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Formatted question description: https://leetcode.ca/all/1446.html

1446. Consecutive Characters (Easy)

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

Return the power of the string.

 

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.

Example 3:

Input: s = "triplepillooooow"
Output: 5

Example 4:

Input: s = "hooraaaaaaaaaaay"
Output: 11

Example 5:

Input: s = "tourist"
Output: 1

 

Constraints:

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

Related Topics:
String

Similar Questions:

• Max Consecutive Ones (Easy)

Solution 1.

• Java
• C++
• Python
• Go

• class Solution {
      public int maxPower(String s) {
          if (s == null || s.length() == 0)
              return 0;
          int length = s.length();
          int maxPower = 1;
          int power = 0;
          char prevC = 'A';
          for (int i = 0; i < length; i++) {
              char c = s.charAt(i);
              if (c == prevC) {
                  power++;
                  maxPower = Math.max(maxPower, power);
              } else {
                  power = 1;
                  prevC = c;
              }
          }
          return maxPower;
      }
  }
  
  ############
  
  class Solution {
      public int maxPower(String s) {
          int ans = 0, t = 0;
          for (int i = 0; i < s.length(); ++i) {
              if (i == 0 || s.charAt(i) == s.charAt(i - 1)) {
                  ++t;
              } else {
                  t = 1;
              }
              ans = Math.max(ans, t);
          }
          return ans;
      }
  }
  

• // OJ: https://leetcode.com/problems/consecutive-characters/
  // Time: O(N)
  // Space: O(1)
  class Solution {
  public:
      int maxPower(string s) {
          int ans = 1, cnt = 1;
          for (int i = 1; i < s.size(); ++i) {
              if (s[i] == s[i - 1]) ans = max(ans, ++cnt);
              else cnt = 1;
          }
          return ans;
      }
  };
  

• class Solution:
      def maxPower(self, s: str) -> int:
          ans = t = 0
          for i, c in enumerate(s):
              if i == 0 or c == s[i - 1]:
                  t += 1
              else:
                  t = 1
              ans = max(ans, t)
          return ans
  
  
  

• func maxPower(s string) int {
  	ans, t := 0, 0
  	for i := range s {
  		if i == 0 || s[i] == s[i-1] {
  			t++
  		} else {
  			t = 1
  		}
  		ans = max(ans, t)
  	}
  	return ans
  }
  
  func max(a, b int) int {
  	if a > b {
  		return a
  	}
  	return b
  }
  

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