Longest Subarray With Sum At Most S in C++

Given an array of positive integers and a number S, find the longest contiguous subarray having the sum at most S.

Return the start and end indices denoting this subarray.

Example

Input: nums = [3, 2, 5, 2, 2, 1, 1, 3, 1 , 2], S = 11
Output: [3, 8]
Explanation:the subarray nums[3...8] of sum 10

Note:

Your algorithm should run in O(n^2) time and use O(1) extra space.

Understanding the Problem

The core challenge of this problem is to find the longest contiguous subarray whose sum does not exceed a given value S. This problem is significant in various applications such as resource allocation, budgeting, and data analysis where constraints on sums are common.

Potential pitfalls include misunderstanding the requirement for the subarray to be contiguous and not considering all possible subarrays.

Approach

To solve this problem, we can use a sliding window approach. The naive solution involves checking all possible subarrays, which is not optimal. Instead, we can use a more efficient approach:

Initialize two pointers to represent the start and end of the subarray.
Expand the end pointer to include more elements until the sum exceeds S.
When the sum exceeds S, move the start pointer to reduce the sum.
Keep track of the longest subarray found that meets the condition.

Algorithm

Here is a step-by-step breakdown of the sliding window algorithm:

Initialize variables to keep track of the current sum, the maximum length of the subarray, and the start and end indices of the longest subarray.
Use a loop to expand the end pointer and add elements to the current sum.
When the current sum exceeds S, use another loop to move the start pointer and subtract elements from the current sum until it is less than or equal to S.
Update the maximum length and the start and end indices if a longer valid subarray is found.

Code Implementation


#include <iostream>
#include <vector>

std::vector<int> longestSubarrayWithSumAtMostS(const std::vector<int>& nums, int S) {
    int n = nums.size();
    int maxLength = 0;
    int start = 0, end = 0;
    int currentSum = 0;
    int bestStart = 0, bestEnd = 0;

    while (end < n) {
        // Add the current element to the current sum
        currentSum += nums[end];

        // If the current sum exceeds S, move the start pointer to reduce the sum
        while (currentSum > S && start <= end) {
            currentSum -= nums[start];
            start++;
        }

        // Update the maximum length and the best start and end indices
        if (end - start + 1 > maxLength) {
            maxLength = end - start + 1;
            bestStart = start;
            bestEnd = end;
        }

        // Move the end pointer to the right
        end++;
    }

    return {bestStart, bestEnd};
}

int main() {
    std::vector<int> nums = {3, 2, 5, 2, 2, 1, 1, 3, 1, 2};
    int S = 11;
    std::vector<int> result = longestSubarrayWithSumAtMostS(nums, S);
    std::cout << "Start: " << result[0] << ", End: " << result[1] << std::endl;
    return 0;
}

Complexity Analysis

The time complexity of this approach is O(n) because each element is processed at most twice (once by the end pointer and once by the start pointer). The space complexity is O(1) as we are using a constant amount of extra space.

Edge Cases

Potential edge cases include:

All elements are greater than S: The function should return an empty subarray or appropriate indices indicating no valid subarray.
The array is empty: The function should handle this gracefully.
S is very large: The function should return the entire array.

Testing

To test the solution comprehensively, consider the following test cases:

Simple cases with small arrays.
Cases where all elements are greater than S.
Cases with varying lengths of subarrays that meet the condition.
Edge cases such as empty arrays and very large S values.

Thinking and Problem-Solving Tips

When approaching such problems, it is helpful to:

Break down the problem into smaller parts and understand the requirements.
Consider different approaches and their trade-offs.
Practice similar problems to improve problem-solving skills.

Conclusion

In this blog post, we discussed how to find the longest subarray with a sum at most S using a sliding window approach. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for improving algorithmic thinking and coding skills.

Additional Resources

For further reading and practice, consider the following resources:

LeetCode - Practice coding problems and challenges.
GeeksforGeeks - Tutorials and articles on various algorithms and data structures.
Coursera - Online courses on algorithms and data structures.

Longest Subarray with Sum at most S in C++ (O(n^2) Time Complexity)