Migratory Birds solution - PHP-

user_19203790 0
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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?

php
3 Answers

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;
    }
user_118445140

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()
user_195435290

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;
user_42903350
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