# Maximum number of characters between any two same character in a string

Given a string, find the maximum number of characters between any two same character in the string. If no character repeats, print -1.
Examples:

```Input : str = "abba"
Output : 2
The maximum number of characters are
between two occurrences of 'a'.

Input : str = "baaabcddc"
Output : 3
The maximum number of characters are
between two occurrences of 'b'.

Input : str = "abc"
Output : -1
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Approach 1 (Simple): Use two nested loops. The outer loop picks character from left to right, the inner loop finds farthest occurrence and keeps track of maximum.

## C++

 `// Simple C++ program to find maximum number ` `// of characters between two occurrences of ` `// same character ` `#include ` `using` `namespace` `std; ` ` `  `int` `maximumChars(string& str) ` `{ ` `    ``int` `n = str.length(); ` `    ``int` `res = -1; ` `    ``for` `(``int` `i = 0; i < n - 1; i++) ` `        ``for` `(``int` `j = i + 1; j < n; j++) ` `            ``if` `(str[i] == str[j]) ` `                ``res = max(res, ``abs``(j - i - 1)); ` `    ``return` `res; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string str = ``"abba"``; ` `    ``cout << maximumChars(str); ` `    ``return` `0; ` `} `

## Java

 `// Simple Java program to find maximum ` `// number of characters between two ` `// occurrences of same character ` `class` `GFG { ` `     `  `    ``static` `int` `maximumChars(String str) ` `    ``{ ` `        ``int` `n = str.length(); ` `        ``int` `res = -``1``; ` `         `  `        ``for` `(``int` `i = ``0``; i < n - ``1``; i++) ` `            ``for` `(``int` `j = i + ``1``; j < n; j++) ` `                ``if` `(str.charAt(i) == str.charAt(j)) ` `                    ``res = Math.max(res,  ` `                         ``Math.abs(j - i - ``1``)); ` `                          `  `        ``return` `res; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``String str = ``"abba"``; ` `         `  `        ``System.out.println(maximumChars(str)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## Python3

 `# Simple Python3 program to find maximum number ` `# of characters between two occurrences of ` `# same character ` `def` `maximumChars(``str``): ` `    ``n ``=` `len``(``str``) ` `    ``res ``=` `-``1` `    ``for` `i ``in` `range``(``0``, n ``-` `1``): ` `        ``for` `j ``in` `range``(i ``+` `1``, n): ` `            ``if` `(``str``[i] ``=``=` `str``[j]): ` `                ``res ``=` `max``(res, ``abs``(j ``-` `i ``-` `1``)) ` `    ``return` `res ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``str` `=` `"abba"` `    ``print``(maximumChars(``str``)) ` ` `  `# This code is contributed by PrinciRaj1992  `

## C#

 `// Simple C# program to find maximum ` `// number of characters between two ` `// occurrences of same character ` `using` `System; ` ` `  `public` `class` `GFG { ` `     `  `    ``static` `int` `maximumChars(``string` `str) ` `    ``{ ` `        ``int` `n = str.Length; ` `        ``int` `res = -1; ` `         `  `        ``for` `(``int` `i = 0; i < n - 1; i++) ` `            ``for` `(``int` `j = i + 1; j < n; j++) ` `                ``if` `(str[i] == str[j]) ` `                    ``res = Math.Max(res,  ` `                         ``Math.Abs(j - i - 1)); ` `                          `  `        ``return` `res; ` `    ``} ` `     `  `    ``// Driver code ` `    ``static` `public` `void` `Main () ` `    ``{ ` `        ``string` `str = ``"abba"``; ` `         `  `        ``Console.WriteLine(maximumChars(str)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

Output:

