LeetCode #2918 — MEDIUM

Minimum Equal Sum of Two Arrays After Replacing Zeros

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[3,2,0,1,0]
[6,5,0]

Example 2

[2,0,2,0]
[1,4]

Step 02

Core Insight

What unlocks the optimal approach

Consider we replace all the 0’s with 1’s on both arrays, the answer will be <code>-1</code> if there was no <code>0</code> in the array with the smaller sum of elements.
Otherwise, how can you update the value of exactly one of these <code>1</code>’s to make the sum of the two arrays equal?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2918: Minimum Equal Sum of Two Arrays After Replacing Zeros
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;
    }
}

// Accepted solution for LeetCode #2918: Minimum Equal Sum of Two Arrays After Replacing Zeros
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
}

# Accepted solution for LeetCode #2918: Minimum Equal Sum of Two Arrays After Replacing Zeros
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

// Accepted solution for LeetCode #2918: Minimum Equal Sum of Two Arrays After Replacing Zeros
/**
 * [2918] Minimum Equal Sum of Two Arrays After Replacing Zeros
 */
pub struct Solution {}

// submission codes start here
use std::cmp::Ordering;

impl Solution {
    pub fn min_sum(nums1: Vec<i32>, nums2: Vec<i32>) -> i64 {
        let mut nums1_sum: i64 = nums1.iter().map(|x| *x as i64).sum();
        let mut nums2_sum: i64 = nums2.iter().map(|x| *x as i64).sum();
        let nums1_zero_count = nums1.iter().filter(|x| **x == 0).count() as i64;
        let nums2_zero_count = nums2.iter().filter(|x| **x == 0).count() as i64;

        match (nums1_sum + nums1_zero_count).cmp(&(nums2_sum + nums2_zero_count)) {
            Ordering::Less => {
                if nums1_zero_count == 0 {
                    -1
                } else {
                    nums2_sum + nums2_zero_count
                }
            }
            Ordering::Greater => {
                if nums2_zero_count == 0 {
                    -1
                } else {
                    nums1_sum + nums1_zero_count
                }
            }
            Ordering::Equal => nums1_sum + nums1_zero_count,
        }
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2918() {
        assert_eq!(12, Solution::min_sum(vec![3, 2, 0, 1, 0], vec![6, 5, 0]));
        assert_eq!(-1, Solution::min_sum(vec![2, 0, 2, 0], vec![1, 4]));
    }
}

// Accepted solution for LeetCode #2918: Minimum Equal Sum of Two Arrays After Replacing Zeros
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.