What is the time complexity of Minimum Equal Sum of Two Arrays After Replacing Zeros?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Minimum Equal Sum of Two Arrays After Replacing Zeros?

This problem uses Array, Greedy. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Equal Sum of Two Arrays After Replacing Zeros?

Insight 1: Replacing a zero with 1 is the minimum possible value, so we should aim to replace each zero with the smallest possible positive integer, which is 1. Insight 2: We need to calculate the sum of each array, along with the number of zeros in each array, to determine the minimum possible sum and to compare the arrays.

Which companies ask LeetCode 2918. Minimum Equal Sum of Two Arrays After Replacing Zeros in interviews?

This problem is frequently asked at Citadel, Palantir Technologies, Twilio.

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Minimum Equal Sum of Two Arrays After Replacing Zeros - LeetCode 2918 Solution

Minimum Equal Sum of Two Arrays After Replacing Zeros - Complete Solution Guide

Minimum Equal Sum of Two Arrays After Replacing Zeros is LeetCode problem 2918, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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,

Detailed Explanation

The problem requires finding the minimum equal sum that can be obtained from two arrays, `nums1` and `nums2`, by replacing all the zeros in both arrays with strictly positive integers. The goal is to make the sum of the elements in both arrays equal after the replacements. If it is impossible to make the sums equal, the function should return -1.

Solution Approach

The solution involves iterating through both arrays to calculate the sum of their elements and count the number of zeros they contain. Then, we calculate the minimum possible sum for each array by adding the number of zeros to the initial sum (because we will replace all the zeros with 1). Finally, we check if making the array sums equal is possible. It's impossible if one array has no zeros and its minimum possible sum is strictly less than the other's. If it is possible, return the maximum of the two minimum possible sums.

Step-by-Step Algorithm

Step 1: Initialize `sum1` and `zeros1` to 0. Iterate through `nums1`. For each element `x`, if `x` is 0, increment `zeros1`. Add `x` to `sum1`.
Step 2: Initialize `sum2` and `zeros2` to 0. Iterate through `nums2`. For each element `x`, if `x` is 0, increment `zeros2`. Add `x` to `sum2`.
Step 3: Calculate `min_sum1` as `sum1 + zeros1` and `min_sum2` as `sum2 + zeros2`. These represent the minimum possible sums after replacing all zeros with 1.
Step 4: Check if it's impossible to equalize the sums. If `zeros1` is 0 and `min_sum1` is less than `min_sum2`, return -1. Similarly, if `zeros2` is 0 and `min_sum2` is less than `min_sum1`, return -1.
Step 5: If it's possible to equalize the sums, return the maximum of `min_sum1` and `min_sum2`. This will be the minimum equal sum that can be achieved.

Key Insights

Insight 1: Replacing a zero with 1 is the minimum possible value, so we should aim to replace each zero with the smallest possible positive integer, which is 1.
Insight 2: We need to calculate the sum of each array, along with the number of zeros in each array, to determine the minimum possible sum and to compare the arrays.
Insight 3: If one array has no zeros and its minimum possible sum is strictly less than the minimum possible sum of the other array, it's impossible to equalize the sums. Return -1 in this case.
Insight 4: The final equal sum will be the maximum of the two minimum possible sums.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy.

Companies

Asked at: Citadel, Palantir Technologies, Twilio.

Minimum Equal Sum of Two Arrays After Replacing Zeros | LeetCode 2918

Medium

Array Greedy

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public long minSum(int[] nums1, int[] nums2) {
        long sum1 = 0;
        long zeros1 = 0;
        for (int x : nums1) {
            if (x == 0) {
                zeros1++;
            }
            sum1 += x;
        }

        long sum2 = 0;
        long zeros2 = 0;
        for (int x : nums2) {
            if (x == 0) {
                zeros2++;
            }
            sum2 += x;
        }
        
        long min_sum1 = sum1 + zeros1;
        long min_sum2 = sum2 + zeros2;
        
        if (zeros1 == 0 && min_sum1 < min_sum2) {
            return -1;
        }
        
        if (zeros2 == 0 && min_sum2 < min_sum1) {
            return -1;
        }
        
        return Math.max(min_sum1, min_sum2);
    }
}

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Problem Info

Description

Problem 2918: Minimum Equal Sum of Two Arrays After Replacing Zeros - Detailed explanation and solution approaches.

