Minimum Sum of Squared Difference LeetCode solution in Java

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class Solution {
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  /**
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   * Algorithm 1. Count the differences between each nums1[i] and nums2[i] and store them into an
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   * int[100_001], as nums is between 0 and 100_000. 2. Let's look at the example of [1,4,10,12],
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   * [4,8,6,7]. k1= 1, k2 =1 Looking at the pairs of abs diff we have 3,4,4,5. So a total of 16 diff
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   * points with k = 2. As we observe, if we use the k operations on the first pair, we can decrease
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   * 3 to 1. but this would only help with 3^2 (9) -> 1. So we decrease the totam sum diff by 8.
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   * However, if we operate on the diff of 5, this would have much more impact. 5 - 1 => (4^2)25 -
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   * 16 . so we save 9 points by using 1 k 5 - 2 => (3^2) 25 - 9. So we save 16 points. 3. As we can
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   * see, we need to operate on the highest diff, lowering them. 4. As we have counted them on step
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   * #1, we would have an array like this [0,0,0,1,2,1] : 1 diff of 3, 2 of 4 and 1 of 5. 5. While k
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   * is > 0 (k1 + k2), start from the back (highest) and decrease it one group at a time. So make
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   * all 5 diffs into 4 diff, only if their cardinal is <= k. If it's greater than k, we can only
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   * lower k diff to diff -1. So [0,0,0,1,2,1] and k = 2 => [0,0,0,1,3,0] and k =1 We have 3 diff of
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   * 4 and just k =1 so we can turn one 4 into a 3. => [0,0,0,2,2,0]. Thus. the diff becomes 2 of 3
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   * and 2 of 4.
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   */
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  public long minSumSquareDiff(int[] nums1, int[] nums2, int k1, int k2) {
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    long minSumSquare = 0;
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    int[] diffs = new int[100_001];
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    long totalDiff = 0;
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    long kSum = k1 + k2;
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    int currentDiff;
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    int maxDiff = 0;
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    for (int i = 0; i < nums1.length; i++) {
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      // get current diff.
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      currentDiff = Math.abs(nums1[i] - nums2[i]);
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      // if current diff > 0, count/store it. If not,then ignore it.
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      if (currentDiff > 0) {
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        totalDiff += currentDiff;
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        diffs[currentDiff]++;
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        maxDiff = Math.max(maxDiff, currentDiff);
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      }
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    }
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    // if kSum (k1 + k2) < totalDifferences, it means we can make all numbers/differences 0s
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    if (totalDiff <= kSum) {
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      return 0;
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    }
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    // starting from the back, from the highest difference, lower that group one by one to the
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    // previous group.
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    // we need to make all n diffs to n-1, then n-2, as long as kSum allows it.
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    for (int i = maxDiff; i > 0 && kSum > 0; i--) {
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      if (diffs[i] > 0) {
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        // if current group has more differences than the totalK, we can only move k of them to the
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        // lower level.
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        if (diffs[i] >= kSum) {
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          diffs[i] -= kSum;
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          diffs[i - 1] += kSum;
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          kSum = 0;
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        } else {
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          // else, we can make this whole group one level lower.
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          diffs[i - 1] += diffs[i];
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          kSum -= diffs[i];
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          diffs[i] = 0;
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        }
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      }
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    }
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    for (int i = 0; i <= maxDiff; i++) {
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      if (diffs[i] > 0) {
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        minSumSquare += (long) (Math.pow((long) i, 2)) * diffs[i];
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      }
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    }
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    return minSumSquare;
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  }
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}