[Solved] Minimum Sum of Squared Difference LeetCode Contest

Minimum Sum of Squared Difference: 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

class Solution {
public:
    long long minSumSquareDiff(vector<int>& nums1, vector<int>& nums2, int k1, int k2) {
        vector<int> difference( (int) 1e5+1);
        
        for(int i = 0; i<nums1.size();i++)
        {
            difference[abs(nums1[i] - nums2[i])]++;
        }
        
        int value = 0, left = k1+k2;
        int acc = 0;
        
        for(int i = (int) 1e5; i>=0; i--)
        {
            acc = difference[i];
            if(acc <= left)
            {
                if(i<=1)
                    return 0;
                
                difference[i-1] += difference[i];
                difference[i]=0;
                left -= acc;
            }
            else
            {
                if(i==0)
                    return 0;
                difference[i] -= left;
                difference[i-1] += left;
                break;
            }
        }
        long long answer = 0;
        
        for(int i = 1ll; i<=1e5; i++)
        {
            answer += (long long)difference[i] * ((long long)i*i);
        }
        
        return answer;
    }
};

from heapq import heapify, heappush, heappop
class Solution:
    def minSumSquareDiff(self, nums1: List[int], nums2: List[int], k1: int, k2: int) -> int:
        heap = [ -abs(x-y) for x, y in zip(nums1, nums2)]
        s = -sum(heap)
        if k1+k2 >= s: return 0
        delta = k1 + k2
        heapify(heap)
        n = len(nums1)
        while delta > 0:
            d = -heappop(heap)
            gap = max(delta//n, 1) if heap else delta
            d -= gap
            heappush(heap, -d)
            delta -= gap
        return sum(pow(e,2) for e in heap)

class Solution {
    public long minSumSquareDiff(int[] nums1, int[] nums2, int k1, int k2) {
        PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> (b - a));
        pq.add(0);
        for (int i = 0; i < nums1.length; i++) {
        	int diff = Math.abs(nums1[i] - nums2[i]);
        	if (diff > 0)
        		pq.add(diff);
        }
        int k = k1 + k2;
        int prev = 0, cnt = 0;
        while (k > 0 && !pq.isEmpty()) {
        	int cur = pq.poll();
        	long x = (prev - cur) * cnt;
        	if (x <= k) {
        		prev = cur;
        		cnt++;
        		k -= x;
        	} else {
        		int a = k / cnt; 
        		int b = k % cnt; 
        		int c = prev - a; 
        		for (int i = 0; i < cnt - b; i++) 
        			pq.add(c);
        		for (int i = 0; i < b; i++) 
        			pq.add(c - 1);
        		pq.add(cur);
        		k = 0;
        	}
        }
        long ans = 0;
        for (long diff : pq)
        	ans += (long)diff * diff;
        return ans;
    }
}

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