Get the sum of all digits in a numeric string - php

How do I find the sum of all the digits in a number in PHP?


array_sum(str_split($number));
This assumes the number is positive (or, more accurately, that the conversion of $number into a string generates only digits).


Artefactos method is obviously unbeatable, but here an version how one could do it "manually":
$number = 1234567890;
$sum = 0;
do {
$sum += $number % 10;
}
while ($number = (int) ($number / 10));
This is actually faster than Artefactos method (at least for 1234567890), because it saves two function calls.


Another way, not so fast, not single line simple
<?php
$n = 123;
$nstr = $n . "";
$sum = 0;
for ($i = 0; $i < strlen($nstr); ++$i)
{
$sum += $nstr[$i];
}
echo $sum;
?>
It also assumes the number is positive.


function addDigits($num) {
if ($num % 9 == 0 && $num > 0) {
return 9;
} else {
return $num % 9;
}
}
only O(n)
at LeetCode submit result：
Runtime: 4 ms, faster than 92.86% of PHP online submissions for Add Digits.
Memory Usage: 14.3 MB, less than 100.00% of PHP online submissions for Add Digits.


<?php
// PHP program to calculate the sum of digits
function sum($num) {
$sum = 0;
for ($i = 0; $i < strlen($num); $i++){
$sum += $num[$i];
}
return $sum;
}
// Driver Code
$num = "925";
echo sum($num);
?>
Result will be 9+2+5 = 16


Try the following code:
<?php
$num = 525;
$sum = 0;
while ($num > 0)
{
$sum= $sum + ($num % 10);
$num= $num / 10;
}
echo "Summation=" . $sum;
?>


If interested with regex:
array_sum(preg_split("//", $number));


<?php
echo"----Sum of digit using php----";
echo"<br/ >";
$num=98765;
$sum=0;
$rem=0;
for($i=0;$i<=$num;$i++)
{
$rem=$num%10;
$sum=$sum+$rem;
$num=$num/10;
}
echo "The sum of digit 98765 is ".$sum;
?>
-----------------Output-------------
----Sum of digit using php----
The sum of digit 98765 is 35


// math before code
// base of digit sums is 9
// the product of all numbers multiplied by 9 equals 9 as digit sum
$nr = 58821.5712; // any number
// Initiallization
$d = array();
$d = explode(".",$nr); // cut decimal digits
$fl = strlen($d[1]); // count decimal digits
$pow = pow(10 ,$fl); // power up for integer
$nr = $nr * $pow; // make float become integer
// The Code
$ds = $nr % 9; // modulo of 9
if($ds == 0) $ds=9; // cancel out zeros
echo $ds;


Assume you want to find the sum of the digits of a number say 2395 the simplest solution would be to first split the digits and find out the sum then concatenate all the numbers into one single number.
<?php
$number=2;
$number1=3;
$number2=9;
$number3=5;
$combine=$number.$number1.$number2.$number3;
$sum=$number+$number1+$number2+$number3;
echo "The sum of $combine is $sum";
?>


One way of getting sum of digit however this is a slowest route.
$n=123;
while(($n=$n-9)>9);
echo "n: $n";


<html>
<head>
<title>detail</title>
</head>
<body>
<?php
$n = 123;
$sum=0; $n1=0;
for ($i =0; $i<=strlen($n);$i++)
{
$n1=$n%10;
$sum += $n1;
$n=$n/10;
}
echo $sum;
?>
</body>
</html>


Here's the code.. Please try this
<?php
$d=0;
$num=12345;
$temp=$num;
$sum=0;
while($temp>1)
{
$temp=$temp/10;
$d++;
}
echo "Digits Are : $d </br>";
for (;$num>1;)
{
$d=$num%10;
$num=$num/10;
$sum=$sum+$d;
}
echo "Sum of Digits is : $sum";
?>

