## Problem

Problem ID: 1775

Title: Equal Sum Arrays With Minimum Number of Operations

Difficulty: Medium

Description:

You are given two arrays of integers nums1 and nums2, possibly of different lengths. The values in the arrays are between 1 and 6, inclusive.

In one operation, you can change any integer’s value in any of the arrays to any value between 1 and 6, inclusive.

Return the minimum number of operations required to make the sum of values in nums1 equal to the sum of values in nums2. Return -1 if it is not possible to make the sum of the two arrays equal.

## Thoughts

This was a hard problem for me and I had to look at the solution.

## Solution

General Idea

Find the bigger and smaller numbers from nums1 and nums2. The num 6 on the bigger and 1 on the smaller. Both of them have same impact on making the sums equal as they both are able to reduce the difference by 5. This also applies to 5 and 2 and so on and so forth.

We can then greedily reduce the difference in sum by changing the pair with the highest impact (1 and 6). If converting all the nums of a pair is insufficient we will then move on the next pair (2 and 5) else we would use all the necessary current pair and if there are extra add extra to it.

class Solution {
public:
int minOperations(vector<int>& nums1, vector<int>& nums2) {
int sum1 = 0;
int sum2 = 0;
for (int num : nums1) sum1 += num;
for (int num : nums2) sum2 += num;

std::array<int, 6> contributions = {0};

vector<int>& bigger = sum1 > sum2 ? nums1 : nums2;
vector<int>& smaller = sum1 > sum2 ? nums2 : nums1;

for (auto num : bigger) contributions[num - 1]++;
for (auto num : smaller) contributions[6 - num]++;

int delta = abs(sum1 - sum2);
int ret = 0;
for (int contrib = 5; delta > 0 && contrib >= 1; contrib--) {
int take = min(contributions[contrib], delta/contrib + (delta % contrib != 0));
bool should_break = take == delta/contrib + (delta % contrib != 0);
delta -= take*contrib;
ret += take;
if (should_break) break;
}

return delta > 0 ? -1 : ret;
}
};