What is the time complexity of Equal Sum Arrays With Minimum Number of Operations?

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

What data structures are used to solve Equal Sum Arrays With Minimum Number of Operations?

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

What are the key insights for solving Equal Sum Arrays With Minimum Number of Operations?

Insight 1: The problem can be simplified by focusing on the difference between the sums of the two arrays. The goal is to reduce this difference to zero with the fewest operations. Insight 2: It's more efficient to greedily apply the largest possible changes in each operation. For `nums1`, the largest reduction is achieved by changing an element to 1 (reduction = element - 1). For `nums2`, the largest increase is achieved by changing an element to 6 (increase = 6 - element).

Which companies ask LeetCode 1775. Equal Sum Arrays With Minimum Number of Operations in interviews?

This problem is frequently asked at American Express.

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Equal Sum Arrays With Minimum Number of Operations - LeetCode 1775 Solution

Equal Sum Arrays With Minimum Number of Operations - Complete Solution Guide

Equal Sum Arrays With Minimum Number of Operations is LeetCode problem 1775, 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 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]

Detailed Explanation

The problem requires finding the minimum number of operations to make the sum of two integer arrays, `nums1` and `nums2`, equal. Each element in the arrays is between 1 and 6. In one operation, you can change any element in either array to any value between 1 and 6. If it's impossible to make the sums equal, return -1. The arrays can have different lengths.

Solution Approach

The solution uses a greedy approach. First, it checks if it is possible to equalize the sums based on array lengths. If possible, it calculates the sums of both arrays. If the sums are already equal, it returns 0. If not, it determines the difference between the sums and stores the counts of possible gains (reduction or increase) in a frequency array `counts`. This array helps in deciding the change amount in each operation. The algorithm then iterates from the largest possible gain (5) to the smallest (1), applying as many operations as possible for each gain until the difference becomes zero.

Step-by-Step Algorithm

Step 1: Check if the sums can be equalized based on array lengths. If len1 * 6 < len2 or len1 > len2 * 6, return -1.
Step 2: Calculate the sum of `nums1` and `nums2`.
Step 3: If the sums are equal, return 0.
Step 4: If sum1 < sum2, swap the arrays and their sums to ensure sum1 is always greater than or equal to sum2. This simplifies the logic as we only need to consider reducing sum1 or increasing sum2.
Step 5: Calculate the difference `diff = sum1 - sum2`.
Step 6: Create a frequency array `counts` of size 6. `counts[i]` represents the number of times a gain of `i+1` can be achieved by either decreasing a number in `nums1` or increasing a number in `nums2`.
Step 7: Populate the `counts` array. For each number in `nums1`, increment `counts[number - 1]`. For each number in `nums2`, increment `counts[6 - number]`.
Step 8: Iterate from `gain = 5` down to 1. For each `gain`, check if `counts[gain]` is greater than 0. If not, continue to the next `gain`.
Step 9: Calculate the total reduction possible with the current `gain`: `total_reduction = counts[gain] * gain`.
Step 10: If `diff <= total_reduction`, then the difference can be reduced to 0 using the current `gain`. Calculate the number of operations needed: `operations += (diff + gain - 1) / gain` and return `operations`.
Step 11: If `diff > total_reduction`, then use all available operations with the current `gain`: `operations += counts[gain]` and reduce the difference: `diff -= total_reduction`.
Step 12: If the loop completes without returning, then it's not possible to equalize the sums (which should be unreachable if initial conditions were checked correctly), return -1.

Key Insights

Insight 1: The problem can be simplified by focusing on the difference between the sums of the two arrays. The goal is to reduce this difference to zero with the fewest operations.
Insight 2: It's more efficient to greedily apply the largest possible changes in each operation. For `nums1`, the largest reduction is achieved by changing an element to 1 (reduction = element - 1). For `nums2`, the largest increase is achieved by changing an element to 6 (increase = 6 - element).
Insight 3: The lengths of the arrays are crucial. If the minimum possible sum of one array (length * 1) is greater than the maximum possible sum of the other array (length * 6), then it's impossible to make the sums equal, and -1 should be returned.

Complexity Analysis

Time Complexity: O(n+m)

Space Complexity: O(1)

Topics

This problem involves: Array, Hash Table, Greedy, Counting.

Companies

Asked at: American Express.

Equal Sum Arrays With Minimum Number of Operations | LeetCode 1775

Medium

Array Hash Table Greedy Counting

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

Solution Code

Available in:

class Solution {
    public int minOperations(int[] nums1, int[] nums2) {
        int len1 = nums1.length;
        int len2 = nums2.length;

        // The sum of nums1 can be in [len1, 6 * len1].
        // The sum of nums2 can be in [len2, 6 * len2].
        // If these ranges don't overlap, it's impossible to make sums equal.
        if (len1 * 6 < len2 || len1 > len2 * 6) {
            return -1;
        }

        int sum1 = 0;
        for (int num : nums1) {
            sum1 += num;
        }

        int sum2 = 0;
        for (int num : nums2) {
            sum2 += num;
        }

        if (sum1 == sum2) {
            return 0;
        }

        // To simplify, we always assume sum1 is greater.
        // So we need to decrease sum1 or increase sum2.
        if (sum1 < sum2) {
            int tempSum = sum1;
            sum1 = sum2;
            sum2 = tempSum;

            int[] tempNums = nums1;
            nums1 = nums2;
            nums2 = tempNums;
        }

        int diff = sum1 - sum2;