Related

PHP random numbers frequency of occurrence

In PHP I want to generate random numbers from 1 to 10 in a loop.
So for example:
$factor="1"; // this will be changed in another area
for ($i=0;$i<10;$i++) {
if ($factor=="1") {$output = rand(1,10);}
else if ($factor=="2") {$output = rand(1,10);}
else {$output = rand(1,10);}
}
Now to explain this - In result I want to receive 10 random numbers, but when $factor = "2", in that case I want to receive numbers from 6 to 10 more frequently as lower numbers.
It means, from 10 numbers I need to have 80% higher random numbers (it means larger than 5) and in 20% lower numbers (5 or lower).
E.g. 1,2,6,7,8,9,7,9,8,6 (1,2 are the only lower numbers = 20%, the rest are higher = 80)
If the $factor will change, then I want to change the percentage, in that case for example 40% lower numbers, 60% higher numbers.
The idea I have is to put each output in the loop to an array, then check each result and somehow calculate, if there is no 80% of larger numbers, then get random numbers again for those, but this seems to be an overkill.
Is there a simplier solution?

Let's go with the percentages you mention and first generate a random number between 1 and 100. Then the lower number, 1 to 20, have to represent outputs 1 to 5 and the higher numbers, 21 to 100, have to represent output 6 to 10. In PHP that would look like this:
function HighMoreOften()
{
$percentage = rand(1, 100);
if ($percentage <= 20) {
return rand(1, 5);
} else {
return rand(6, 10);
}
}
That should do the trick. You can also convert the percentage you got into the output, this would probably be slightly faster:
function HighMoreOften()
{
$percentage = rand(1, 100);
if ($percentage <= 20) {
return ceil($percentage / 5);
} else {
return 6 + ceil(($percentage - 20) / 20);
}
}
but personally I think the first version is easier to understand and change.

To change frequency you gonna need an array of numbers. And a sum to this direction. frequency is the relation of something between an array of things.
$t = 0;
// $factor = 1; // let's say that this means 20%
$factor = 2; // let's say that this means 40%
if ($factor === 1) {
for ($i = 1; $i <= 80; $i++) {
$t += rand(1,10);
}
for ($i = 1; $i <= 20; $i++) {
$t += rand(6,10);
}
} else if ($factor === 2) {
for ($i = 1; $i <= 60; $i++) {
$t += rand(1,10);
}
for ($i = 1; $i <= 40; $i++) {
$t += rand(6,10);
}
} else {
for ($i = 1; $i <= 100; $i++) {
$t += rand(1,10);
}
}
echo round(($t/100), 0);
Something like that! :)

I came with a very simple (maybe creepy) solution, but this works as I wanted:
echo print_r(generate("2"));
function generate($factor) {
$nums=array();
for ($i=0;$i<10;$i++) {
if ($i<$factor) {$rnd = rand(1,5);}
else {$rnd = rand(6,10);}
array_push($nums,$rnd);
}
return $nums;
}
I can also shuffle the final array results, as the lower numbers will be on the beginning always, but in my case it doesn't matter.

PHP Add numbers 1-10 and only even numbers 1-10. Display sums separately with one loop

I'm trying to use ONE PHP for loop to echo the sum of the numbers 1-10 as well as echo the sum of only the even numbers. I seem to have a problem as these iterations won't be "parallel"
Code:
<?php
$sum = 0; $evensum = 0;
for($x = 1, $y=2; $x<=10, $y<=6; $x++, $y += 2) {
$sum = $sum + $x; $evensum = $evensum + $y;
}
echo "total sum= ". $sum, ", even sum=" . $evensum;
?>
total sum should reflect 55 (1+2+3+4+5+6+7+8+9+10) and even sum should reflect 30 (2+4+6+8+10)

Just Use
<?php
$sum = 0; $evensum = 0;
for($x = 1; $x<=10; $x++) {
// sum all the number
$sum = $sum + $x;
// check the number is even
if( $x % 2 === 0 ) {
// sum only the even numbers
$evensum = $evensum + $x;
}
}
// output
echo "total sum= ". $sum, ", even sum=" . $evensum;
?>

Another approach is using range() and array_functions
$arr=range(1,10);
echo $sum=array_sum($arr);
function even($var)
{
return(!($var & 1));
}
echo $even=array_sum(array_filter($arr, "even"));

Sum all ranges
$all_sum=array_sum(range(1,10));
Sum even number (0,2,4,6,8,10)
$even_sum = array_sum(range(0,10,2));

