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  Formatted question description: https://leetcode.ca/all/2333.html
  
  # 2333. Minimum Sum of Squared Difference
  
  • Difficulty: Medium.
  • Related Topics: Array, Math, Sorting, Heap (Priority Queue).
  • Similar Questions: Minimum Absolute Sum Difference, Partition Array Into Two Arrays to Minimize Sum Difference.
  
  ## Problem
  
  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
  
  ## Solution (Java, C++, Python)
  
  • 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;
  }
  }
  
  ############
  
  class Solution {
  public long minSumSquareDiff(int[] nums1, int[] nums2, int k1, int k2) {
  int n = nums1.length;
  int[] d = new int[n];
  long s = 0;
  int mx = 0;
  int k = k1 + k2;
  for (int i = 0; i < n; ++i) {
  d[i] = Math.abs(nums1[i] - nums2[i]);
  s += d[i];
  mx = Math.max(mx, d[i]);
  }
  if (s <= k) {
  return 0;
  }
  int left = 0, right = mx;
  while (left < right) {
  int mid = (left + right) >> 1;
  long t = 0;
  for (int v : d) {
  t += Math.max(v - mid, 0);
  }
  if (t <= k) {
  right = mid;
  } else {
  left = mid + 1;
  }
  }
  for (int i = 0; i < n; ++i) {
  k -= Math.max(0, d[i] - left);
  d[i] = Math.min(d[i], left);
  }
  for (int i = 0; i < n && k > 0; ++i) {
  if (d[i] == left) {
  --k;
  --d[i];
  }
  }
  long ans = 0;
  for (int v : d) {
  ans += (long) v * v;
  }
  return ans;
  }
  }
  
  • class Solution:
  def minSumSquareDiff(
  self, nums1: List[int], nums2: List[int], k1: int, k2: int
  ) -> int:
  d = [abs(a - b) for a, b in zip(nums1, nums2)]
  k = k1 + k2
  if sum(d) <= k:
  return 0
  left, right = 0, max(d)
  while left < right:
  mid = (left + right) >> 1
  if sum(max(v - mid, 0) for v in d) <= k:
  right = mid
  else:
  left = mid + 1
  for i, v in enumerate(d):
  d[i] = min(left, v)
  k -= max(0, v - left)
  for i, v in enumerate(d):
  if k == 0:
  break
  if v == left:
  k -= 1
  d[i] -= 1
  return sum(v * v for v in d)
  
  ############
  
  # 2333. Minimum Sum of Squared Difference
  # https://leetcode.com/problems/minimum-sum-of-squared-difference/
  
  class Solution:
  def minSumSquareDiff(self, nums1: List[int], nums2: List[int], k1: int, k2: int) -> int:
  pq = []
  
  for a, b in zip(nums1, nums2):
  heappush(pq, ((-abs(a - b), 1)))
  
  orders = k1 + k2
  
  while pq and orders > 0:
  flag = True
  top, count = heappop(pq)
  top = -top
  
  while pq and top == -pq[0][0]:
  count += heappop(pq)[1]
  nextTop = 0
  if pq:
  nextTop = -pq[0][0]
  
  delta = top - nextTop
  if delta * count <= orders:
  orders -= (delta * count)
  else:
  flag = False
  heappush(pq, (-top, count))
  
  while pq and orders > 0:
  x, cnt = heappop(pq)
  take = min(orders, cnt)
  orders -= take
  heappush(pq, (x + 1, take))
  if cnt != take:
  heappush(pq, (x, cnt - take))
  
  if flag and nextTop:
  heappush(pq, (-nextTop, count))
  
  total = 0
  for x, count in pq:
  total += x * x * count
  
  
  • using ll = long long;
  
  class Solution {
  public:
  long long minSumSquareDiff(vector<int>& nums1, vector<int>& nums2, int k1, int k2) {
  int n = nums1.size();
  vector<int> d(n);
  ll s = 0;
  int mx = 0;
  int k = k1 + k2;
  for (int i = 0; i < n; ++i) {
  d[i] = abs(nums1[i] - nums2[i]);
  s += d[i];
  mx = max(mx, d[i]);
  }
  if (s <= k) return 0;
  int left = 0, right = mx;
  while (left < right) {
  int mid = (left + right) >> 1;
  ll t = 0;
  for (int v : d) t += max(v - mid, 0);
  if (t <= k)
  right = mid;
  else
  left = mid + 1;
  }
  for (int i = 0; i < n; ++i) {
  k -= max(0, d[i] - left);
  d[i] = min(d[i], left);
  }
  for (int i = 0; i < n && k; ++i) {
  if (d[i] == left) {
  --k;
  --d[i];
  }
  }
  ll ans = 0;
  for (int v : d) ans += 1ll * v * v;
  return ans;
  }
  };
  
  • func minSumSquareDiff(nums1 []int, nums2 []int, k1 int, k2 int) int64 {
  k := k1 + k2
  s, mx := 0, 0
  n := len(nums1)
  d := make([]int, n)
  for i, v := range nums1 {
  d[i] = abs(v - nums2[i])
  s += d[i]
  mx = max(mx, d[i])
  }
  if s <= k {
  return 0
  }
  left, right := 0, mx
  for left < right {
  mid := (left + right) >> 1
  t := 0
  for _, v := range d {
  t += max(v-mid, 0)
  }
  if t <= k {
  right = mid
  } else {
  left = mid + 1
  }
  }
  for i, v := range d {
  k -= max(v-left, 0)
  d[i] = min(v, left)
  }
  for i, v := range d {
  if k <= 0 {
  break
  }
  if v == left {
  d[i]--
  k--
  }
  }
  ans := 0
  for _, v := range d {
  ans += v * v
  }
  return int64(ans)
  }
  
  func abs(x int) int {
  if x < 0 {
  return -x
  }
  return x
  }
  
  func max(a, b int) int {
  if a > b {
  return a
  }
  return b
  }
  
  func min(a, b int) int {
  if a < b {
  return a
  }
  return b
  }
  
  
  Explain:
  
  nope.
  
  Complexity:
  
  • Time complexity : O(n).
  • Space complexity : O(n).