# Length of longest substring having all characters as K

Given a string S and a character K. The task is to find the length of the longest substring of S having all characters the same as character K.
Examples:

Input: S = “abcd1111aabc”, K = ‘1’
Output:
Explanation:
1111 is the largest substring of length 4.
Input: S = “#1234#@@abcd”, K = ‘@’
Output:
Explanation:
@@ is the largest substring of length 2.

Approach: The idea is to iterate over the string and check the following two conditions:

• If the current character is the same as character K then increase the value of the counter by one.
• If the current character is not the same as K then update the previous count and reinitialize the counter to 0.
• Repeat the steps above till the length of the string.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to find the length of` `// longest sub-string having all` `// characters same as character K` `int` `length_substring(string S, ``char` `K)` `{` `    ``// Initialize variables` `    ``int` `curr_cnt = 0, prev_cnt = 0, max_len;`   `    ``// Iterate till size of string` `    ``for` `(``int` `i = 0; i < S.size(); i++) {`   `        ``// Check if current character is K` `        ``if` `(S[i] == K) {` `            ``curr_cnt += 1;` `        ``}`   `        ``else` `{` `            ``prev_cnt = max(prev_cnt, curr_cnt);` `            ``curr_cnt = 0;` `        ``}` `    ``}`   `    ``prev_cnt = max(prev_cnt, curr_cnt);`   `    ``// Assingning the max` `    ``// value to max_len` `    ``max_len = prev_cnt;`   `    ``return` `max_len;` `}`   `// Driver code` `int` `main()` `{` `    ``string S = ``"abcd1111aabc"``;` `    ``char` `K = ``'1'``;`   `    ``// Function call` `    ``cout << length_substring(S, K);` `    ``return` `0;` `}`

## Java

 `// Java program for ` `// the above approach` `import` `java.util.*;` `class` `GFG {`   `// Function to find the length of` `// longest sub-string having all` `// characters same as character K` `static` `int` `length_substring(String S, ` `                            ``char` `K)` `{` `  ``// Initialize variables` `  ``int` `curr_cnt = ``0``, prev_cnt = ``0``, ` `      ``max_len;`   `  ``// Iterate till size of string` `  ``for` `(``int` `i = ``0``; i < S.length(); i++) ` `  ``{` `    ``// Check if current character is K` `    ``if` `(S.charAt(i) == K) ` `    ``{` `      ``curr_cnt += ``1``;` `    ``}` `    ``else` `    ``{` `      ``prev_cnt = Math.max(prev_cnt, ` `                          ``curr_cnt);` `      ``curr_cnt = ``0``;` `    ``}` `  ``}`   `  ``prev_cnt = Math.max(prev_cnt, ` `                      ``curr_cnt);`   `  ``// Assingning the max` `  ``// value to max_len` `  ``max_len = prev_cnt;`   `  ``return` `max_len;` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `  ``String S = ``"abcd1111aabc"``;` `  ``char` `K = ``'1'``;`   `  ``// Function call` `  ``System.out.print(length_substring(S, K));` `}` `}`   `// This code is contributed by Chitranayal`

## Python3

 `# Python3 program for the above approach`   `# Function to find the length of` `# longest sub-string having all` `# characters same as character K` `def` `length_substring(S, K):` `    `  `    ``# Initialize variables` `    ``curr_cnt ``=` `0` `    ``prev_cnt ``=` `0` `    ``max_len ``=` `0`   `    ``# Iterate till size of string` `    ``for` `i ``in` `range``(``len``(S)):`   `        ``# Check if current character is K` `        ``if` `(S[i] ``=``=` `K):` `            ``curr_cnt ``+``=` `1` `        ``else``:` `            ``prev_cnt ``=` `max``(prev_cnt, ` `                           ``curr_cnt)` `            ``curr_cnt ``=` `0`   `    ``prev_cnt ``=` `max``(prev_cnt, curr_cnt)`   `    ``# Assingning the max` `    ``# value to max_len` `    ``max_len ``=` `prev_cnt`   `    ``return` `max_len`   `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `    `  `    ``S ``=` `"abcd1111aabc"` `    ``K ``=` `'1'`   `    ``# Function call` `    ``print``(length_substring(S, K))`   `# This code is contributed by mohit kumar 29`

Output:

```4

```

Time Complexity: O(N)
Auxiliary Space: O(1)

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