Longest Subarray with Sum at most S II in O(n) Time Complexity using 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

Your algorithm should run in O(n) 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 (or two-pointer) technique. This approach allows us to efficiently find the longest subarray with a sum at most S in O(n) time complexity.

Here’s a step-by-step breakdown of the approach:

1. Initialize two pointers, start and end, both set to the beginning of the array.
2. Maintain a variable current_sum to store the sum of the current subarray.
3. Iterate through the array using the end pointer. Add the value at the end pointer to current_sum.
4. If current_sum exceeds S, increment the start pointer to reduce the sum until it is less than or equal to S.
5. Keep track of the maximum length of the subarray and the corresponding start and end indices.

Algorithm

Here is the step-by-step algorithm:

1. Initialize start, end, current_sum, max_length, best_start, and best_end.
2. Iterate through the array with the end pointer.
3. Add the current element to current_sum.
4. While current_sum is greater than S, increment the start pointer and subtract the value at start from current_sum.
5. If the current subarray length is greater than max_length, update max_length, best_start, and best_end.
6. Return the indices best_start and best_end.

Code Implementation

#include <iostream>
#include <vector>
using namespace std;

pair longestSubarrayWithSumAtMostS(const vector<int>& nums, int S) {
    int start = 0, end = 0, current_sum = 0;
    int max_length = 0, best_start = 0, best_end = 0;

    while (end < nums.size()) {
        // Add the current element to the current_sum
        current_sum += nums[end];

        // While current_sum exceeds S, move the start pointer to the right
        while (current_sum > S) {
            current_sum -= nums[start];
            start++;
        }

        // Update the maximum length and best indices if we found a longer subarray
        if (end - start + 1 > max_length) {
            max_length = end - start + 1;
            best_start = start;
            best_end = end;
        }

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

    return {best_start, best_end};
}

int main() {
    vector<int> nums = {3, 2, 5, 2, 2, 1, 1, 3, 1, 2};
    int S = 11;
    pair<int, int> result = longestSubarrayWithSumAtMostS(nums, S);
    cout << "Longest subarray with sum at most " << S << " is from index " << result.first << " to " << result.second << 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

Consider the following edge cases:

• All elements are greater than S: The algorithm should return an empty subarray or handle it gracefully.
• The array is empty: The algorithm should return an empty subarray.
• S is very large: The algorithm should return the entire array if the sum is within the limit.

Testing

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

• Simple cases with small arrays.
• Cases where the entire array is the longest subarray.
• Cases with varying values of S.
• Edge cases as mentioned above.

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

• Understand the problem requirements and constraints thoroughly.
• Think about different approaches and their trade-offs.
• Use diagrams or pseudo-code to visualize the problem and solution.
• 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 technique. 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 - A platform for practicing coding problems.
• GeeksforGeeks - A comprehensive resource for learning algorithms and data structures.
• Coursera - Online courses on algorithms and data structures.