In this blog post, we will discuss a popular algorithmic problem: finding the largest subarray within a given Java array that contains an equal number of 0s and 1s. This problem can be solved efficiently using a technique called prefix sum.

Table of Contents

1. Problem Overview
2. Approach
3. Implementation in Java
4. Example
5. Conclusion
6. References

Problem Overview

Given a binary (0s and 1s) array of length n, our task is to find the largest subarray that contains an equal number of 0s and 1s.

For example, consider the following input array:

[0, 1, 0, 1, 1, 0, 1, 0]

In this case, the largest subarray with an equal number of 0s and 1s is [0, 1, 1, 0, 1] with a length of 5.

Approach

To solve this problem, we can use the concept of prefix sum. We iterate through the given array and maintain a count variable.

• Initialize a variable maxLen to store the maximum length of the subarray found so far, and sum to 0.
• Create a HashMap to store the cumulative sums as keys and their corresponding indexes as values.
• Traverse the input array and for every element, add 1 to the sum if the element is 1, and −1 if the element is 0.
• Check if the sum value is 0. If it is, update maxLen to the index of the current element + 1 (to include all elements up to the current position).
• If the sum value exists as a key in the HashMap, update maxLen to the maximum value between maxLen and the difference between the current index and the index stored in the HashMap for that sum value.
• If the sum value does not exist as a key in the HashMap, add the sum value with the current index as the value.
• Finally, return maxLen as the length of the largest subarray with an equal number of 0s and 1s.

Implementation in Java

Here is the Java implementation of the approach:

import java.util.HashMap;

class LargestSubarrayEqualZeroOne {

    public static int findMaxLength(int[] nums) {
        int maxLen = 0;
        int sum = 0;
        HashMap<Integer, Integer> map = new HashMap<>();
        map.put(0, -1);
        
        for (int i = 0; i < nums.length; i++) {
            sum += (nums[i] == 0 ? -1 : 1);
            
            if (map.containsKey(sum)) {
                maxLen = Math.max(maxLen, i - map.get(sum));
            } else {
                map.put(sum, i);
            }
        }
        
        return maxLen;
    }
    
    public static void main(String[] args) {
        int[] nums = {0, 1, 0, 1, 1, 0, 1, 0};
        System.out.println("Length of the largest subarray: " + findMaxLength(nums));
    }
}

Example

Let’s consider the same input array as before: [0, 1, 0, 1, 1, 0, 1, 0].

When we execute the above Java program, it will print:

Length of the largest subarray: 5

This means that the largest subarray with an equal number of 0s and 1s in the given array is [0, 1, 1, 0, 1] with a length of 5.

Conclusion

In this blog post, we discussed how to find the largest subarray with an equal number of 0s and 1s in a Java array. We used the technique of prefix sum to solve this problem efficiently. By using a HashMap to store the cumulative sums and their indexes, we were able to find the largest subarray in linear time complexity.

I hope you found this blog post helpful. Happy coding!

References

• GeeksforGeeks - Largest subarray with equal number of 0s and 1s