Dividing a integer equally in X parts

I'm looking for a efficient way in PHP to divide a number in equal part. Number will always be integer (no float).
Let's say that I have an array $hours with values from "1" to "24" ($hours['1'], etc) and a variable $int containing an integer. What I want to acheive is spreading the value of $int equally in 24 parts so I can assing the value to each corresponding array entries. (Should the number be odd, the remaining would be added to the last or first values in the 24).
Regards,

Here's the algorithm you're looking for; it evenly spreads an integer N over K cells:
for i = 0 to K
array[i] = N / K # integer division
# divide up the remainder
for i = 0 to N mod K
array[i] += 1

Try this code
<?php
$num = 400;
$val = floor($num/24);
for($i=0;$i<24;$i++) {
$arr[$i] = $val;
}
$arr[0] += $num - array_sum($arr);
?>

function split($x, $n)
{
// If we cannot split the
// number into exactly 'N' parts
if($x < $n)
echo (-1);
// If x % n == 0 then the minimum
// difference is 0 and all
// numbers are x / n
else if ($x % $n == 0)
{
for($i = 0; $i < $n; $i++)
{
echo ($x / $n);
echo (" ");
}
}
else
{
// upto n-(x % n) the values
// will be x / n
// after that the values
// will be x / n + 1
$zp = $n - ($x % $n);
$pp = $x / $n;
for ($i = 0; $i < $n; $i++)
{
if($i >= $zp)
{
echo (int)$pp + 1;
echo (" ");
}
else
{
echo (int)$pp;
echo (" ");
}
}
}
}
// Driver code
$x = 5;
$n = 3;
split( $x, $n);

how to find the sum of all the multiples of 3 or 5 below 1000 in php, issue?

i have an small issue with the way this problem is resolved.
some would say: println((0 /: ((0 until 1000).filter(x => x % 3 == 0 || x % 5 == 0))) (_+_)) is the solution witch adds to 233168
my way was to do:
$maxnumber = 1000;
for ($i = 3; $i < $maxnumber; $i += 3)
{
$t += $i;
echo $i.',';
}
echo '<br>';
for ($j = 5; $j < $maxnumber; $j += 5)
{
$d += $j;
echo $j.',';
}
echo '<br>';
echo $t;
echo '<br>';
echo $d;
echo '<br>';
echo $t+$d;
this will give me :
3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96,99,102,105,108,111,114,117,120,123,126,129,132,135,138,141,144,147,150,153,156,159,162,165,168,171,174,177,180,183,186,189,192,195,198,201,204,207,210,213,216,219,222,225,228,231,234,237,240,243,246,249,252,255,258,261,264,267,270,273,276,279,282,285,288,291,294,297,300,303,306,309,312,315,318,321,324,327,330,333,336,339,342,345,348,351,354,357,360,363,366,369,372,375,378,381,384,387,390,393,396,399,402,405,408,411,414,417,420,423,426,429,432,435,438,441,444,447,450,453,456,459,462,465,468,471,474,477,480,483,486,489,492,495,498,501,504,507,510,513,516,519,522,525,528,531,534,537,540,543,546,549,552,555,558,561,564,567,570,573,576,579,582,585,588,591,594,597,600,603,606,609,612,615,618,621,624,627,630,633,636,639,642,645,648,651,654,657,660,663,666,669,672,675,678,681,684,687,690,693,696,699,702,705,708,711,714,717,720,723,726,729,732,735,738,741,744,747,750,753,756,759,762,765,768,771,774,777,780,783,786,789,792,795,798,801,804,807,810,813,816,819,822,825,828,831,834,837,840,843,846,849,852,855,858,861,864,867,870,873,876,879,882,885,888,891,894,897,900,903,906,909,912,915,918,921,924,927,930,933,936,939,942,945,948,951,954,957,960,963,966,969,972,975,978,981,984,987,990,993,996,999
5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100,105,110,115,120,125,130,135,140,145,150,155,160,165,170,175,180,185,190,195,200,205,210,215,220,225,230,235,240,245,250,255,260,265,270,275,280,285,290,295,300,305,310,315,320,325,330,335,340,345,350,355,360,365,370,375,380,385,390,395,400,405,410,415,420,425,430,435,440,445,450,455,460,465,470,475,480,485,490,495,500,505,510,515,520,525,530,535,540,545,550,555,560,565,570,575,580,585,590,595,600,605,610,615,620,625,630,635,640,645,650,655,660,665,670,675,680,685,690,695,700,705,710,715,720,725,730,735,740,745,750,755,760,765,770,775,780,785,790,795,800,805,810,815,820,825,830,835,840,845,850,855,860,865,870,875,880,885,890,895,900,905,910,915,920,925,930,935,940,945,950,955,960,965,970,975,980,985,990,995
$t - 166833
$d - 99500
and total:
266333
why am i wrong?

