1775

1775. Equal Sum Arrays With Minimum Number of Operations (Medium)

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.

 

Example 1:

Input: nums1 = [1,2,3,4,5,6], nums2 = [1,1,2,2,2,2]
Output: 3
Explanation: You can make the sums of nums1 and nums2 equal with 3 operations. All indices are 0-indexed.
- Change nums2[0] to 6. nums1 = [1,2,3,4,5,6], nums2 = [6,1,2,2,2,2].
- Change nums1[5] to 1. nums1 = [1,2,3,4,5,1], nums2 = [6,1,2,2,2,2].
- Change nums1[2] to 2. nums1 = [1,2,2,4,5,1], nums2 = [6,1,2,2,2,2].

Example 2:

Input: nums1 = [1,1,1,1,1,1,1], nums2 = [6]
Output: -1
Explanation: There is no way to decrease the sum of nums1 or to increase the sum of nums2 to make them equal.

Example 3:

Input: nums1 = [6,6], nums2 = [1]
Output: 3
Explanation: You can make the sums of nums1 and nums2 equal with 3 operations. All indices are 0-indexed. 
- Change nums1[0] to 2. nums1 = [2,6], nums2 = [1].
- Change nums1[1] to 2. nums1 = [2,2], nums2 = [1].
- Change nums2[0] to 4. nums1 = [2,2], nums2 = [4].

 

Constraints:

• 1 <= nums1.length, nums2.length <= 105
• 1 <= nums1[i], nums2[i] <= 6

Related Topics:
Greedy

Similar Questions:

• Number of Dice Rolls With Target Sum (Medium)

Solution 1. Brute force

// OJ: https://leetcode.com/problems/equal-sum-arrays-with-minimum-number-of-operations/
// Author: github.com/lzl124631x
// Time: O(M + N)
// Space: O(1)
class Solution {
    int getOp(int cnt[7], int diff) {
        int ans = 0;
        if (diff > 0) {
            for (int i = 1; i < 6; ++i) {
                int d = 6 - i;
                int q = min((diff + d - 1) / d, cnt[i]);
                diff -= q * d;
                ans += q;
                if (diff <= 0) break;
            }
        } else {
            diff = -diff;
            for (int i = 6; i > 1; --i) {
                int d = i - 1;
                int q = min((diff + d - 1) / d, cnt[i]);
                diff -= q * d;
                ans += q;
                if (diff <= 0) break;
            }
        }
        return ans;
    }
public:
    int minOperations(vector<int>& A, vector<int>& B) {
        int M = A.size(), N = B.size();
        if (M > 6 * N || N > 6 * M) return -1;
        int ans = INT_MAX, cnta[7] = {}, cntb[7] = {}, sa = accumulate(begin(A), end(A), 0), sb = accumulate(begin(B), end(B), 0);
        for (int a : A) cnta[a]++;
        for (int b : B) cntb[b]++;
        for (int i = max(M, N), mx = 6 * min(M, N); i <= mx; ++i) {
            int op = getOp(cnta, i - sa) + getOp(cntb, i - sb);
            ans = min(ans, op);
        }
        return ans;
    }
};

Solution 2. Greedy

// OJ: https://leetcode.com/problems/equal-sum-arrays-with-minimum-number-of-operations/
// Author: github.com/lzl124631x
// Time: O(M + N)
// Space: O(1)
class Solution {
    int solve(int A[7], int B[7], int diff) {
        int ans = 0;
        for (int d = 5; d >= 1 && diff > 0; --d) {
            int q = min((diff + d - 1) / d, A[d + 1]);
            diff -= q * d;
            ans += q;
            q = min((diff + d - 1) / d, B[6 - d]);
            diff -= q * d;
            ans += q;
        }
        return ans;
    }
public:
    int minOperations(vector<int>& A, vector<int>& B) {
        int M = A.size(), N = B.size();
        if (M > 6 * N || N > 6 * M) return -1;
        int ans = INT_MAX, cnta[7] = {}, cntb[7] = {}, sa = accumulate(begin(A), end(A), 0), sb = accumulate(begin(B), end(B), 0);
        if (sa == sb) return 0;
        for (int a : A) cnta[a]++;
        for (int b : B) cntb[b]++;
        return sa > sb ? solve(cnta, cntb, sa - sb) : solve(cntb, cnta, sb - sa);
    }
};