Description
Minimum Sum of Squared Difference
MediumYou 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.length1 <= n <= 1050 <= nums1[i], nums2[i] <= 1050 <= k1, k2 <= 109
Solutions
Best approach: difference frequency counting approach.
Language: javascript()DONE
CPU Performance100.00%
x
Memory Performance80.00%
x
/**
* @param {number[]} nums1
* @param {number[]} nums2
* @param {number} k1
* @param {number} k2
* @return {number}
*/
var minSumSquareDiff = function (nums1, nums2, k1, k2) {
const n = nums1.length;
let max = 0;
let total = 0;
let k = k1 + k2;
for (let i = 0; i < n; i++) {
const diff = Math.abs(nums1[i] - nums2[i]);
max = Math.max(max, diff);
total += diff;
}
if (k >= total) return 0;
const freq = new Int32Array(max + 1);
for (let i = 0; i < n; i++) {
freq[Math.abs(nums1[i] - nums2[i])]++;
}
for (let d = max; d > 0 && k > 0; d--) {
const reductions = Math.min(k, freq[d]);
freq[d] -= reductions;
freq[d - 1] += reductions;
k -= reductions;
}
let result = 0;
for (let d = 1; d <= max; d++) {
result += d * d * freq[d];
}
return result;
};