Some numbers are multiples of both 3 and 5. (Your algorithm adds these numbers to the total twice.)

Because 6 * 5 == 30 and 10 * 3 == 30, you're adding the some numbers up twice.
$sum = 0;
$i = 0;
foreach(range(0, 999) as $i) {
if($i % 3 == 0 || $i % 5 == 0) $sum += $i;
}

Because you double-count numbers that are multiple of both 3 and 5, i.e. multiples of 15.
You can account for this naively by subtracting all multiples of 15.
for ($j = 15; $j < $maxnumber; $j += 15)
{
$e += $j;
echo $j.',';
}
$total = $total - $d;

In your case, if it is 15, you will add the number twice.
Try this:
$t = 0;
$d = 0;
for ($i = 0; $i <= $maxnumber; $i++){
if ($i % 3 == 0)
$t+= $i;
else if ($i % 5 == 0)
$d += $i;
}
echo $t.'<br>'.$d;

I think that in your code, if a number is a multiple of 3 and 5, it is added twice. Take 15 for example. It's in your list of multiples of 3 and in the list of multiples of 5. Is this the behaviour you want?

One of the best approach to this solution (to achieve optimum time complexity), run an Arithmetic Progression series and find the number of terms in all series by using AP formula: T=a+(n-1)d, then find sum by : S=n/2[2*a+(n-1)d]
where : a=first term ,n=no. of term , d=common deference, T=nth term
The code solution below has been implemented to suit the question above - so the values 3 and 5 are hard-coded. However, the function can modified such that values are passed in as variable parameters.
function solution($number){
$val1 = 3;
$val2 = 5;
$common_term = $val1 * $val2;
$sum_of_terms1 = calculateSumofMulitples($val1,$number);
$sum_of_terms2 = calculateSumofMulitples($val2,$number);
$sum_of_cterms = calculateSumofMulitples($common_term,$number);
$final_result = $sum_of_terms1 + $sum_of_terms2 - $sum_of_cterms;
return $final_result;
}
function calculateSumofMulitples($val, $number)
{
//first, we begin by running an aithmetic prograssion for $val up to $number say N to get the no of terms [using T=a +(n-1)d]
$no_of_terms = (int) ($number / $val);
if($number % $val == 0) $no_of_terms = (int) ( ($number-1)/$val ); //since we are computing for a no of terms below N and not up to/exactly/up to N. if N divides val with no remainder, use no of terms = (N-1)/val
//second, we compute the sum of terms [using Sn = n/2[2a + (n-1)d]
$sum_of_terms = ($no_of_terms * 0.5) * ( (2*$val) + ($no_of_terms - 1) * $val );
// sum of multiples
return $sum_of_terms;
}

You can run a single loop checking whether the number is multiple of 3 OR 5:
for ($i = 0; $i < $maxnumber; $i++)
{
if($i%3 || $i%5){
$t += $i;
echo $i.',';}
}

I think the original code is not including numbers which are multiples of both 3 and 5 in the total: if the test for multiple of 3 matches, it takes that and goes on.
If you total the multiples of 15 up to 1000, you get 33165, which is exactly the difference between your total, 266333, and the original total, 233168.

Here's my solution to the question:
<?php
$sum = 0;
$arr = [];
for($i = 1; $i < 1000; $i++){
if((int)$i % 3 === 0 || (int)$i % 5 === 0)
{
$sum += $i;
array_push($arr,$i);
}
}
echo $sum;
echo '<br>';
print_r($arr);//Displays the values meeting the criteria as an array of values

php random x digit number

I need to create a random number with x amount of digits.
So lets say x is 5, I need a number to be eg. 35562
If x is 3, then it would throw back something like; 463
Could someone show me how this is done?

