What is the time complexity of Ways to Make a Fair Array?

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

What data structures are used to solve Ways to Make a Fair Array?

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

What are the key insights for solving Ways to Make a Fair Array?

Insight 1: We can't actually modify the array for each removal since that would lead to O(n^2) time complexity. Instead, we can calculate the prefix sums of even and odd indices to efficiently compute the even and odd sums after removing an element. Insight 2: We can calculate total even and odd sums of the original array beforehand. Then, while iterating, we can maintain left even and left odd sums and use these to derive the right even and right odd sums after a removal.

Which companies ask LeetCode 1664. Ways to Make a Fair Array in interviews?

This problem is frequently asked at DoorDash, Dunzo, Nvidia, PhonePe, Twilio.

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Ways to Make a Fair Array - LeetCode 1664 Solution

Q: How to solve Ways to Make a Fair Array?

The solution calculates the total even and odd sums of the input array. It then iterates through the array, simulating the removal of each element. For each index, it calculates the sums of the even and odd indices to the left and to the right of the removed element. By combining the left and right sums, it determines if the resulting array would be fair. It maintains counters for the left even and odd sums, updating them with each iteration. The final count of indices that, when removed, make the array fair is returned.

Ways to Make a Fair Array - Complete Solution Guide

Ways to Make a Fair Array is LeetCode problem 1664, 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 an integer array nums . You can choose exactly one index ( 0-indexed ) and remove the element. Notice that the index of the elements may change after the removal. For example, if nums = [6,1,7,4,1] : Choosing to remove index 1 results in nums = [6,7,4,1] . Choosing to remove index 2 results in nums = [6,1,4,1] . Choosing to remove index 4 results in nums = [6,1,7,4] . An array is fair if the sum of the odd-indexed values equals the sum of the even-indexed values. Return the number

Detailed Explanation

The problem asks us to find the number of indices in a given integer array `nums` that, when removed, will make the array 'fair'. An array is 'fair' if the sum of its elements at even indices equals the sum of its elements at odd indices. We need to iterate through the array, hypothetically removing each element, and check if the resulting array is fair. The input is an integer array `nums`, and the output is the number of indices that can be removed to make the array fair. The constraints are 1 <= nums.length <= 10^5 and 1 <= nums[i] <= 10^4.

Solution Approach

The solution calculates the total even and odd sums of the input array. It then iterates through the array, simulating the removal of each element. For each index, it calculates the sums of the even and odd indices to the left and to the right of the removed element. By combining the left and right sums, it determines if the resulting array would be fair. It maintains counters for the left even and odd sums, updating them with each iteration. The final count of indices that, when removed, make the array fair is returned.

Step-by-Step Algorithm

Step 1: Calculate the total sum of elements at even indices (`total_even`) and the total sum of elements at odd indices (`total_odd`) in the original array.
Step 2: Initialize `left_even` and `left_odd` to 0. These will store the sums of even and odd indices to the left of the current element being considered for removal.
Step 3: Initialize a `count` variable to 0. This will store the number of indices that, when removed, make the array fair.
Step 4: Iterate through the array using a `for` loop.
Step 5: Inside the loop, for each index `i`, calculate the `right_even` and `right_odd` sums. If `i` is even, `right_even` is `total_even - left_even - nums[i]` and `right_odd` is `total_odd - left_odd`. If `i` is odd, `right_even` is `total_even - left_even` and `right_odd` is `total_odd - left_odd - nums[i]`.
Step 6: Check if `left_even + right_odd == left_odd + right_even`. If they are equal, increment the `count`.
Step 7: Update `left_even` and `left_odd` before moving to the next index. If `i` is even, add `nums[i]` to `left_even`. If `i` is odd, add `nums[i]` to `left_odd`.
Step 8: After the loop finishes, return the `count`.

Key Insights

Insight 1: We can't actually modify the array for each removal since that would lead to O(n^2) time complexity. Instead, we can calculate the prefix sums of even and odd indices to efficiently compute the even and odd sums after removing an element.
Insight 2: We can calculate total even and odd sums of the original array beforehand. Then, while iterating, we can maintain left even and left odd sums and use these to derive the right even and right odd sums after a removal.
Insight 3: The index parity shifts after removing an element. Elements to the right of the removed element will have their index parity flipped (even becomes odd, odd becomes even).

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Prefix Sum.

Companies

Asked at: DoorDash, Dunzo, Nvidia, PhonePe, Twilio.

