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525. Contiguous Array

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.

 

Constraints:

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

Solutions

  • class Solution {
        public int findMaxLength(int[] nums) {
            Map<Integer, Integer> mp = new HashMap<>();
            mp.put(0, -1);
            int s = 0, ans = 0;
            for (int i = 0; i < nums.length; ++i) {
                s += nums[i] == 1 ? 1 : -1;
                if (mp.containsKey(s)) {
                    ans = Math.max(ans, i - mp.get(s));
                } else {
                    mp.put(s, i);
                }
            }
            return ans;
        }
    }
    
  • class Solution {
    public:
        int findMaxLength(vector<int>& nums) {
            unordered_map<int, int> mp;
            int s = 0, ans = 0;
            mp[0] = -1;
            for (int i = 0; i < nums.size(); ++i) {
                s += nums[i] == 1 ? 1 : -1;
                if (mp.count(s))
                    ans = max(ans, i - mp[s]);
                else
                    mp[s] = i;
            }
            return ans;
        }
    };
    
  • class Solution:
        def findMaxLength(self, nums: List[int]) -> int:
            s = ans = 0
            mp = {0: -1}
            for i, v in enumerate(nums):
                s += 1 if v == 1 else -1
                if s in mp:
                    ans = max(ans, i - mp[s])
                else:
                    mp[s] = i
            return ans
    
    
  • func findMaxLength(nums []int) int {
    	mp := map[int]int{0: -1}
    	s, ans := 0, 0
    	for i, v := range nums {
    		if v == 0 {
    			v = -1
    		}
    		s += v
    		if j, ok := mp[s]; ok {
    			ans = max(ans, i-j)
    		} else {
    			mp[s] = i
    		}
    	}
    	return ans
    }
    
  • /**
     * @param {number[]} nums
     * @return {number}
     */
    var findMaxLength = function (nums) {
        const mp = new Map();
        mp.set(0, -1);
        let s = 0;
        let ans = 0;
        for (let i = 0; i < nums.length; ++i) {
            s += nums[i] == 0 ? -1 : 1;
            if (mp.has(s)) ans = Math.max(ans, i - mp.get(s));
            else mp.set(s, i);
        }
        return ans;
    };
    
    
  • 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;
    }
    
    

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