Leetcode 2333 Minimum Sum of Squared Difference Solution in c++ | Hindi Coding Community

 

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

C++ Code :

class Solution {

public:

long long minSumSquareDiff(vector<int>& nums1, vector<int>& nums2, int k1, int k2) {

unordered_map<long, int> table;

for(int i = 0; i != nums1.size(); i++)

table[abs((long)nums1[i] - nums2[i])]++;

int k = k1 + k2, id = 1;

long ans = 0, dif;

vector<pair<long, int>> t(table.size()+1);

for(auto [d, v] : table)t[id++] = {d, v};

sort(t.begin(), t.end());

for(id = table.size(); id != 0 && k != 0; id--)

if( (dif = (t[id].first - t[id-1].first) * t[id].second) <= k){

t[id-1].second += t[id].second;

k -= dif;

}

else{

long n = t[id].first - k/t[id].second;

ans += n * n * (t[id].second - k%t[id].second) + (n-1) * (n-1) * (k%t[id].second);

k = 0;

}

for(; id >= 0; id--) ans += t[id].first * t[id].first * t[id].second;

return ans;

}

};