Migratory Birds solution - PHP- - php

Question

I wrote this solution for this practice in PHP but it's not work for all case:
Given an array of bird sightings where every element represents a bird type id, determine the id of the most frequently sighted type. If more than 1 type has been spotted that maximum amount, return the smallest of their ids.
arr=[1,1,2,2,3]
Example
There are two each of types 1 and 2 , and one sighting of type .3 Pick the lower of the two types seen twice: type 1.
Function Description
Complete the migratoryBirds function in the editor below.
migratoryBirds has the following parameter(s):
int arr[n]: the types of birds sighted
Returns
int: the lowest type id of the most frequently sighted birds
Input Format
The first line contains an integer,n , the size of arr .
The second line describes arr as n space-separated integers, each a type number of the bird sighted.
Constraints
5 < n < 2 X 10 2
It is guaranteed that each type is 1,2 ,3 ,4 , or 5 .
Sample Input 0
6
1 4 4 4 5 3
Sample Output 0
4
this is my code
function migratoryBirds($arr) {
// Write your code here
$length=count($arr);
$a1=0;$a2=0;$a3=0;$a4=0;$a5=0;
for($i=0; $i < $length; $i++){
if($arr[$i]==1){
$a1++;
}
if($arr[$i]==2){
$a2++;
}
if($arr[$i]==3){
$a3++;
}
if($arr[$i]==4){
$a4++;
}
if($arr[$i]==5){
$a5++;
}
}
if($a1>=$a2 && $a1>=$a3 && $a1>=$a4 && $a1>=$a5){
$result=1;
}
if($a2>=$a1 && $a2>=$a3 && $a2>=$a4 && $a2>=$a5){
$result=2;
}
if($a3>=$a2 && $a3>=$a1 && $a3>=$a4 && $a3>=$a5){
$result=3;
}
if($a4>=$a2 && $a4>=$a3 && $a4>=$a1 && $a4>=$a5){
$result=4;
}
if($a5>=$a2 && $a5>=$a3 && $a5>=$a4 && $a5>=$a1){
$result=5;
}
return $result;
}
How can I solve it?

Answer 1

write your condition like this
if($a1>=$a2 && $a1>=$a3 && $a1>=$a4 && $a1>=$a5){
$result=1;
}
else if($a2>=$a1 && $a2>=$a3 && $a2>=$a4 && $a2>=$a5){
$result=2;
}
else if($a3>=$a2 && $a3>=$a1 && $a3>=$a4 && $a3>=$a5){
$result=3;
}
else if($a4>=$a2 && $a4>=$a3 && $a4>=$a1 && $a4>=$a5){
$result=4;
}
else if($a5>=$a2 && $a5>=$a3 && $a5>=$a4 && $a5>=$a1){
$result=5;
}
else{
$result=1;
}

Answer 2

Given an array of bird sightings where every element represents a bird type id, determine the id of the most frequently sighted type. If more than 1 type has been spotted that maximum amount, return the smallest of their ids.
For this question I wrote a code.
It passed all the test cases.
My code is as follows:
import math
import os
import random
import re
import sys
def migratoryBirds(arr):
x = [0] *(max(arr)+1)
for i in arr:
x[i] += 1
return x.index(max(x))
if name == 'main':
fptr = open(os.environ['OUTPUT_PATH'], 'w')
arr_count = int(input().strip())
arr = list(map(int, input().rstrip().split()))
result = migratoryBirds(arr)
fptr.write(str(result) + '\n')
fptr.close()

Answer 3

Here try this one:
$array = array(1,2 ,3, 4, 5, 4, 3, 2, 2, 1, 3, 4);
$counted = array_count_values($array);
$n = null;
$v = 0;
foreach($counted as $nm => $val) {
if ($val > $v) {
$n = $nm;
$v = $val;
}
}
echo $n;

Answer 4

Hope this will help you.
function migratoryBirds($arr) {
// Write your code here
foreach($arr as $value){
if(isset($new_arr[$value])){
$new_arr[$value] += 1;
}else{
$new_arr[$value] = 1;
}
}
$min_key = '';
$max_val = 0;
for($i=(count($new_arr)); $i>=0; $i--){
if(isset($new_arr[$i])){
if($new_arr[$i] >= $max_val){
$min_key = $i;
$max_val = $new_arr[$i];
}
}
}
return $min_key;
}