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Find indices of all occurrence of one string in other

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  • Difficulty Level : Medium
  • Last Updated : 30 Nov, 2022
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Given two strings, str1 and str2, the task is to print the indices(Consider, indices starting from 0) of the occurrence of str2 in str1. If no such index occurs, print “NONE”.

Examples:

Input : GeeksforGeeks
        Geeks
Output : 0 8

Input : GFG
        g
Output : NONE

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

Implementation:

C++




// C++ program to find indices of all
// occurrences of one string in other.
#include <iostream>
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




// 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


Python3




# 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#




// 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


PHP




<?php
// PHP program to find indices of all
// occurrences of one string in other.
function printIndex($str, $s)
{
    $flag = false;
    for ($i = 0; $i < strlen($str); $i++)
    {
        if (substr($str,$i, strlen($s)) == $s)
        {
            echo $i . " ";
            $flag = true;
        }
    }
 
    if ($flag == false)
        echo "NONE";
}
 
// Driver Code
$str1 = "GeeksforGeeks";
$str2 = "Geeks";
printIndex($str1, $str2);
 
// This code is contributed by mits
?>


Javascript




<script>
      // JavaScript program to find indices of all
      // occurrences of one String in other.
      function printIndex(str, s) {
        var flag = false;
        for (var i = 0; i < str.length - s.length + 1; i++) {
          if (str.substring(i, s.length + i) == s) {
            document.write(i + " ");
            flag = true;
          }
        }
 
        if (flag === false) {
          document.write("NONE");
        }
      }
 
      // Driver code
      var str1 = "GeeksforGeeks";
      var str2 = "Geeks";
      printIndex(str1, str2);
</script>


Output

0 8 

Time Complexity: O(n * n)
Auxiliary Space: O(1), no extra space is required, so it is a constant.

An efficient solution is to KMP string matching algorith


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