Fluctuating Treasury Balance — Problem Statement & Solution Guide

ArraysMediumRange Sum Query using Prefix Sum

TimeO(N + Q)

SpaceO(N)

Quick Answer & Algorithm Key Takeaway

Master coding challenges related to Arrays and solve the Fluctuating Treasury Balance problem optimally.

TopicArrays

PatternRange Sum Query using Prefix Sum

TimeO(N + Q)

SpaceO(N)

Problem Description

A financial analyst is tracking a company's daily treasury fluctuations. You are given an integer array arr representing daily transaction values. The analyst wants to evaluate the net performance of several campaigns. Each campaign is defined by a range [L, R] (0-indexed). The net performance of a campaign is calculated by alternatingly adding and subtracting the daily transaction values starting from the first day of the campaign.

DSA Pattern Breakdown

"Fluctuating Treasury Balance"

medium

WHY DOES IT MATTER?

Efficient range‑query patterns turn quadratic brute‑force into linear‑time solutions, crucial for high‑throughput systems.

OPTIMIZATION CHALLENGE

The key is reducing per‑query work from O(length) to O(1) by pre‑aggregating with parity‑aware prefix sums.

REAL-WORLD CONNECTION

Think of a ledger where debits and credits alternate; computing net balance over periods must be fast for real‑time reporting.

Store both the signed prefix and the original parity prefix; a one‑liner answer emerges by checking L%2.

COMPLEXITY AT A GLANCE

⏱ Time:O(N + Q)

💾 Space:O(N)

Core Theory — Why This Approach?

The query asks for an alternating sum over a sub‑array, i.e., arr[L] − arr[L+1] + arr[L+2] − … . A naïve solution recomputes this sum for each query, leading to O(N·Q) time which is prohibitive when N and Q reach 10^5 or more. The optimal paradigm leverages prefix sums combined with parity handling: by storing two cumulative arrays—one for elements at even indices and one for odd indices—we can retrieve the required alternating sum in O(1) per query. Alternatively, we can pre‑multiply each element by (+1) or (‑1) depending on its index parity, build a single signed prefix sum, and adjust the sign of the result based on the parity of L, achieving the same O(N+Q) overall complexity.

Interview Questions on This Problem

Q1How can you answer alternating‑sum range queries in O(1) after O(N) preprocessing?

Create a signed array where sign = (+1) for even indices and (‑1) for odd indices, build its prefix sum, and multiply the difference by sign(L).

Q2Why does a single prefix sum not suffice without considering index parity?

Because the sign of each element in the alternating sum depends on its position relative to L, not just its absolute index.

Q3What data type should you use for the prefix sums and why?

Use 64‑bit integers (long long) to avoid overflow when N·max|arr[i]| exceeds 32‑bit limits.

Examples

Example 1

Input

[1, -2, 3, -4, 5]

Output

Explanation: Step-by-step: with input [1, -2, 3, -4, 5], we alternate between adding and subtracting the daily transaction values. Starting with 0, we add 1, subtract 2, add 3, subtract 4, and add 5. The final result is 3.

Example 2

Input

[-1, 2, -3, 4, -5]

Output

-3

Explanation: Step-by-step: with input [-1, 2, -3, 4, -5], we alternate between adding and subtracting the daily transaction values. Starting with 0, we subtract 1, add 2, subtract 3, add 4, and subtract 5. The final result is -3.

Constraints

1 <= arr.length <= 10^5
1 <= queries.length <= 10^5
queries[i].length == 2
0 <= queries[i][0] <= queries[i][1] < arr.length
-10^4 <= arr[i] <= 10^4

Optimal Approach & Strategy

Pre‑compute a signed prefix sum (or two parity‑specific prefixes) and answer each query with a constant‑time arithmetic expression.

Brute Force Approach

Iterate from L to R for each query, adding and subtracting elements alternately, which costs O(R‑L+1) per query.

Code Solutions

Language:

JavaScript Solution

Time: O(N + Q)

function solution(nums) {
   let result = 0;
   let add = true;
   for (let num of nums) {
       if (add) {
           result += num;
       } else {
           result -= num;
       }
       add = !add;
   }
   return result;
}

class Solution {
   public:
       int solution(vector<int>& nums) {
           int result = 0;
           bool add = true;
           for (int num : nums) {
               if (add) {
                   result += num;
               } else {
                   result -= num;
               }
               add = !add;
           }
           return result;
       }
   };

class Solution {
   public int solution(int[] nums) {
       int result = 0;
       boolean add = true;
       for (int num : nums) {
           if (add) {
               result += num;
           } else {
               result -= num;
           }
           add = !add;
       }
       return result;
   }
}

def solution(nums):
   result = 0
   add = True
   for num in nums:
       if add:
           result += num
       else:
           result -= num
       add = not add
   return result

function solution(nums) {
   let result = 0;
   let add = true;
   for (let num of nums) {
       if (add) {
           result += num;
       } else {
           result -= num;
       }
       add = !add;
   }
   return result;
}

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