Equal Sum Arrays With Minimum Number of Operations LeetCode solution in JavaScript

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var minOperations = function (nums1, nums2) {
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  /*
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    Two impossible cases:
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    min1...max1 ... min2...max2 -> we can't make them equal
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    min2...max2 ... min1...max1 -> we can't make them equal
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    */
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  let min1 = nums1.length * 1;
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  let max1 = nums1.length * 6;
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  let min2 = nums2.length * 1;
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  let max2 = nums2.length * 6;
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  if (min2 > max1 || min1 > max2) {
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    return -1;
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  }
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  let sum1 = nums1.reduce((acc, cur) => acc + cur);
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  let sum2 = nums2.reduce((acc, cur) => acc + cur);
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  if (sum1 === sum2) return 0;
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  if (sum1 < sum2) {
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    return helper(nums1, nums2, sum1, sum2);
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  } else {
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    return helper(nums2, nums1, sum2, sum1);
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  }
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  // T.C: O(M + N), M = # of nums1, N = # of nums2
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  // S.C: O(1)
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};
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// Condition: sum of A < sum of B
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// the idea is to add the maximum possible value to sumA and
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// subtract the maximum possible value from sumB so that
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// we make sumA >= sumB as soon as possible
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function helper(A, B, sumA, sumB) {
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  let freqA = new Array(7).fill(0);
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  let freqB = new Array(7).fill(0);
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  for (let i = 0; i < A.length; i++) {
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    freqA[A[i]]++;
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  }
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  for (let i = 0; i < B.length; i++) {
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    freqB[B[i]]++;
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  }
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  let count = 0;
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  for (let i = 1; i <= 6; i++) {
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    // increase sumA
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    while (freqA[i] > 0 && sumA < sumB) {
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      sumA += 6 - i;
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      freqA[i]--;
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      count++;
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    }
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    let j = 7 - i;
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    // decrease sumB
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    while (freqB[j] > 0 && sumA < sumB) {
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      sumB -= j - 1;
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      freqB[j]--;
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      count++;
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    }
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    if (sumA >= sumB) break;
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  }
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  return count;
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}