Welcome to Subscribe On Youtube

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

• Java
• C++
• Python
• Go
• Javascript

• 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;
  };
  
  

All Problems

All Solutions