LeetCode 0525. Contiguous Array Solution in Java, C++, Python & More | Explanation + Code

Description

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.

Example 3:

Input: nums = [0,1,1,1,1,1,0,0,0]
Output: 6
Explanation: [1,1,1,0,0,0] is the longest contiguous subarray with equal number of 0 and 1.

 

Constraints:

• 1 <= nums.length <= 105
• nums[i] is either 0 or 1.

Solutions

Solution 1: Prefix Sum + Hash Table

According to the problem description, we can treat 0s in the array as -1. In this way, when encountering a 0, the prefix sum s will decrease by one, and when encountering a 1, the prefix sum s will increase by one. Therefore, suppose the prefix sum s is equal at indices j and i, where j < i, then the subarray from index j + 1 to i has an equal number of 0s and 1s.

We use a hash table to store all prefix sums and their first occurrence indices. Initially, we map the prefix sum of 0 to -1.

As we iterate through the array, we calculate the prefix sum s. If s is already in the hash table, then we have found a subarray with a sum of 0, and its length is i - d[s], where d[s] is the index where s first appeared in the hash table. If s is not in the hash table, we store s and its index i in the hash table.

The time complexity is O(n), and the space complexity is O(n), where n is the length of the array.

PythonJavaC++GoTypeScriptJavaScript

class Solution: def findMaxLength(self, nums: List[int]) -> int: d = {0: -1} ans = s = 0 for i, x in enumerate(nums): s += 1 if x else -1 if s in d: ans = max(ans, i - d[s]) else: d[s] = i return ans(code-box)

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

class Solution { public: int findMaxLength(vector<int>& nums) { unordered_map<int, int> d{{0, -1}}; int ans = 0, s = 0; for (int i = 0; i < nums.size(); ++i) { s += nums[i] ? 1 : -1; if (d.contains(s)) { ans = max(ans, i - d[s]); } else { d[s] = i; } } return ans; } };(code-box)

func findMaxLength(nums []int) int { d := map[int]int{0: -1} ans, s := 0, 0 for i, x := range nums { if x == 0 { x = -1 } s += x if j, ok := d[s]; ok { ans = max(ans, i-j) } else { d[s] = i } } return ans }(code-box)

function findMaxLength(nums: number[]): number { const d: Record<number, number> = { 0: -1 }; let ans = 0; let s = 0; for (let i = 0; i < nums.length; ++i) { s += nums[i] ? 1 : -1; if (d.hasOwnProperty(s)) { ans = Math.max(ans, i - d[s]); } else { d[s] = i; } } return ans; }(code-box)

/** * @param {number[]} nums * @return {number} */ var findMaxLength = function (nums) { const d = { 0: -1 }; let ans = 0; let s = 0; for (let i = 0; i < nums.length; ++i) { s += nums[i] ? 1 : -1; if (d.hasOwnProperty(s)) { ans = Math.max(ans, i - d[s]); } else { d[s] = i; } } return ans; };(code-box)