Topics

Array Greedy

Companies

CitadelPalantir TechnologiesTwilio

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:2918

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Initialize `sum1` and `zeros1` to 0. Iterate through `nums1`. For each element `x`, if `x` is 0, increment `zeros1`. Add `x` to `sum1`.
2
Step 2: Initialize `sum2` and `zeros2` to 0. Iterate through `nums2`. For each element `x`, if `x` is 0, increment `zeros2`. Add `x` to `sum2`.
3
Step 3: Calculate `min_sum1` as `sum1 + zeros1` and `min_sum2` as `sum2 + zeros2`. These represent the minimum possible sums after replacing all zeros with 1.
4
Step 4: Check if it's impossible to equalize the sums. If `zeros1` is 0 and `min_sum1` is less than `min_sum2`, return -1. Similarly, if `zeros2` is 0 and `min_sum2` is less than `min_sum1`, return -1.
5
Step 5: If it's possible to equalize the sums, return the maximum of `min_sum1` and `min_sum2`. This will be the minimum equal sum that can be achieved.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the two `for` loops, each iterating through one of the arrays. The space complexity is constant as no additional data structures are created with size relative to the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using streams in Java (for summing)

Instead of explicit for loops, Java streams could be used to calculate the sum and count the zeros. This may be more readable for some, but does not affect the complexity.

Time: O(n)Space: O(1)

Trade-offs: Slightly different syntax, but same underlying performance. Preference based on code style.

✨

Key Insights

1Insight 1: Replacing a zero with 1 is the minimum possible value, so we should aim to replace each zero with the smallest possible positive integer, which is 1.
2Insight 2: We need to calculate the sum of each array, along with the number of zeros in each array, to determine the minimum possible sum and to compare the arrays.
3Insight 3: If one array has no zeros and its minimum possible sum is strictly less than the minimum possible sum of the other array, it's impossible to equalize the sums. Return -1 in this case.
4Insight 4: The final equal sum will be the maximum of the two minimum possible sums.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: One or both arrays contain only zeros. If both arrays only have zeros and the length is different, it is not possible to make them equal. The solution correctly accounts for cases where zeros are present.
Edge case 2: One array has no zeros, and the other does. If the array with no zeros has a smaller sum than the minimum possible sum of the other array (after replacing its zeros with ones), then it's impossible to equalize the sums. The solution handles this situation by returning -1.
Edge case 3: Empty arrays. While the prompt specifies the lengths are at least 1, if an empty array were allowed the algorithm would still work correctly returning 0 if both were empty and -1 in all other cases.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the minimum value that can replace zero (which is 1). Replacing zeros with a larger value might lead to a larger sum than necessary.

✗Common mistake 2: Forgetting to check for the impossible case where one array has no zeros and a smaller sum than the minimum possible sum of the other array.

✗Common mistake 3: Using `int` instead of `long` for the sums in Java and C++. The sum of the arrays could exceed the range of `int`, causing overflow errors. Using `long` prevents this.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Balancing resource allocation. Imagine `nums1` and `nums2` represent two systems needing to be balanced. Zeros represent unallocated resources, which must be allocated to ensure a minimum equal distribution.

🏦Real-world scenario 2: Data normalization. When dealing with datasets with missing values (represented as zeros), this algorithm can be adapted to determine the minimum adjustments needed to equalize certain metrics across different datasets while replacing missing values with a minimum plausible value.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 9, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Equal Sum of Two Arrays After Replacing Zeros"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?