Last updated on October 5th, 2024 at 04:21 pm

Here, We see Contiguous Array LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Contiguous Array LeetCode Solution

Problem Statement

Given a binary array nums, return the maximum length of a contiguous subarray with an equal number of 0 and 1.

Example 1:
Input: nums = [0,1]
Output: 2
Explanation: [0, 1] is the longest contiguous subarray with an equal number of 0 and 1.

Example 2:
Input: nums = [0,1,0]
Output: 2
Explanation: [0, 1] (or [1, 0]) is a longest contiguous subarray with equal number of 0 and 1.

1. Contiguous Array Leetcode Solution C++

class Solution {
public:
    int findMaxLength(vector<int>& nums) {
		int maxlen = 0;
        for (int i = 0; i < nums.size(); i++) {
            int zeroes = 0, ones = 0;
            for (int j = i; j < nums.size(); j++) {
                if (nums[j] == 0) {
                    zeroes++;
                } else {
                    ones++;
                }
                if (zeroes == ones) {
                    maxlen = max(maxlen, j - i + 1);
                }
            }
        }
        return maxlen;
    }
};

2. Contiguous Array Leetcode Solution Java

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

3. Contiguous Array Leetcode Solution JavaScript

var findMaxLength = function(nums) {
    let mp = new Map();
    let sum = 0;
    let subArrayLength = 0;
    for (let i = 0; i < nums.length; i++) {
        sum += nums[i] === 0 ? -1 : 1;
        if (sum === 0) {
            subArrayLength = i + 1;
        } else if (mp.has(sum)) {
            subArrayLength = Math.max(subArrayLength, i - mp.get(sum));
        } else {
            mp.set(sum, i);
        }
    }
    return subArrayLength;
};

4. Contiguous Array Leetcode Solution Python

class Solution(object):
    def findMaxLength(self, nums):
        mp = {}
        sum_val = 0
        max_len = 0
        for i, num in enumerate(nums):
            sum_val += 1 if num == 1 else -1
            if sum_val == 0:
                max_len = i + 1
            elif sum_val in mp:
                max_len = max(max_len, i - mp[sum_val])
            else:
                mp[sum_val] = i
        return max_len