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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

Solution 1: Traversal and Counting

We define a variable \(\textit{t}\) to represent the length of the current consecutive characters, initially \(\textit{t}=1\).

Next, we traverse the string \(s\) starting from the second character. If the current character is the same as the previous character, then \(\textit{t} = \textit{t} + 1\), and update the answer \(\textit{ans} = \max(\textit{ans}, \textit{t})\); otherwise, set \(\textit{t} = 1\).

Finally, return the answer \(\textit{ans}\).

The time complexity is \(O(n)\), where \(n\) is the length of the string \(s\). The space complexity is \(O(1)\).

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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