Odd Occurrences In Array exercise from Codility (JavaScript/NodeJS)

OddOccurrencesInArray.js

/**
 * Finds element without a pair in an unsorted array of integers
 * @param {array} A
 * @return {int} element without a pair
 */
function solution(A) {
    let result = 0;
  
    for (let element of A) {
        // Apply Bitwise XOR to the current and next element
        result ^= element
    }

    return result
}


// =======================
// SLOW
// =======================
function slowSolution(A) {
    // Convert the array to a map with element as a key and number of occurrences as a value
    // Example: { '3': 2, '7': 1, '9': 4 }
    let map = A.reduce((map, obj) => {
        map[obj] = ++map[obj] || 1
        return map
    }, {})
    
    // Then filter the map by checking if the key's value is equal to 1
    // Returns array with key
    let filteredArr = Object.keys(map).filter(el => map[el] === 1)
    
    return  +filteredArr[0] || null
}

Fast and understandable 🤯 solution.

function solution(A) {
  A = A.sort();

  let val = A[A.length - 1];
  for (let i = 0; i < (A.length - 1) / 2; i++) {
    if (A[i * 2] !== A[i * 2 + 1]) {
      val = A[i * 2];
      break;
    }
  }
  return val;
}

@k0ff33 I can't figure out how Bitwise XOR operator does work.
@RomaSto It is still O(N**2).