Fluctuating Treasury Balance — Problem Statement & Solution Guide
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.
Examples
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 add 1, subtract 2, add 3, subtract 4, and add 5. The final result is 3.
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
We precompute an alternating prefix sum array where each element at index i adds arr[i] if i is even, and subtracts arr[i] if i is odd. For each query [L, R], we compute the difference between prefix[R+1] and prefix[L]. If L is odd, the query starts with a negative term relative to our absolute prefix sum, so we simply negate the difference. This allows us to answer each query in O(1) time.
Brute Force Approach
Iterate through each query range [L, R] and manually simulate the alternating addition and subtraction element by element. This requires a nested loop where the outer loop iterates over all queries and the inner loop travels from index L to R, resulting in an O(Q * N) time complexity which will time out for large arrays.
Verified Code Solutions
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 resultfunction solution(nums) {
let result = 0;
let add = true;
for (let num of nums) {
if (add) {
result += num;
} else {
result -= num;
}
add = !add;
}
return result;
}Solve in Interative Editor
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