Lazy Segment Query Validator 4 — Problem Statement & Solution Guide

Binary SearchHardBinary Search on Answer Matrix

TimeO(N)

SpaceO(1)

Problem Description

You are given an array of N integers. Compute the sum of all elements. The array may contain negative numbers. The sum should be output as a 64-bit signed integer.

The input begins with an integer N, followed by a line containing N space-separated integers. The output is a single integer representing the sum.

Constraints guarantee that the sum fits within a signed 64-bit integer.

DSA Pattern Breakdown

"Lazy Segment Query Validator 4"

hard

WHY DOES IT MATTER?

The single‑pass accumulation pattern is essential because it guarantees linear time and constant space, making it suitable for large data streams and real‑time analytics where memory and latency are critical constraints.

OPTIMIZATION CHALLENGE

The key insight is that addition is associative and commutative, allowing you to accumulate the result incrementally without needing to store intermediate results. This reduces both time and space overhead.

REAL-WORLD CONNECTION

Think of a bank that needs to compute the net balance of all accounts each day. The bank processes millions of transactions, and a single‑pass sum ensures the daily reconciliation completes quickly without storing all transaction details in memory.

When presenting this solution in an interview, emphasize the importance of choosing the correct data type (long long) and explain how a single loop eliminates unnecessary overhead, making the code both simple and efficient.

COMPLEXITY AT A GLANCE

⏱ Time:O(N)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem reduces to computing the sum of a sequence of integers, a classic example of a linear-time aggregation problem. A naive approach might involve storing all elements and then iterating over them, but this introduces unnecessary memory overhead and can be problematic when the input size is large. The optimal paradigm is a single-pass scan that accumulates the sum in a 64‑bit signed variable, ensuring that the intermediate and final results stay within the bounds of a signed long long. This approach leverages the associative property of addition and the fact that addition is a constant‑time operation, yielding an O(N) time complexity with O(1) auxiliary space. Importantly, using a 64‑bit accumulator prevents overflow even when the individual elements are large or negative, as guaranteed by the problem constraints.

In many real‑world systems, such as financial transaction processing or log aggregation, the ability to compute running totals efficiently is critical. The linear scan pattern is a building block for more complex algorithms like prefix sums, sliding windows, and streaming analytics, where the data may not fit entirely in memory. By mastering this simple pattern, engineers can confidently extend it to distributed settings, ensuring correctness and performance across large data sets.

Interview Questions on This Problem

Q1At Google, how would you handle summing a massive stream of integers that may not fit in memory, ensuring no overflow occurs?

I would use a 64‑bit accumulator and process the stream in chunks, updating the sum incrementally. If the stream is distributed, I would aggregate partial sums in each worker and then combine them, still using 64‑bit integers to avoid overflow.

Q2In a fintech startup, you need to compute daily transaction totals for millions of users. What algorithmic pattern would you use and why?

I would use a single-pass aggregation with a 64‑bit accumulator, possibly parallelized across shards. This pattern is efficient, uses minimal memory, and scales horizontally, which is essential for high‑throughput transaction processing.

Q3At a high‑growth engineering startup, you’re asked to implement a feature that reports the sum of user scores in real time. What pitfalls should you avoid?

Avoid using 32‑bit integers to store the sum, as overflow can silently corrupt results. Also, ensure that the input parsing is robust to negative values and that the accumulation happens in a single loop to keep the algorithm O(N) and O(1) space.

Examples

Example 1

Input

5
1 2 3 4 5

Output

Explanation: 1+2+3+4+5=15.

Example 2

Input

4
-1 -2 -3 -4

Output

-10

Explanation: -1-2-3-4=-10.

Constraints

1 <= N <= 2 * 10^5
-10^9 <= arr[i] <= 10^9
Time Complexity: O(N log N) or O(N log^2 N)
Space Complexity: O(N)

Optimal Approach & Strategy

Read each integer one by one and immediately add it to a 64‑bit accumulator, avoiding any extra storage.

Brute Force Approach

Read all N integers into an array, then loop over the array to add each element to a running total.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N)

const readline = require('readline');

const rl = readline.createInterface({
    input: process.stdin,
    terminal: false
});

let lines = [];
let lineCount = 0;

rl.on('line', (line) => {
    lines.push(line);
    lineCount++;
    
    if (lineCount === 2) {
        const N = parseInt(lines[0]);
        const arr = lines[1].split(' ').map(Number);
        
        let sum = 0;
        for (let i = 0; i < N; i++) {
            sum += arr[i];
        }
        
        console.log(sum);
        rl.close();
    }
});

#include <iostream>
#include <vector>

using namespace std;

int main() {
    int N;
    cin >> N;
    vector<long long> arr(N);
    for (int i = 0; i < N; i++) {
        cin >> arr[i];
    }
    
    long long sum = 0;
    for (int i = 0; i < N; i++) {
        sum += arr[i];
    }
    
    cout << sum << endl;
    
    return 0;
}

import java.util.*;
import java.io.*;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        
        int N = Integer.parseInt(br.readLine().trim());
        String[] parts = br.readLine().trim().split(" ");
        long[] arr = new long[N];
        for (int i = 0; i < N; i++) {
            arr[i] = Long.parseLong(parts[i]);
        }
        
        long sum = 0;
        for (int i = 0; i < N; i++) {
            sum += arr[i];
        }
        
        System.out.println(sum);
        br.close();
    }
}

import sys

def main():
    input = sys.stdin.read
    data = input().split()
    
    N = int(data[0])
    arr = list(map(int, data[1:N+1]))
    
    total = sum(arr)
    print(total)

if __name__ == "__main__":
    main()

const readline = require('readline');

const rl = readline.createInterface({
    input: process.stdin,
    terminal: false
});

let lines = [];
let lineCount = 0;

rl.on('line', (line) => {
    lines.push(line);
    lineCount++;
    
    if (lineCount === 2) {
        const N = parseInt(lines[0]);
        const arr = lines[1].split(' ').map(Number);
        
        let sum = 0;
        for (let i = 0; i < N; i++) {
            sum += arr[i];
        }
        
        console.log(sum);
        rl.close();
    }
});

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Lazy Segment Query Validator 4 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Binary Search

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Lazy Segment Query Validator 4 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor