Given an array of integers, find the one that appears an odd number of times. There will always be only one integer that appears an odd number of times code example

Example 1: you are given an array of integers. your task is to print the sum of numbers that occurs for an even number of times in the array.

import collections
def fun(arr):
    mp = collections.defaultdict(int)
      
    for i in range(len(arr)):
        mp[arr[i]] += 1 
    sum = 0 
    for i in mp.keys(): 
          
        
        if (mp[i] % 2 == 0): 
            sum += i
    return sum
n= int(input())
arr = list(map(int,input().split()))
print(fun(arr))

Example 2: Given an array of integers, find the one that appears an odd number of times. There will always be only one integer that appears an odd number of times

function findOdd(A) {
    var countOccurencesOfInt = 0;
    for (let i = 0; i < A.length; i++) {
        var currentIterationInt = A[i];
        for (let j = 0; j < A.length; j++) {
            if (currentIterationInt == A[j]) {
                countOccurencesOfInt++;
            }
        }
        if (countOccurencesOfInt % 2 != 0) {
            return currentIterationInt;
        }
    }
}
//or
function findOdd(arr) {
  var result, num = 0;

  arr = arr.sort();
  for (var i = 0; i < arr.length; i++) {
    if (arr[i] === arr[i+1]) {
      num++;
    } else {
      num++;
      if (num % 2 != 0) {
        result = arr[i];
        break;
      }
    }
  }
  return result;
}