525. Contiguous Array LeetCode Solution
In this guide, you will get 525. Contiguous Array LeetCode Solution with the best time and space complexity. The solution to Contiguous Array problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Table of Contents
- Problem Statement
- Complexity Analysis
- Contiguous Array solution in C++
- Contiguous Array solution in Java
- Contiguous Array solution in Python
- Additional Resources

Problem Statement of Contiguous Array
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.
Constraints:
1 <= nums.length <= 105
nums[i] is either 0 or 1.
Complexity Analysis
- Time Complexity: O(n)
- Space Complexity: O(n)
525. Contiguous Array LeetCode Solution in C++
class Solution {
public:
int findMaxLength(vector<int>& nums) {
int ans = 0;
int prefix = 0;
unordered_map<int, int> prefixToIndex{{0, -1}};
for (int i = 0; i < nums.size(); ++i) {
prefix += nums[i] ? 1 : -1;
if (const auto it = prefixToIndex.find(prefix);
it != prefixToIndex.cend())
ans = max(ans, i - it->second);
else
prefixToIndex[prefix] = i;
}
return ans;
}
};
/* code provided by PROGIEZ */
525. Contiguous Array LeetCode Solution in Java
class Solution {
public int findMaxLength(int[] nums) {
int ans = 0;
int prefix = 0;
Map<Integer, Integer> prefixToIndex = new HashMap<>();
prefixToIndex.put(0, -1);
for (int i = 0; i < nums.length; ++i) {
prefix += nums[i] == 1 ? 1 : -1;
if (prefixToIndex.containsKey(prefix))
ans = Math.max(ans, i - prefixToIndex.get(prefix));
else
prefixToIndex.put(prefix, i);
}
return ans;
}
}
// code provided by PROGIEZ
525. Contiguous Array LeetCode Solution in Python
class Solution:
def findMaxLength(self, nums: list[int]) -> int:
ans = 0
prefix = 0
prefixToIndex = {0: -1}
for i, num in enumerate(nums):
prefix += 1 if num else -1
ans = max(ans, i - prefixToIndex.setdefault(prefix, i))
return ans
# code by PROGIEZ
Additional Resources
- Explore all LeetCode problem solutions at Progiez here
- Explore all problems on LeetCode website here
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