2918: Minimum-Equal-Sum-of-Two-Arrays-After-Replacing-Zeros
Medium

table of contents

This question a fake medium or what?!

You can basically reduce the original question to:

If you change all the 0s present in both arrays to 1s and sum the elements of both arrays, which array has a larger sum?

Then, add on a follow-up question:

Given that you know which array has the LARGER sum after changing all the 0s present to 1s, does the array with the SMALLER sum have any 0s present in its INITIAL STATE?

Finally, after answering this question:

This is because, we are allowed to change all the 0s to ANY STRICTLY POSITIVE NUMBER, meaning the minimum sum we can possibly find has to occur when all 0s are swapped to 1s.

Therefore that implies that if the smaller sum has no 0s present, it is simply impossible to set both sums to be equal.

code

class Solution {
public:
    long long minSum(vector<int>& nums1, vector<int>& nums2) {
        long long sum1 = 0, sum2 = 0;
        int count1 = 0, count2 = 0;
        for (int& num : nums1) {
            sum1 += num;
            if (num == 0) ++count1;
        }
        for (int& num : nums2) {
            sum2 += num;
            if (num == 0) ++count2;
        }

        if (sum1 + count1 > sum2 + count2 && count2 == 0) {
            return -1;
        } else if (sum1 + count1 < sum2 + count2 && count1 == 0) {
            return -1;
        }
        return max(sum1 + count1, sum2 + count2);
    }
};

complexity

time taken