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

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) 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 network data buffering, memory management, and financial analysis where we need to find optimal subarrays under certain constraints.

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 the sliding window technique, which is efficient for problems involving contiguous subarrays. The idea is to maintain a window that expands and contracts to ensure the sum of its elements is at most S.

1. **Naive Solution**: A naive approach would involve checking all possible subarrays and their sums, which would be inefficient with a time complexity of O(n^2).

2. **Optimized Solution**: The sliding window technique allows us to achieve the desired result in O(n) time complexity. We maintain two pointers (start and end) to represent the window and adjust them based on the sum of the elements within the window.

Algorithm

1. Initialize two pointers, start and end, both set to 0.

2. Initialize variables to keep track of the current sum and the maximum length of the subarray found.

3. Iterate through the array using the end pointer to expand the window.

4. Add the current element to the current sum.

5. If the current sum exceeds S, move the start pointer to the right until the sum is less than or equal to S.

6. Update the maximum length and the indices of the longest subarray whenever a longer valid subarray is found.

Code Implementation

public class LongestSubarrayWithSumAtMostS {
    public static int[] findLongestSubarray(int[] nums, int S) {
        int start = 0, end = 0, currentSum = 0;
        int maxLength = 0;
        int[] result = new int[2];

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

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

            // Update the maximum length and the result indices
            if (end - start + 1 > maxLength) {
                maxLength = end - start + 1;
                result[0] = start;
                result[1] = end;
            }

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

        return result;
    }

    public static void main(String[] args) {
        int[] nums = {3, 2, 5, 2, 2, 1, 1, 3, 1, 2};
        int S = 11;
        int[] result = findLongestSubarray(nums, S);
        System.out.println("Start index: " + result[0] + ", End index: " + result[1]);
    }
}

Complexity Analysis

The time complexity of the sliding window 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 only a few extra variables.

Edge Cases

1. If all elements are greater than S, the function should return an empty subarray or appropriate indices indicating no valid subarray.

2. If the array is empty, the function should handle it gracefully.

3. If S is very large, the entire array might be the longest subarray.

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 criteria.
Edge cases with empty arrays or very large S values.

Thinking and Problem-Solving Tips

When approaching such problems, it is crucial to understand the constraints and requirements clearly. Using efficient techniques like the sliding window can significantly reduce the complexity. Practice similar problems to get a better grasp of these techniques.

Conclusion

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