Maximum Fruits Harvested After at Most K Steps LeetCode solution in JavaScript

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/**
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 * @param {number[][]} fruits
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 * @param {number} startPos
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 * @param {number} k
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 * @return {number}
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 */
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var maxTotalFruits = function (fruits, startPos, k) {
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  let n = Math.max(fruits[fruits.length - 1][0], startPos) + 1;
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  let numFruits = new Array(n).fill(0);
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  let sums = new Array(n).fill(0);
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  for (let obj of fruits) {
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    let [pos, num] = obj;
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    numFruits[pos] = num;
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  }
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  sums[startPos] = numFruits[startPos];
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  for (let i = startPos + 1; i < n && i <= startPos + k; i++) {
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    sums[i] = sums[i - 1] + numFruits[i];
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  }
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  for (let i = startPos - 1; i >= 0 && i >= startPos - k; i--) {
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    sums[i] = sums[i + 1] + numFruits[i];
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  }
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  //console.log(sums);
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  let output = 0;
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  for (let leftMoves = k; leftMoves >= 0; leftMoves--) {
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    let rightMoves = Math.max(k - 2 * leftMoves, 0);
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    let leftPos = Math.max(0, startPos - leftMoves);
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    let rightPos = Math.min(n - 1, startPos + rightMoves);
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    let count = sums[leftPos] + sums[rightPos] - sums[startPos];
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    output = Math.max(output, count);
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  }
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  for (let rightMoves = k; rightMoves >= 0; rightMoves--) {
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    let leftMoves = Math.max(k - 2 * rightMoves, 0);
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    let leftPos = Math.max(0, startPos - leftMoves);
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    let rightPos = Math.min(n - 1, startPos + rightMoves);
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    let count = sums[leftPos] + sums[rightPos] - sums[startPos];
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    output = Math.max(output, count);
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  }
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  return output;
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};