Finding the maximum subarray length with given sum.

Below is a Java program that finds the maximum subarray length with a given sum:

    public class MaximumSubarrayLength {

    public static int findMaxLength(int[] nums, int targetSum) {

        int maxLength = 0;

        int sum = 0;

        HashMap<Integer, Integer> sumMap = new HashMap<>();

        sumMap.put(0, -1); // To handle cases where the subarray starts from index 0

      for (int i = 0; i < nums.length; i++) {

            sum += nums[i];

            if (sumMap.containsKey(sum - targetSum)) {

                int subarrayLength = i - sumMap.get(sum - targetSum);

                maxLength = Math.max(maxLength, subarrayLength);

            }

            if (!sumMap.containsKey(sum)) {

                sumMap.put(sum, i);

            }
        }

        return maxLength;

    }

    public static void main(String[] args) {

        int[] nums = {1, -2, 3, 4, -5, 6, 7};

        int targetSum = 5;

        int maxLength = findMaxLength(nums, targetSum);

        System.out.println("Maximum subarray length with sum " + targetSum + ": " + maxLength);

    }

}

In this program, we iterate through the given array `nums` and maintain a running sum `sum` of the elements encountered so far. We also use a HashMap `sumMap` to store the sum and its corresponding index.

At each iteration, we check if `sum – targetSum` exists in `sumMap`. If it does, it means we have found a subarray with the target sum. We calculate the length of the subarray and update `maxLength` if the current length is greater than the previous maximum length.

Finally, we return `maxLength` as the maximum subarray length with the given sum.

In the `main` method, we provide a sample input array `nums` and the target sum `targetSum`. The program then calls the `findMaxLength` method and prints the result.