Ways to Make a Fair Array | LeetCode 1664

Medium

Array Prefix Sum

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

Solution Code

Available in:

class Solution {
    public int waysToMakeFair(int[] nums) {
        int n = nums.length;
        
        int total_even = 0;
        int total_odd = 0;
        for (int i = 0; i < n; i++) {
            if (i % 2 == 0) {
                total_even += nums[i];
            } else {
                total_odd += nums[i];
            }
        }
        
        int left_even = 0;
        int left_odd = 0;
        int count = 0;
        
        for (int i = 0; i < n; i++) {
            int right_even;
            int right_odd;
            
            if (i % 2 == 0) {
                right_even = total_even - left_even - nums[i];
                right_odd = total_odd - left_odd;
            } else {
                right_even = total_even - left_even;
                right_odd = total_odd - left_odd - nums[i];
            }
            
            if (left_even + right_odd == left_odd + right_even) {
                count++;
            }
            
            if (i % 2 == 0) {
                left_even += nums[i];
            } else {
                left_odd += nums[i];
            }
        }
        
        return count;
    }
}

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Problem Info

Description

Problem 1664: Ways to Make a Fair Array - Detailed explanation and solution approaches.

Topics

Array Prefix Sum

Companies

DoorDashDunzoNvidiaPhonePeTwilio

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:1664

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

8 steps

1
Step 1: Calculate the total sum of elements at even indices (`total_even`) and the total sum of elements at odd indices (`total_odd`) in the original array.
2
Step 2: Initialize `left_even` and `left_odd` to 0. These will store the sums of even and odd indices to the left of the current element being considered for removal.
3
Step 3: Initialize a `count` variable to 0. This will store the number of indices that, when removed, make the array fair.
4
Step 4: Iterate through the array using a `for` loop.
5
Step 5: Inside the loop, for each index `i`, calculate the `right_even` and `right_odd` sums. If `i` is even, `right_even` is `total_even - left_even - nums[i]` and `right_odd` is `total_odd - left_odd`. If `i` is odd, `right_even` is `total_even - left_even` and `right_odd` is `total_odd - left_odd - nums[i]`.
6
Step 6: Check if `left_even + right_odd == left_odd + right_even`. If they are equal, increment the `count`.
7
Step 7: Update `left_even` and `left_odd` before moving to the next index. If `i` is even, add `nums[i]` to `left_even`. If `i` is odd, add `nums[i]` to `left_odd`.
8
Step 8: After the loop finishes, return the `count`.

💡

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 number of iterations through the array. The space complexity is determined by the number of variables used, which remains constant regardless of 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

Prefix Sum Array

Instead of calculating `right_even` and `right_odd` during the loop, we could precompute prefix sums for even and odd indices into separate arrays. Then, we could calculate right side sums using prefix sum differences. This approach has the same time complexity and space complexity.

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

Trade-offs: Slightly less efficient than the current O(1) space solution but may improve readability for some.

✨

Key Insights

1Insight 1: We can't actually modify the array for each removal since that would lead to O(n^2) time complexity. Instead, we can calculate the prefix sums of even and odd indices to efficiently compute the even and odd sums after removing an element.
2Insight 2: We can calculate total even and odd sums of the original array beforehand. Then, while iterating, we can maintain left even and left odd sums and use these to derive the right even and right odd sums after a removal.
3Insight 3: The index parity shifts after removing an element. Elements to the right of the removed element will have their index parity flipped (even becomes odd, odd becomes even).

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: Empty array (nums.length == 0). The problem statement doesn't explicitly state the behavior if the input array is empty. The given code handles an empty array gracefully as it won't enter the for loop and returns 0. It satisfies the requirements because an empty array would have even and odd indexed sums of 0, and thus be fair.
Edge case 2: Single element array (nums.length == 1). When the array has only one element, removing it results in an empty array, which is considered fair, so the algorithm should return 1. The provided solution handles this case correctly.
Edge case 3: Array with all same elements. For example, [1,1,1,1]. The solution works correctly because it checks the fairness condition for each possible removal.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Modifying the original array. Creating a new array after each removal would result in an O(n^2) solution, which is inefficient and likely to exceed the time limit.

✗Common mistake 2: Incorrectly calculating even/odd sums after removal. It is crucial to consider the parity change to the right of the removed element. Failing to do so will lead to incorrect results.

✗Common mistake 3: Not accounting for the fact the array is fair when it contains no elements if we remove the last element.

✅

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: Load Balancing. In a system with distributed servers, we might want to remove a server to perform maintenance. This algorithm helps in identifying which server removal would result in the most balanced workload distribution among the remaining servers.

🏦Real-world scenario 2: Data Analysis. When analyzing a dataset with time series data, we might want to remove certain data points to analyze the impact on trends. This algorithm helps in identifying data points whose removal would lead to a more 'fair' or balanced representation of the underlying patterns.

💡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 15, 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

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

@TechLeadJul 13

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

@DevExpert92Jul 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.

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 "Ways to Make a Fair Array"?

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

🤖 Need help with this problem?