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2918. Minimum Equal Sum of Two Arrays After Replacing Zeros
Description
You are given two arrays nums1
and nums2
consisting of positive integers.
You have to replace all the 0
's in both arrays with strictly positive integers such that the sum of elements of both arrays becomes equal.
Return the minimum equal sum you can obtain, or -1
if it is impossible.
Example 1:
Input: nums1 = [3,2,0,1,0], nums2 = [6,5,0] Output: 12 Explanation: We can replace 0's in the following way: - Replace the two 0's in nums1 with the values 2 and 4. The resulting array is nums1 = [3,2,2,1,4]. - Replace the 0 in nums2 with the value 1. The resulting array is nums2 = [6,5,1]. Both arrays have an equal sum of 12. It can be shown that it is the minimum sum we can obtain.
Example 2:
Input: nums1 = [2,0,2,0], nums2 = [1,4] Output: -1 Explanation: It is impossible to make the sum of both arrays equal.
Constraints:
1 <= nums1.length, nums2.length <= 105
0 <= nums1[i], nums2[i] <= 106
Solutions
Solution 1: Case Analysis
We consider the case where we treat all $0$s in the array as $1$s, and calculate the sum of the two arrays separately, denoted as $s_1$ and $s_2$. Without loss of generality, we assume that $s_1 \le s_2$.
- If $s_1 = s_2$, then the answer is $s_1$.
- If $s_1 \lt s_2$, then there must exist a $0$ in $nums1$ to make the sum of the two arrays equal. In this case, the answer is $s_2$. Otherwise, it means that it is impossible to make the sum of the two arrays equal, and we return $-1$.
The time complexity is $O(n + m)$, where $n$ and $m$ are the lengths of the arrays $nums1$ and $nums2$, respectively. The space complexity is $O(1)$.
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class Solution { public long minSum(int[] nums1, int[] nums2) { long s1 = 0, s2 = 0; boolean hasZero = false; for (int x : nums1) { hasZero |= x == 0; s1 += Math.max(x, 1); } for (int x : nums2) { s2 += Math.max(x, 1); } if (s1 > s2) { return minSum(nums2, nums1); } if (s1 == s2) { return s1; } return hasZero ? s2 : -1; } }
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class Solution { public: long long minSum(vector<int>& nums1, vector<int>& nums2) { long long s1 = 0, s2 = 0; bool hasZero = false; for (int x : nums1) { hasZero |= x == 0; s1 += max(x, 1); } for (int x : nums2) { s2 += max(x, 1); } if (s1 > s2) { return minSum(nums2, nums1); } if (s1 == s2) { return s1; } return hasZero ? s2 : -1; } };
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class Solution: def minSum(self, nums1: List[int], nums2: List[int]) -> int: s1 = sum(nums1) + nums1.count(0) s2 = sum(nums2) + nums2.count(0) if s1 > s2: return self.minSum(nums2, nums1) if s1 == s2: return s1 return -1 if nums1.count(0) == 0 else s2
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func minSum(nums1 []int, nums2 []int) int64 { s1, s2 := 0, 0 hasZero := false for _, x := range nums1 { if x == 0 { hasZero = true } s1 += max(x, 1) } for _, x := range nums2 { s2 += max(x, 1) } if s1 > s2 { return minSum(nums2, nums1) } if s1 == s2 { return int64(s1) } if hasZero { return int64(s2) } return -1 } func max(a, b int) int { if a > b { return a } return b }
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function minSum(nums1: number[], nums2: number[]): number { let [s1, s2] = [0, 0]; let hasZero = false; for (const x of nums1) { if (x === 0) { hasZero = true; } s1 += Math.max(x, 1); } for (const x of nums2) { s2 += Math.max(x, 1); } if (s1 > s2) { return minSum(nums2, nums1); } if (s1 === s2) { return s1; } return hasZero ? s2 : -1; }