You can use rand() together with pow() to make this happen:
$digits = 3;
echo rand(pow(10, $digits-1), pow(10, $digits)-1);
This will output a number between 100 and 999. This because 10^2 = 100 and 10^3 = 1000 and then you need to subtract it with one to get it in the desired range.
If 005 also is a valid example you'd use the following code to pad it with leading zeros:
$digits = 3;
echo str_pad(rand(0, pow(10, $digits)-1), $digits, '0', STR_PAD_LEFT);

I usually just use RAND() http://php.net/manual/en/function.rand.php
e.g.
rand ( 10000 , 99999 );
for your 5 digit random number

Here is a simple solution without any loops or any hassle which will
allow you to create random string with characters, numbers or even with special symbols.
$randomNum = substr(str_shuffle("0123456789"), 0, $x);
where $x can be number of digits
Eg.
substr(str_shuffle("0123456789"), 0, 5);
Results after a couple of executions
98450
79324
23017
04317
26479
You can use the same code to generate random string also, like this
$randomNum=substr(str_shuffle("0123456789abcdefghijklmnopqrstvwxyzABCDEFGHIJKLMNOPQRSTVWXYZ"), 0, $x);
Results with $x = 11
FgHmqpTR3Ox
O9BsNgcPJDb
1v8Aw5b6H7f
haH40dmAxZf
0EpvHL5lTKr

You can use rand($min, $max) for that exact purpose.
In order to limit the values to values with x digits you can use the following:
$x = 3; // Amount of digits
$min = pow(10,$x);
$max = pow(10,$x+1)-1);
$value = rand($min, $max);

Treat your number as a list of digits and just append a random digit each time:
function n_digit_random($digits) {
$temp = "";
for ($i = 0; $i < $digits; $i++) {
$temp .= rand(0, 9);
}
return (int)$temp;
}
Or a purely numerical solution:
function n_digit_random($digits)
return rand(pow(10, $digits - 1) - 1, pow(10, $digits) - 1);
}

the simplest way i can think of is using rand function with str_pad
<?php
echo str_pad(rand(0,999), 5, "0", STR_PAD_LEFT);
?>
In above example , it will generate random number in range 0 to 999.
And having 5 digits.

function random_numbers($digits) {
$min = pow(10, $digits - 1);
$max = pow(10, $digits) - 1;
return mt_rand($min, $max);
}
Tested here.

rand(1000, 9999); works more faster than x4 times rand(0,9);
benchmark:
rand(1000, 9999) : 0.147 sec.
rand(0,9)x4 times : 0.547 sec.
both functions was running in 100000 iterations to make results more explicit

Well you can use as simple php function mt_rand(2000,9000) which can generate a 4 digit random number
mt_rand(2000,9000)

You can generate any x-digit random number with mt_rand() function.
mt_rand() is faster than rand().
Syntax : mt_rand() or mt_rand($min , $max).
Example : <?php echo mt_rand(); ?>
read more

do it with a loop:
function randomWithLength($length){
$number = '';
for ($i = 0; $i < $length; $i++){
$number .= rand(0,9);
}
return (int)$number;
}

rand or mt_rand will do...
usage:
rand(min, max);
mt_rand(min, max);

function random_number($size = 5)
{
$random_number='';
$count=0;
while ($count < $size )
{
$random_digit = mt_rand(0, 9);
$random_number .= $random_digit;
$count++;
}
return $random_number;
}

Following is simple method to generate specific length verification code. Length can be specified, by default, it generates 4 digit code.
function get_sms_token($length = 4) {
return rand(
((int) str_pad(1, $length, 0, STR_PAD_RIGHT)),
((int) str_pad(9, $length, 9, STR_PAD_RIGHT))
);
}
echo get_sms_token(6);

this simple script will do
$x = 4;//want number of digits for the random number
$sum = 0;
for($i=0;$i<$x;$i++)
{
$sum = $sum + rand(0,9)*pow(10,$i);
}
echo $sum;

