In this blog post, we will explore how to find the largest subarray in a Java array whose sum is less than or equal to a given value. This problem can be efficiently solved using the sliding window technique.

Table of Contents

• Introduction to the Problem
• Approach
• Implementation in Java
• Conclusion
• Hashtags

Introduction to the Problem

Given an array of integers and a target value, we need to find the length of the largest subarray whose sum is less than or equal to the target value.

For example, given the array {2, 4, 5, 8, 10} and target value 15, the subarray with the largest sum less than or equal to 15 is {2, 4, 5, 8}, and its length is 4.

Approach

To solve this problem, we can use the sliding window technique. We will maintain two pointers, start and end, that define the current subarray. We will also keep track of the current sum of the subarray.

1. Initialize start and end pointers to the beginning of the array.
2. Initialize maxLen to 0, which will store the length of the largest subarray.
3. Iterate over the array while the end pointer is less than the array length.
   1. Calculate the current sum of the subarray from start to end.
   2. If the sum is less than or equal to the target value and the length of the current subarray is greater than maxLen, update maxLen.
   3. If the sum is greater than the target value, move the start pointer to the right to shrink the subarray.
   4. Move the end pointer to the right to expand the subarray.
4. Return maxLen as the length of the largest subarray.

Implementation in Java

Here’s the Java code implementing the above approach:

class Solution {
    public int findLargestSubarray(int[] nums, int target) {
        int start = 0;
        int end = 0;
        int maxLen = 0;
        int currentSum = 0;

        while (end < nums.length) {
            currentSum += nums[end];

            while (currentSum > target) {
                currentSum -= nums[start];
                start++;
            }

            if (currentSum <= target) {
                int currentLen = end - start + 1;
                maxLen = Math.max(maxLen, currentLen);
            }

            end++;
        }

        return maxLen;
    }
}

Conclusion

By using the sliding window technique, we can efficiently find the largest subarray in a Java array whose sum is less than or equal to a given target value. This approach has a time complexity of O(n), where n is the length of the array.

In this blog post, we discussed the problem, the approach, and provided an implementation in Java. Feel free to use this code as a reference for your own projects.

Hashtags

#Java #SlidingWindow