Most Stones Removed with Same Row or Column LeetCode solution in JavaScript

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/**
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 * @param {number[][]} stones
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 * @return {number}
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 */
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var removeStones = function (stones) {
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  const n = stones.length;
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  // initial number of components(a.k.a island)
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  let numComponents = n;
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  //initial forest for union find
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  let forest = new Array(n).fill(0).map((ele, index) => index);
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  // recursively finding the root of rarget node
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  const find = (a) => {
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    if (forest[a] === a) {
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      return a;
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    }
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    return find(forest[a]);
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  };
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  // function for uniting two stones
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  const union = (a, b) => {
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    const rootA = find(a);
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    const rootB = find(b);
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    if (rootA != rootB) {
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      //connect their roots if currently they are not connected
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      forest[rootA] = rootB;
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      //subtract the number of islands by one since two islands are connected now
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      numComponents -= 1;
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    }
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  };
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  for (let i = 0; i < stones.length - 1; i++) {
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    for (let j = i + 1; j < stones.length; j++) {
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      // if two stones are connected(i.e. share same row or column), unite them
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      if (stones[i][0] === stones[j][0] || stones[i][1] === stones[j][1]) {
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        union(i, j);
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      }
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    }
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  }
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  // this is the most confusing part.
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  // The number of stones can be removed is not equal to the number of islands.
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  // e.g. we have two islands with total of 10 stores, each island will leave one extra stone after the
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  // removal, therefore we can remove 10 - 2 = 8 stones in total
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  return n - numComponents;
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};