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1446. Consecutive Characters

Description

The power of the string is the maximum length of a non-empty substring that contains only one unique character.

Given a string s, return the power of s.

 

Example 1:

Input: s = "leetcode"
Output: 2
Explanation: The substring "ee" is of length 2 with the character 'e' only.

Example 2:

Input: s = "abbcccddddeeeeedcba"
Output: 5
Explanation: The substring "eeeee" is of length 5 with the character 'e' only.

 

Constraints:

  • 1 <= s.length <= 500
  • s consists of only lowercase English letters.

Solutions

  • class Solution {
        public int maxPower(String s) {
            int ans = 1, t = 1;
            for (int i = 1; i < s.length(); ++i) {
                if (s.charAt(i) == s.charAt(i - 1)) {
                    ans = Math.max(ans, ++t);
                } else {
                    t = 1;
                }
            }
            return ans;
        }
    }
    
  • class Solution {
    public:
        int maxPower(string s) {
            int ans = 1, t = 1;
            for (int i = 1; i < s.size(); ++i) {
                if (s[i] == s[i - 1]) {
                    ans = max(ans, ++t);
                } else {
                    t = 1;
                }
            }
            return ans;
        }
    };
    
  • class Solution:
        def maxPower(self, s: str) -> int:
            ans = t = 1
            for a, b in pairwise(s):
                if a == b:
                    t += 1
                    ans = max(ans, t)
                else:
                    t = 1
            return ans
    
    
  • func maxPower(s string) int {
    	ans, t := 1, 1
    	for i := 1; i < len(s); i++ {
    		if s[i] == s[i-1] {
    			t++
    			ans = max(ans, t)
    		} else {
    			t = 1
    		}
    	}
    	return ans
    }
    
  • function maxPower(s: string): number {
        let ans = 1;
        let t = 1;
        for (let i = 1; i < s.length; ++i) {
            if (s[i] === s[i - 1]) {
                ans = Math.max(ans, ++t);
            } else {
                t = 1;
            }
        }
        return ans;
    }
    
    

All Problems

All Solutions