in PHP.
Since numbers and strings are scalars in PHP, you cannot call methods on them, so you cannot do tmp1.substring but would have to do substr($tmp1, …).
septet and octet would be arrays in PHP. Again, there is no methods on arrays in PHP. To add to an array, you simply do $octet[] = $tmp. To count the elements you do count($septet).
There is no SWAP in PHP, but you can derive the rather trivial function from Is there a built in swap function in C?
These hints should get you closer to a working implementation.

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