LeetCode Entry

2918. Minimum Equal Sum of Two Arrays After Replacing Zeros

10.05.2025 medium 2025 kotlin rust

Min equal array sum, fill zeros

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Problem TLDR

Min equal array sum, fill zeros #medium

Intuition

Any zero place can act as any number. Compare minimum sums by filling zeros with ones, then equalize.

Approach

  • the interesting golf is how to make it CPU-branchless: if (x == 0) 1 else 0 can be written as ((x | -x) >> 31) + 1 where x | -x would will everything but a sign bit, transforming into -1 for any non-zero value

Complexity

  • Time complexity: \(O(n)\)

  • Space complexity: \(O(1)\)

Code

Kotlin Rust C++
image.png image.png image.png 

// 523ms
    fun minSum(n1: IntArray, n2: IntArray): Long {
        val s1 = n1.sumOf { 1L * max(1, it) }; val s2 = n2.sumOf { 1L * max(1, it) }
        return if (s1 < s2 && 0 !in n1 || s1 > s2 && 0 !in n2) -1 else max(s1, s2)
    }



// 411ms https://leetcode.com/problems/minimum-equal-sum-of-two-arrays-after-replacing-zeros/submissions/1630261946
    fun minSum(nums1: IntArray, nums2: IntArray): Long {
        var s1 = 0L; var s2 = 0L; var z1 = 0; var z2 = 0
        for (x in nums1) { s1 += x + ((x or -x).ushr(31) xor 1); z1 = z1 or ((x or -x).ushr(31) xor 1) }
        for (x in nums2) { s2 += x + ((x or -x).ushr(31) xor 1); z2 = z2 or ((x or -x).ushr(31) xor 1) }
        return if (s1 < s2 && z1 < 1 || s1 > s2 && z2 < 1) -1 else max(s1, s2)
    }



// 8ms https://leetcode.com/problems/minimum-equal-sum-of-two-arrays-after-replacing-zeros/submissions/1630002606
    pub fn min_sum(n1: Vec<i32>, n2: Vec<i32>) -> i64 {
        let s1: i64 = n1.iter().map(|&n| n.max(1) as i64).sum();
        let s2: i64 = n2.iter().map(|&n| n.max(1) as i64).sum();
        let z1 = n1.contains(&0); let z2 = n2.contains(&0);
        if (s1 < s2 && !z1) || (s1 > s2 && !z2) { -1 } else { s1.max(s2) }
    }



// 43ms https://leetcode.com/problems/minimum-equal-sum-of-two-arrays-after-replacing-zeros/submissions/1630272912
    long long minSum(vector<int>& n1, vector<int>& n2) {
        long long s1 = size(n1), s2 = size(n2); int z1 = s1, z2 = s2;
        for (int& n: n1) s1 += n + ((n|-n)>>31), z1 += (n|-n)>>31;
        for (int& n: n2) s2 += n + ((n|-n)>>31), z2 += (n|-n)>>31;
        return s1 < s2 && !z1 || s1 > s2 && !z2 ? -1 : max(s1, s2);
    }