Min heap in javascript

Question

Hey I tried to implement a min heap in javascript, but i had a question regarding the algorithm for remove min. I'm using an array to represent the heap internally. When I'm percolating downwards, what should be the stopping condition? In my code I have it at 2 * k <= this.size so it would travel down to potentially the very last element, but it doesn't feel "correct", is there a better stopping condition? Thanks in advance!

this.removeMin = function () {
    //replace root with last element and percolate downwards
    var min = this._heap[1],
        k,
        left,
        right;

    this._heap[1] = this._heap.pop();
    this.size--;
    k = 1;

    while ( ( 2 * k ) <= this.size) {
        left = 2 * k;
        right = 2 * k + 1;

        if (this._heap[k] > this._heap[left] && this._heap[k] > this._heap[right]) {
            if (this._heap[left] <= this._heap[right]) {
                swap(this._heap, k, left);
                k = left;
            } else {
                swap(this._heap, k, right);
                k = right;
            }
        } else if (this._heap[k] > this._heap[left]) {
            swap(this._heap, k, left);
            k = left;
        } else {
            swap(this._heap, k, right);
            k = right;
        }
    }

    return min;
};

Why? The index basically doubles every step, so you should get there in O(log n) — Stefan Haustein, Mar 14 '16 at 01:51
I think this is the correct condition. As you are checking with a number of elements inserted into the array, not with array length. — Prasanna, Jun 06 '20 at 17:17

Wesam Mustafa · Answer 1 · 2022-06-21T18:27:23.917

This is a much easier implementation, I hope this can help.

class MinHeap {
  constructor(array) {
    this.heap = this.buildHeap(array);
  }

  // O(n) time | O(1) space
  buildHeap(array) {
    const firstParentIdx = Math.floor((array.length - 2) / 2);
    for (let currentIdx = firstParentIdx; currentIdx >= 0; currentIdx--) {
      this.siftDown(currentIdx, array.length - 1, array);
    }
    return array;
  }

  // O(log(n)) time | O(1) space
  siftDown(currentIdx, endIdx, heap) {
    let childOneIdx = currentIdx * 2 + 1;
    while (childOneIdx <= endIdx) {
      const childTwoIdx = currentIdx * 2 + 2 <= endIdx ? currentIdx * 2 + 2 : -1;
      let idxToSwap;
      if (childTwoIdx !== -1 && heap[childTwoIdx] < heap[childOneIdx]) {
        idxToSwap = childTwoIdx;
      } else {
        idxToSwap = childOneIdx;
      }

      if (heap[idxToSwap] < heap[currentIdx]) {
        this.swap(currentIdx, idxToSwap, heap);
        currentIdx = idxToSwap;
        childOneIdx = currentIdx * 2 + 1;
      } else {
        return;
      }
    }
  }

  // O(log(n)) time | O(1) space
  siftUp(currentIdx, heap) {
    let parentIdx = Math.floor((currentIdx - 1) / 2);
    while (currentIdx > 0 && heap[currentIdx] < heap[parentIdx]) {
     this.swap(currentIdx, parentIdx, heap);
      currentIdx = parentIdx;
      parentIdx = Math.floor((currentIdx - 1) / 2);
    }
  }

  // O(1)  time | O(1) space
  peek() {
    return this.heap[0];
  }

  // O(log(n)) time | O(1) space
  remove() {
    this.swap(0, this.heap.length - 1, this.heap);
    const valueToRemove = this.heap.pop();
    this.siftDown(0, this.heap.length - 1, this.heap);
    return valueToRemove;
  }

  // O(log(n)) time | O(1) space
  insert(value) {
    this.heap.push(value);
    this.siftUp(this.heap.length - 1, this.heap);
  }

  swap(i, j, heap) {
    [heap[i], heap[j]] = [heap[j], heap[i]];
  }
}

score 1 · Answer 2 · answered Mar 14 '16 at 02:22

I think you miss one if condition. When the k element is both less than the right and the left, the downwards must stop. It must be:

   if (this._heap[k] > this._heap[left] && this._heap[k] > this._heap[right]) {
        if (this._heap[left] <= this._heap[right]) {
            swap(this._heap, k, left);
            k = left;
        } else {
            swap(this._heap, k, right);
            k = right;
        }
    } else if (this._heap[k] > this._heap[left]) {
        swap(this._heap, k, left);
        k = left;
    } else if(this._heap[k] < this._heap[right]) {
        swap(this._heap, k, right);
        k = right;
    }else{
        break;
    }

score 0 · Answer 3 · answered Jun 06 '20 at 17:30

Why you are writing compare code again and again. Sharing below code for reference which will optimize your code, you can replace it with your while loop.

.....
while (2 * k <= this.size) {
        let j = 2 * key;
        if (j < this.size && less(j, j + 1)) j++; // find smallest child
        if (!less(k, j)) break; // check parent is lesser than smallest child or not
        exch(k, j); //if parent is bigger then exchange
        k = j; //keep checking untill reaches to the end(this.size)
    }
.....
function less(i, j){
    if(this._heap[j] < this._heap[i] < 0) return true;
    else return false;
}
function exch(i, j){
    let temp = this._heap[i];
    this._heap[i] = this._heap[j];
    this._heap[j] = temp;
}

Hope this works.

Min heap in javascript

3 Answers3