```2
```

Time Complexity : O(n^2)

Approach 2 (Efficient) : Initialize an array”FIRST” of length 26 in which we have to store first occurrence of an alphabet in the string and another array “LAST” of length 26 in which we will store last occurrence of the alphabet in the string here index 0 is correspond to alphabet a, 1 for b and so on . After that we will take the difference of last and first arrays to find max difference if they are not at same position.

## C++

 `// Efficient C++ program to find maximum number ` `// of characters between two occurrences of ` `// same character ` `#include ` `using` `namespace` `std; ` ` `  `const` `int` `MAX_CHAR = 256; ` ` `  `int` `maximumChars(string& str) ` `{ ` `    ``int` `n = str.length(); ` `    ``int` `res = -1; ` ` `  `    ``// Initialize all indexes as -1. ` `    ``int` `firstInd[MAX_CHAR]; ` `    ``for` `(``int` `i = 0; i < MAX_CHAR; i++) ` `        ``firstInd[i] = -1; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``int` `first_ind = firstInd[str[i]]; ` ` `  `        ``// If this is first occurrence  ` `        ``if` `(first_ind == -1) ` `            ``firstInd[str[i]] = i; ` ` `  `        ``// Else find distance from previous ` `        ``// occurrence and update result (if ` `        ``// required).  ` `        ``else` `            ``res = max(res, ``abs``(i - first_ind - 1)); ` `    ``} ` `    ``return` `res; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string str = ``"abba"``; ` `    ``cout << maximumChars(str); ` `    ``return` `0; ` `} `

## Java

 `// Efficient java program to find maximum ` `// number of characters between two ` `// occurrences of same character ` `import` `java.io.*; ` ` `  `public` `class` `GFG { ` ` `  `    ``static` `int` `MAX_CHAR = ``256``; ` `     `  `    ``static` `int` `maximumChars(String str) ` `    ``{ ` `        ``int` `n = str.length(); ` `        ``int` `res = -``1``; ` `     `  `        ``// Initialize all indexes as -1. ` `        ``int` `[]firstInd = ``new` `int``[MAX_CHAR]; ` `         `  `        ``for` `(``int` `i = ``0``; i < MAX_CHAR; i++) ` `            ``firstInd[i] = -``1``; ` `     `  `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` `            ``int` `first_ind = firstInd[str.charAt(i)]; ` `     `  `            ``// If this is first occurrence  ` `            ``if` `(first_ind == -``1``) ` `                ``firstInd[str.charAt(i)] = i; ` `     `  `            ``// Else find distance from previous ` `            ``// occurrence and update result (if ` `            ``// required).  ` `            ``else` `                ``res = Math.max(res, Math.abs(i ` `                                  ``- first_ind - ``1``)); ` `        ``} ` `         `  `        ``return` `res; ` `    ``} ` `     `  `    ``// Driver code ` `     `  `    ``static` `public` `void` `main (String[] args) ` `    ``{ ` `        ``String str = ``"abba"``; ` `         `  `        ``System.out.println(maximumChars(str)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## Python3

 `# Efficient Python3 program to find maximum  ` `# number of characters between two occurrences  ` `# of same character ` ` `  `MAX_CHAR ``=` `256` ` `  `def` `maximumChars(str1): ` ` `  `    ``n ``=` `len``(str1) ` `    ``res ``=` `-``1` ` `  `    ``# Initialize all indexes as -1. ` `    ``firstInd ``=` `[``-``1` `for` `i ``in` `range``(MAX_CHAR)] ` ` `  `    ``for` `i ``in` `range``(n):  ` `        ``first_ind ``=` `firstInd[``ord``(str1[i])] ` ` `  `        ``# If this is first occurrence  ` `        ``if` `(first_ind ``=``=` `-``1``): ` `            ``firstInd[``ord``(str1[i])] ``=` `i ` ` `  `        ``# Else find distance from previous ` `        ``# occurrence and update result (if ` `        ``# required).  ` `        ``else``: ` `            ``res ``=` `max``(res, ``abs``(i ``-` `first_ind ``-` `1``)) ` `     `  `    ``return` `res ` ` `  `# Driver code ` `str1 ``=` `"abba"` `print``(maximumChars(str1)) ` ` `  `# This code is contributed by Mohit kumar 29 `

## C#

 `// Efficient C# program to find maximum ` `// number of characters between two ` `// occurrences of same character ` `using` `System; ` ` `  `public` `class` `GFG { ` ` `  `    ``static` `int` `MAX_CHAR = 256; ` `     `  `    ``static` `int` `maximumChars(``string` `str) ` `    ``{ ` `        ``int` `n = str.Length; ` `        ``int` `res = -1; ` `     `  `        ``// Initialize all indexes as -1. ` `        ``int` `[]firstInd = ``new` `int``[MAX_CHAR]; ` `         `  `        ``for` `(``int` `i = 0; i < MAX_CHAR; i++) ` `            ``firstInd[i] = -1; ` `     `  `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``int` `first_ind = firstInd[str[i]]; ` `     `  `            ``// If this is first occurrence  ` `            ``if` `(first_ind == -1) ` `                ``firstInd[str[i]] = i; ` `     `  `            ``// Else find distance from previous ` `            ``// occurrence and update result (if ` `            ``// required).  ` `            ``else` `                ``res = Math.Max(res, Math.Abs(i  ` `                              ``- first_ind - 1)); ` `        ``} ` `         `  `        ``return` `res; ` `    ``} ` `     `  `    ``// Driver code ` ` `  `    ``static` `public` `void` `Main () ` `    ``{ ` `        ``string` `str = ``"abba"``; ` `         `  `        ``Console.WriteLine(maximumChars(str)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

Output:

```2
```

Time Complexity : O(n)

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