Day 3 - Largest subarray of 0's and 1's.md

Difficulty	Easy
Source	160 Days of Problem Solving
Tags	prefix-sum sliding-window Hash

🚀 Day 3. Largest Subarray of 0s and 1s 🧠

The problem can be found at the following link: Problem Link

💡 Problem Description:

You are given a binary array arr[] consisting of 0s and 1s. Your task is to find the length of the largest subarray that contains an equal number of 0s and 1s.

🔍 Example Walkthrough:

Input:
arr[] = [1, 0, 1, 1, 1, 0, 0]
Output:
6

Explanation: The subarray [0, 1, 1, 1, 0, 0] has an equal number of 0s and 1s (three 0s and three 1s).

Input:
arr[] = [0, 0, 1, 1, 0]
Output:
4

Explanation: Both [0, 0, 1, 1] and [0, 1, 1, 0] are valid subarrays with an equal number of 0s and 1s.

Input:
arr[] = [0]
Output:
0

Explanation: No subarray has an equal number of 0s and 1s.

Constraints:

• $1 <= arr.size() <= 10^5$
• arr[i] is either 0 or 1.

🎯 My Approach:

HashMap and Prefix Sum Technique

To solve this problem efficiently, we use a hashmap to store the first occurrence of prefix sums. This helps us determine the length of subarrays with equal numbers of 0s and 1s:

1. Treat 0 as -1 to convert the problem into finding a subarray with sum 0.
2. Maintain a prefix sum while iterating over the array.
3. Use a hashmap to store the first index where each prefix sum occurs.
4. If the same prefix sum is encountered again, the subarray between these two indices has a sum of 0 (indicating equal numbers of 0s and 1s).
5. Update the maximum length for each valid subarray.

Steps:

1. Initialize a hashmap to store prefix sums and their first occurrence index.
2. Replace all 0s with -1 in the array.
3. Traverse the array while maintaining a prefix sum:
   • If the prefix sum is 0, the subarray from the start to the current index is valid.
   • If the prefix sum has been seen before, calculate the length of the subarray and update the maximum length.
   • Otherwise, store the prefix sum with its index.
4. Return the maximum length.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(n), where n is the size of the array. Each element is processed once, and hashmap operations (insert and lookup) are O(1) on average.
• Expected Auxiliary Space Complexity: O(n), as the hashmap stores at most n unique prefix sums.

📝 Solution Code

Code (C++)

class Solution {
public:
    int maxLen(vector<int>& arr) {
        unordered_map<int, int> hM;
        int sum = 0, max_len = 0;
        for (int i = 0; i < arr.size(); i++) {
            sum += (arr[i] == 0) ? -1 : 1;
            if (sum == 0) max_len = i + 1;
            if (hM.count(sum)) max_len = max(max_len, i - hM[sum]);
            else hM[sum] = i;
        }
        return max_len;
    }
};

Code (Java)

class Solution {
    public int maxLen(int[] arr) {
        Map<Integer, Integer> map = new HashMap<>();
        int sum = 0, maxLen = 0;
        for (int i = 0; i < arr.length; i++) {
            sum += (arr[i] == 0) ? -1 : 1;
            if (sum == 0) maxLen = i + 1;
            else if (map.containsKey(sum)) maxLen = Math.max(maxLen, i - map.get(sum));
            else map.put(sum, i);
        }
        return maxLen;
    }
}

Code (Python)

class Solution:
    def maxLen(self, arr):
        hmap = {}
        sum, max_len = 0, 0
        for i in range(len(arr)):
            sum += -1 if arr[i] == 0 else 1
            if sum == 0:
                max_len = i + 1
            elif sum in hmap:
                max_len = max(max_len, i - hmap[sum])
            else:
                hmap[sum] = i
        return max_len

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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