g2301_2400.s2333_minimum_sum_of_squared_difference.Solution Maven / Gradle / Ivy

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package g2301_2400.s2333_minimum_sum_of_squared_difference;

// #Medium #Array #Math #Sorting #Heap_Priority_Queue
// #2022_07_12_Time_15_ms_(95.13%)_Space_106.4_MB_(46.91%)

/**
 * 2333 - Minimum Sum of Squared Difference\.
 *
 * Medium
 *
 * You are given two positive **0-indexed** integer arrays `nums1` and `nums2`, both of length `n`.
 *
 * The **sum of squared difference** of arrays `nums1` and `nums2` is defined as the **sum** of (nums1[i] - nums2[i])² for each `0 <= i < n`.
 *
 * You are also given two positive integers `k1` and `k2`. You can modify any of the elements of `nums1` by `+1` or `-1` at most `k1` times. Similarly, you can modify any of the elements of `nums2` by `+1` or `-1` at most `k2` times.
 *
 * Return _the minimum **sum of squared difference** after modifying array_ `nums1` _at most_ `k1` _times and modifying array_ `nums2` _at most_ `k2` _times_.
 *
 * **Note**: You are allowed to modify the array elements to become **negative** integers.
 *
 * **Example 1:**
 *
 * **Input:** nums1 = [1,2,3,4], nums2 = [2,10,20,19], k1 = 0, k2 = 0
 *
 * **Output:** 579
 *
 * **Explanation:** The elements in nums1 and nums2 cannot be modified because k1 = 0 and k2 = 0.
 *
 * The sum of square difference will be: (1 - 2)² \+ (2 - 10)² \+ (3 - 20)² \+ (4 - 19)² = 579. 
 *
 * **Example 2:**
 *
 * **Input:** nums1 = [1,4,10,12], nums2 = [5,8,6,9], k1 = 1, k2 = 1
 *
 * **Output:** 43
 *
 * **Explanation:** One way to obtain the minimum sum of square difference is:
 *
 * - Increase nums1[0] once.
 *
 * - Increase nums2[2] once.
 *
 * The minimum of the sum of square difference will be: (2 - 5)² \+ (4 - 8)² \+ (10 - 7)² \+ (12 - 9)² = 43.
 *
 * Note that, there are other ways to obtain the minimum of the sum of square difference, but there is no way to obtain a sum smaller than 43.
 *
 * **Constraints:**
 *
 * *   `n == nums1.length == nums2.length`
 * *   1 <= n <= 10⁵
 * *   0 <= nums1[i], nums2[i] <= 10⁵
 * *   0 <= k1, k2 <= 10⁹
**/
public class Solution {
    public long minSumSquareDiff(int[] nums1, int[] nums2, int k1, int k2) {
        long minSumSquare = 0;
        int[] diffs = new int[100_001];
        long totalDiff = 0;
        long kSum = (long) k1 + k2;
        int currentDiff;
        int maxDiff = 0;
        for (int i = 0; i < nums1.length; i++) {
            // get current diff.
            currentDiff = Math.abs(nums1[i] - nums2[i]);
            // if current diff > 0, count/store it. If not,then ignore it.
            if (currentDiff > 0) {
                totalDiff += currentDiff;
                diffs[currentDiff]++;
                maxDiff = Math.max(maxDiff, currentDiff);
            }
        }
        // if kSum (k1 + k2) < totalDifferences, it means we can make all numbers/differences 0s
        if (totalDiff <= kSum) {
            return 0;
        }
        // starting from the back, from the highest difference, lower that group one by one to the
        // previous group.
        // we need to make all n diffs to n-1, then n-2, as long as kSum allows it.
        for (int i = maxDiff; i > 0 && kSum > 0; i--) {
            if (diffs[i] > 0) {
                // if current group has more differences than the totalK, we can only move k of them
                // to the lower level.
                if (diffs[i] >= kSum) {
                    diffs[i] -= kSum;
                    diffs[i - 1] += kSum;
                    kSum = 0;
                } else {
                    // else, we can make this whole group one level lower.
                    diffs[i - 1] += diffs[i];
                    kSum -= diffs[i];
                    diffs[i] = 0;
                }
            }
        }
        for (int i = 0; i <= maxDiff; i++) {
            if (diffs[i] > 0) {
                minSumSquare += (long) (Math.pow(i, 2)) * diffs[i];
            }
        }
        return minSumSquare;
    }
}