Question

2333. Minimum Sum of Squared Difference

中文文档

Description

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

Solutions

Solution 1

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)

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;
  }
}

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
}