What is the time complexity of Sum of Even Numbers After Queries?

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

What data structures are used to solve Sum of Even Numbers After Queries?

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

What are the key insights for solving Sum of Even Numbers After Queries?

Insight 1: The key is to efficiently maintain the sum of even numbers. We don't need to recompute the entire sum after each query. We can update the sum incrementally based on whether the updated element was even before and after the query. Insight 2: Modifying the original array `nums` in place allows us to directly see the effect of the query on the array and thus on the even sum.

Which companies ask LeetCode 985. Sum of Even Numbers After Queries in interviews?

This problem is frequently asked at Indeed.

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Sum of Even Numbers After Queries - LeetCode 985 Solution

Sum of Even Numbers After Queries - Complete Solution Guide

Sum of Even Numbers After Queries is LeetCode problem 985, 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 and an array queries where queries[i] = [val i , index i ] . For each query i , first, apply nums[index i ] = nums[index i ] + val i , then print the sum of the even values of nums . Return an integer array answer where answer[i] is the answer to the i th query . Example 1: Input: nums = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]] Output: [8,6,2,4] Explanation: At the beginning, the array is [1,2,3,4]. After adding 1 to nums[0], the array is [2,2,3,4], an

Detailed Explanation

The problem asks us to process a series of queries on an integer array `nums`. Each query consists of a value `val` and an index `index`. For each query, we must update the element at `nums[index]` by adding `val` to it, i.e., `nums[index] = nums[index] + val`. After each update, we need to compute the sum of all even numbers in the updated `nums` array and store it in the `answer` array. Finally, we return the `answer` array containing the sum of even numbers after each query.

Solution Approach

The provided solution calculates the initial sum of even numbers in the input array `nums`. Then, for each query, it checks if the element at the specified index was even before the update. If it was, it subtracts it from the current even sum. After updating the element, it checks if the updated element is even. If it is, it adds it to the current even sum. Finally, it appends the updated even sum to the `answer` array. This incremental update approach avoids recalculating the entire even sum after each query, leading to a more efficient solution.

Step-by-Step Algorithm

Step 1: Initialize `even_sum` to 0. Iterate through `nums` and add each even number to `even_sum`.
Step 2: Initialize an empty array `answer` to store the results.
Step 3: Iterate through the `queries` array. For each query `[val, index]`:
Step 4: Check if `nums[index]` is even. If it is, subtract `nums[index]` from `even_sum`.
Step 5: Update `nums[index]` by adding `val` to it: `nums[index] += val`.
Step 6: Check if the updated `nums[index]` is even. If it is, add `nums[index]` to `even_sum`.
Step 7: Append `even_sum` to the `answer` array.
Step 8: Return the `answer` array.

Key Insights

Insight 1: The key is to efficiently maintain the sum of even numbers. We don't need to recompute the entire sum after each query. We can update the sum incrementally based on whether the updated element was even before and after the query.
Insight 2: Modifying the original array `nums` in place allows us to directly see the effect of the query on the array and thus on the even sum.
Insight 3: We only need to track whether the element at the given index was even before and after the query. This helps avoid iterating through the entire array after each modification.

Complexity Analysis

Time Complexity: O(n + q)

Space Complexity: O(q)

Topics

This problem involves: Array, Simulation.

Companies

Asked at: Indeed.

Sum of Even Numbers After Queries | LeetCode 985

Medium

Array Simulation

Time: O(n + q)Space: O(q)

Solution Code

Available in:

class Solution {
    public int[] sumEvenAfterQueries(int[] nums, int[][] queries) {
        int even_sum = 0;
        for (int x : nums) {
            if (x % 2 == 0) {
                even_sum += x;
            }
        }
        
        int[] answer = new int[queries.length];
        
        for (int i = 0; i < queries.length; i++) {
            int val = queries[i][0];
            int index = queries[i][1];
            
            if (nums[index] % 2 == 0) {
                even_sum -= nums[index];
            }
            
            nums[index] += val;
            
            if (nums[index] % 2 == 0) {
                even_sum += nums[index];
            }
            
            answer[i] = even_sum;
        }
        
        return answer;
    }
}

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Problem Info

Description

Problem 985: Sum of Even Numbers After Queries - Detailed explanation and solution approaches.