This is another simple solution to generate random number of N digits:
$number_of_digits = 10;
echo substr(number_format(time() * mt_rand(),0,'',''),0,$number_of_digits);
Check it here: http://codepad.org/pyVvNiof

function rand_number_available($already_mem_array,$boundary_min,$boundary_max,$digits_num)
{
$already_mem_array_dim = count($already_mem_array); // dimension of array, that contain occupied elements
// --- creating Boundaries and possible Errors
if( empty($digits_num) ){
$boundary_dim = $boundary_max - $boundary_min;
if($boundary_dim <= 0){
$error = -1; // Error that might happen. Difference between $boundary_max and $boundary_min must be positive
}else{
$error = -2; // Error that might happen. All numbers between, $boundary_min and $boundary_max , are occupied, by $already_mem_array
}
}else{
if($digits_num < 0){ // Error. If exist, $digits_num must be, 1,2,3 or higher
$error = -3;
}elseif($digits_num == 1){ // if 'one-figure' number
$error = -4; // Error that might happen. All 'one-figure' numbers are occupied, by $already_mem_array
$boundary_min = 0;
$boundary_max = 9;
$boundary_dim = $boundary_max-$boundary_min;
}elseif($digits_num == 2){ // if 'two-figure' number
$error = -5; // Error that might happen. All 'two-figure' numbers are occupied, by $already_mem_array
$boundary_min = 10;
$boundary_max = 99;
$boundary_dim = $boundary_max-$boundary_min;
}elseif($digits_num>2){ // if 'X-figure' number. X>2
$error = -6; // Error that might happen. All 'X-figure' numbers are occupied, by $already_mem_array. Unlikely to happen
$boundary_min = pow(10, $digits_num-1); // stepenovanje - graduation
$boundary_max = pow(10, $digits_num)-1;
$boundary_dim = $boundary_max-$boundary_min;
}
}
// -------------------------------------------------------------------
// --- creating response ---------------------------------------------
if( ($already_mem_array_dim <= $boundary_dim) && $boundary_dim>0 ){ // go here only if, there are AVAILABLE numbers to extract, and [difference] $boundary_dim , is positive
do{
$num = rand($boundary_min,$boundary_max);
}while( in_array($num, $already_mem_array) );
$result = $num;
}else{
$result = $error; // Limit that happened
}
return $result;
// -------------------------------------------------------------------
}

This function works perfectly with no repeats and desired number of digits.
$digits = '';
function randomDigits($length){
$numbers = range(0,9);
shuffle($numbers);
for($i = 0; $i < $length; $i++){
global $digits;
$digits .= $numbers[$i];
}
return $digits;
}
You can call the function and pass the number of digits for example:
randomDigits(4);
sample results:
4957 8710 6730 6082 2987 2041 6721
Original script got from this gist

Please not that rand() does not generate a cryptographically secure value according to the docs:
http://php.net/manual/en/function.rand.php
This function does not generate cryptographically secure values, and should not be used for cryptographic purposes. If you need a cryptographically secure value, consider using random_int(), random_bytes(), or openssl_random_pseudo_bytes() instead.
Instead it is better to use random_int(), available on PHP 7 (See: http://php.net/manual/en/function.random-int.php).
So to extend #Marcus's answer, you should use:
function generateSecureRandomNumber($digits): int {
return random_int(pow(10, $digits - 1), pow(10, $digits) - 1);
}
function generateSecureRandomNumberWithPadding($digits): string {
$randomNumber = random_int(0, pow(10, $digits) - 1);
return str_pad($randomNumber, $digits, '0', STR_PAD_LEFT);
}
Note that using rand() is fine if you don't need a secure random number.

The following code generates a 4 digits random number:
echo sprintf( "%04d", rand(0,9999));

you people really likes to complicate things :)
the real problem is that the OP wants to, probably, add that to the end of some really big number. if not, there is no need I can think of for that to be required. as left zeros in any number is just, well, left zeroes.
so, just append the larger portion of that number as a math sum, not string.
e.g.
$x = "102384129" . complex_3_digit_random_string();
simply becomes
$x = 102384129000 + rand(0, 999);
done.

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