Maximum Trailing Zeros in a Cornered Path LeetCode solution in JavaScript

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var maxTrailingZeros = function (grid) {
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  const m = grid.length;
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  const n = grid[0].length;
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  const postfixCols = [];
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  for (let i = 0; i < m; ++i) {
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    for (let j = 0; j < n; ++j) {
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      const num = grid[i][j];
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      if (postfixCols[j] == null) postfixCols[j] = { 2: 0, 5: 0 };
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      postfixCols[j]["2"] += getCount(num, 2);
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      postfixCols[j]["5"] += getCount(num, 5);
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    }
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  }
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  const prefixCols = [];
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  for (let j = 0; j < n; ++j) {
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    prefixCols[j] = { 0: 0, 2: 0, 5: 0 };
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  }
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  let maxZeros = 0;
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  for (let i = 0; i < m; ++i) {
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    const postfixRow = { 0: 0, 2: 0, 5: 0 };
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    for (let j = n - 1; j >= 0; --j) {
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      const num = grid[i][j];
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      postfixRow["2"] += getCount(num, 2);
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      postfixRow["5"] += getCount(num, 5);
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    }
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    let prefixRow = { 0: 0, 2: 0, 5: 0 };
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    for (let j = 0; j < n; ++j) {
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      const num = grid[i][j];
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      const twoCount = getCount(num, 2);
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      const fiveCount = getCount(num, 5);
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      postfixRow["2"] -= twoCount;
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      postfixCols[j]["2"] -= twoCount;
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      postfixRow["5"] -= fiveCount;
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      postfixCols[j]["5"] -= fiveCount;
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      // down-right => prefixCol + postfixRow
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      const downRight = calculateTrailingZeros(prefixCols[j], postfixRow, num);
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      // down-left => prefixCol + prefixRow
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      const downLeft = calculateTrailingZeros(prefixCols[j], prefixRow, num);
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      // up-right => postfixCols + postfixRow
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      const upRight = calculateTrailingZeros(postfixCols[j], postfixRow, num);
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      // up-left => postfixCols + prefixRow
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      const upLeft = calculateTrailingZeros(postfixCols[j], prefixRow, num);
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      maxZeros = Math.max(maxZeros, downRight, downLeft, upRight, upLeft);
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      prefixRow["2"] += twoCount;
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      prefixCols[j]["2"] += twoCount;
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      prefixRow["5"] += fiveCount;
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      prefixCols[j]["5"] += fiveCount;
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    }
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  }
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  return maxZeros;
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  function calculateTrailingZeros(col, row, currNum) {
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    let twosCount = 0;
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    let fivesCount = 0;
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    let zerosCount = 0;
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    twosCount = row["2"] + col["2"];
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    fivesCount = row["5"] + col["5"];
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    twosCount += getCount(currNum, 2);
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    fivesCount += getCount(currNum, 5);
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    return Math.min(twosCount, fivesCount);
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  }
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  function getCount(num, divisor) {
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    let count = 0;
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    while (num % divisor === 0) {
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      ++count;
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      num /= divisor;
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    }
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    return count;
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  }
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};