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Counting Sort in Javascript

Counting Sort — Perfect when you have psoitive integers to sort

Counting Sort for Unique Integer Values

function countingUniqueSort(arr)
{
  // set 1 for the index which is there in the input array
  const count = [];
  for (let i=0; i < arr.length; i++) {
    count[arr[i]] = 1;
  }
  // set indices over input array in order
  let j = 0;
  for (i=0; i<=count.length; i++) {
      if (count[i] === 1) {
        arr[j] = i;
        j++;
      }
  }
  return arr;
}

Counting Sort for possibly duplicate Integer Values

function countingSort(arr)
{
  // increase the value at the index which is there in the input array
  const count = [];
  for (let i=0; i < arr.length; i++) {
    if (count[arr[i]] > 0) {
      count[arr[i]]++;
    } else {
      count[arr[i]] = 1;
    }
  }
  // set indices over input array in order
  let j = 0;
  let k = 0;
  for (i=0; i<=count.length; i++) {
      k = count[i];
      while (k > 0) {
        arr[j] = i;
        j++;
        k--;
      }
  }
  return arr;
}

Test Input

console.log(`Original Array of Unique Elements: ${[3, 2, 5, 4, 8, 7]}`); 
console.log(`Sorted Array of Unique Elements: ${countingUniqueSort([3, 2, 5, 4, 8, 7])}`);

console.log(`Original Array Elements: ${[3, 2, 5, 5, 4, 8, 7]}`); 
console.log(`Sorted Array Elements: ${countingSort([3, 2, 5, 5, 4, 8, 7])}`);

Output

Original Array of Unique Elements: 3,2,5,4,8,7
Sorted Array of Unique Elements: 2,3,4,5,7,8
Original Array Elements: 3,2,5,5,4,8,7
Sorted Array Elements: 2,3,4,5,5,7,8

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