Topics

Array Simulation

Companies

Indeed

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:985

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: Initialize `even_sum` to 0. Iterate through `nums` and add each even number to `even_sum`.
2
Step 2: Initialize an empty array `answer` to store the results.
3
Step 3: Iterate through the `queries` array. For each query `[val, index]`:
4
Step 4: Check if `nums[index]` is even. If it is, subtract `nums[index]` from `even_sum`.
5
Step 5: Update `nums[index]` by adding `val` to it: `nums[index] += val`.
6
Step 6: Check if the updated `nums[index]` is even. If it is, add `nums[index]` to `even_sum`.
7
Step 7: Append `even_sum` to the `answer` array.
8
Step 8: Return the `answer` array.

💡

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 + q)

💾

Space Complexity

O(q)

📝

Why?

The time complexity is determined by the initial O(n) loop to calculate the initial `even_sum` and the O(q) loop to process the queries. The space complexity comes from storing the results in the `answer` array, which has the same size as the number of queries.

📖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

Recompute Even Sum After Each Query

After each query, iterate through the entire `nums` array and compute the sum of even numbers. This is a straightforward approach, but less efficient.

Time: O(n * q), where n is the length of the `nums` array and q is the length of the `queries` array.Space: O(q), for storing the results in the answer array.

Trade-offs: This approach is simpler to understand but significantly less efficient than the main solution, especially when dealing with a large number of queries. It should only be used if performance is not a concern.

✨

Key Insights

1Insight 1: The key is to efficiently maintain the sum of even numbers. We don't need to recompute the entire sum after each query. We can update the sum incrementally based on whether the updated element was even before and after the query.
2Insight 2: Modifying the original array `nums` in place allows us to directly see the effect of the query on the array and thus on the even sum.
3Insight 3: We only need to track whether the element at the given index was even before and after the query. This helps avoid iterating through the entire array after each modification.

📚

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: `nums` array is empty. The algorithm will still work correctly, as the initial `even_sum` will be 0, and the queries will still be processed according to the problem statement.
Edge case 2: All numbers in the `nums` array are odd. The initial `even_sum` will be 0, and the updates will be performed based on whether the numbers become even after adding the `val` from the queries.
Edge case 3: All `val` values in the `queries` are 0. The `nums` array will not change, but the algorithm will still correctly calculate the sum of even numbers after each query (which will be the same after each query).
Edge case 4: A query results in nums[index] becoming a large number (positive or negative). The algorithm handles this correctly because the even/odd check (`nums[index] % 2 == 0`) will work for any integer value.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Recomputing the entire even sum after each query instead of incrementally updating it. This leads to a time complexity of O(n * q) instead of O(n + q).

✗Common mistake 2: Not considering the case where a number becomes even after being updated, or vice versa (even becomes odd). Incorrectly handling this leads to incorrect calculation of the even sum.

✗Common mistake 3: Not understanding the modulo operator (`%`) and how it can be used to efficiently check if a number is even or odd.

✅

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: Processing real-time data streams where you need to maintain aggregate statistics (like sum of even numbers) based on updates. For example, tracking even transaction amounts in a financial system.

🏦Real-world scenario 2: Implementing a system that tracks performance metrics based on user actions. Each query could represent a user action that modifies the data and requires a recalculation of certain statistics.

🎮Real-world scenario 3: In game development, dynamically adjusting game state based on player actions and maintaining statistics of certain attributes (e.g., sum of even-numbered power-ups collected).

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: August 25, 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 14

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

@AlgoMasterJul 7

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

@InterviewProJun 30

❓

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 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 "Sum of Even Numbers After Queries"?

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

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