# Find indices of all occurrence of one string in other

Given two strings, str1 and str2, the task is to print the indices(Consider, indices starting from 0) of occurrence of str2 in str1. If no such index occurs, print “NONE”.

Examples:

```Input : GeeksforGeeks
Geeks
Output : 0 8

Input : GFG
g
Output : NONE
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to check all substrings of given string one by one. If a substring matches, print its index.

 `// C++ program to find indices of all ` `// occurrences of one string in other. ` `#include ` `using` `namespace` `std; ` `void` `printIndex(string str, string s) ` `{ ` ` `  `    ``bool` `flag = ``false``; ` `    ``for` `(``int` `i = 0; i < str.length(); i++) { ` `        ``if` `(str.substr(i, s.length()) == s) { ` `            ``cout << i << ``" "``; ` `            ``flag = ``true``; ` `        ``} ` `    ``} ` ` `  `    ``if` `(flag == ``false``) ` `        ``cout << ``"NONE"``; ` `} ` `int` `main() ` `{ ` `    ``string str1 = ``"GeeksforGeeks"``; ` `    ``string str2 = ``"Geeks"``; ` `    ``printIndex(str1, str2); ` `    ``return` `0; ` `} `

 `// Java program to find indices of all ` `// occurrences of one String in other. ` `class` `GFG { ` ` `  `    ``static` `void` `printIndex(String str, String s) ` `    ``{ ` ` `  `        ``boolean` `flag = ``false``; ` `        ``for` `(``int` `i = ``0``; i < str.length() - s.length() + ``1``; i++) { ` `            ``if` `(str.substring(i, i + s.length()).equals(s)) { ` `                ``System.out.print(i + ``" "``); ` `                ``flag = ``true``; ` `            ``} ` `        ``} ` ` `  `        ``if` `(flag == ``false``) { ` `            ``System.out.println(``"NONE"``); ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``String str1 = ``"GeeksforGeeks"``; ` `        ``String str2 = ``"Geeks"``; ` `        ``printIndex(str1, str2); ` `    ``} ` `} ` ` `  `// This code is contributed by Rajput-JI `

 `# Python program to find indices of all ` `# occurrences of one String in other. ` `def` `printIndex(``str``, s): ` ` `  `    ``flag ``=` `False``; ` `    ``for` `i ``in` `range``(``len``(``str``)): ` `        ``if` `(``str``[i:i ``+` `len``(s)] ``=``=` `s): ` `             `  `            ``print``( i, end ``=``" "``); ` `            ``flag ``=` `True``; ` ` `  `    ``if` `(flag ``=``=` `False``): ` `        ``print``(``"NONE"``); ` `         `  `# Driver code         ` `str1 ``=` `"GeeksforGeeks"``; ` `str2 ``=` `"Geeks"``; ` `printIndex(str1, str2); ` ` `  `# This code contributed by PrinciRaj1992  ` ` `

 `// C# program to find indices of all ` `// occurrences of one String in other. ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``static` `void` `printIndex(String str, String s) ` `    ``{ ` ` `  `        ``bool` `flag = ``false``; ` `        ``for` `(``int` `i = 0; i < str.Length - s.Length + 1; i++) { ` `            ``if` `(str.Substring(i, ` `                              ``s.Length) ` `                    ``.Equals(s)) { ` `                ``Console.Write(i + ``" "``); ` `                ``flag = ``true``; ` `            ``} ` `        ``} ` ` `  `        ``if` `(flag == ``false``) { ` `            ``Console.WriteLine(``"NONE"``); ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``String str1 = ``"GeeksforGeeks"``; ` `        ``String str2 = ``"Geeks"``; ` `        ``printIndex(str1, str2); ` `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

 ` `

Output:
```0 8
```

Time Complexity : O(n * n)

An efficient solution is to KMP string matching algorithm.

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