        // Use a frequency map for all possible gains.
        // Gain from nums1 (decrease x): x - 1
        // Gain from nums2 (increase y): 6 - y
        // The gains can range from 0 to 5.
        int[] counts = new int[6];

        for (int num : nums1) {
            counts[num - 1]++;
        }

        for (int num : nums2) {
            counts[6 - num]++;
        }

        int operations = 0;
        // Greedily apply the largest gains first (from 5 down to 1).
        // A gain of 0 is useless.
        for (int gain = 5; gain > 0; gain--) {
            int num_available = counts[gain];
            if (num_available == 0) {
                continue;
            
            }

            // Total reduction possible with this gain value.
            int total_reduction = num_available * gain;

            if (diff <= total_reduction) {
                // We need ceil(diff / gain) more operations.
                // In integer arithmetic, this is (diff + gain - 1) // gain.
                operations += (diff + gain - 1) / gain;
                return operations;
            } else {
                // Use all available operations for this gain and move to the next smaller gain.
                operations += num_available;
                diff -= total_reduction;
            }
        }

        return -1; // Should be unreachable if the initial checks are correct
    }
}

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Problem Info

Description

Problem 1775: Equal Sum Arrays With Minimum Number of Operations - Detailed explanation and solution approaches.

Topics

Array Hash Table Greedy Counting

Companies

American Express

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1775

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

12 steps

1
Step 1: Check if the sums can be equalized based on array lengths. If len1 * 6 < len2 or len1 > len2 * 6, return -1.
2
Step 2: Calculate the sum of `nums1` and `nums2`.
3
Step 3: If the sums are equal, return 0.
4
Step 4: If sum1 < sum2, swap the arrays and their sums to ensure sum1 is always greater than or equal to sum2. This simplifies the logic as we only need to consider reducing sum1 or increasing sum2.
5
Step 5: Calculate the difference `diff = sum1 - sum2`.
6
Step 6: Create a frequency array `counts` of size 6. `counts[i]` represents the number of times a gain of `i+1` can be achieved by either decreasing a number in `nums1` or increasing a number in `nums2`.
7
Step 7: Populate the `counts` array. For each number in `nums1`, increment `counts[number - 1]`. For each number in `nums2`, increment `counts[6 - number]`.
8
Step 8: Iterate from `gain = 5` down to 1. For each `gain`, check if `counts[gain]` is greater than 0. If not, continue to the next `gain`.
9
Step 9: Calculate the total reduction possible with the current `gain`: `total_reduction = counts[gain] * gain`.
10
Step 10: If `diff <= total_reduction`, then the difference can be reduced to 0 using the current `gain`. Calculate the number of operations needed: `operations += (diff + gain - 1) / gain` and return `operations`.
11
Step 11: If `diff > total_reduction`, then use all available operations with the current `gain`: `operations += counts[gain]` and reduce the difference: `diff -= total_reduction`.
12
Step 12: If the loop completes without returning, then it's not possible to equalize the sums (which should be unreachable if initial conditions were checked correctly), return -1.

💡

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+m)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the linear scans of the input arrays. The space complexity is constant because the `counts` array has a fixed size, independent of the input array sizes.

📖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

Dynamic Programming

Dynamic programming can be used to solve this problem by building a table of minimum operations required to reach a specific difference between the sums. However, this is more complex and less efficient than the greedy approach.

Time: O(n*m*k) where n and m are the lengths of the arrays and k is the range of values(1..6), which would translate to O(n*m) since k is constant. This is due to calculating all possible combinations.Space: O(n*m). Space would be needed to construct the DP table.

Trade-offs: The greedy approach is much more efficient in both time and space complexity. Dynamic programming is unnecessarily complex for this problem.

✨

Key Insights

1Insight 1: The problem can be simplified by focusing on the difference between the sums of the two arrays. The goal is to reduce this difference to zero with the fewest operations.
2Insight 2: It's more efficient to greedily apply the largest possible changes in each operation. For `nums1`, the largest reduction is achieved by changing an element to 1 (reduction = element - 1). For `nums2`, the largest increase is achieved by changing an element to 6 (increase = 6 - element).
3Insight 3: The lengths of the arrays are crucial. If the minimum possible sum of one array (length * 1) is greater than the maximum possible sum of the other array (length * 6), then it's impossible to make the sums equal, and -1 should be returned.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: When the lengths of the arrays make it impossible to equalize the sums. The code correctly returns -1 in this case due to the initial check `if (len1 * 6 < len2 || len1 > len2 * 6)`.
Edge case 2: When the sums are already equal. The code correctly returns 0 in this case after calculating the sums.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the lengths of the arrays and incorrectly concluding that the sums can always be equalized.

✗Common mistake 2: Not greedily applying the largest possible changes in each operation, leading to a suboptimal solution or a more complex algorithm.

✅

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: Resource allocation - minimizing the number of transfers needed to balance resources between two groups.

🏦Real-world scenario 2: Load balancing - finding the fewest adjustments to distribute load evenly across servers.

💡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.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: November 3, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Equal Sum Arrays With Minimum Number of Operations"?

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 hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?