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Check if a string contains two non overlapping sub-strings “geek” and “keeg”

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  • Last Updated : 21 May, 2021

Given a string str, the task is to check whether the string contains two non-overlapping sub-strings s1 = “geek” and s2 = “keeg” such that s2 starts after s1 ends.
Examples: 
 

Input: str = “geekeekeeg” 
Output: Yes 
“geek” and “keeg” both are present in the 
given string without overlapping.
Input: str = “geekeeg” 
Output: No 
“geek” and “keeg” both are present but they overlap. 
 

 

Approach: Check if the sub-string “geek” occurs before “keeg” in the given string. This problem is simpler when we use a predefined function strstr in order to find the occurrence of a sub-string in the given string.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true
// if s contains two non overlapping
// sub strings "geek" and "keeg"
bool isValid(char s[])
{
    char* p;
 
    // If "geek" and "keeg" are both present
    // in s without over-lapping and "keeg"
    // starts after "geek" ends
    if ((p = strstr(s, "geek")) && (strstr(p + 4, "keeg")))
        return true;
 
    return false;
}
 
// Driver code
int main()
{
    char s[] = "geekeekeeg";
 
    if (isValid(s))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
 
// Function that returns true
// if s contains two non overlapping
// sub Strings "geek" and "keeg"
static boolean isValid(String s)
{
    // If "geek" and "keeg" are both present
    // in s without over-lapping and "keeg"
    // starts after "geek" ends
    if ((s.indexOf( "geek")!=-1) &&
        (s.indexOf( "keeg",s.indexOf( "geek") + 4)!=-1))
        return true;
 
    return false;
}
 
// Driver code
public static void main(String args[])
{
    String s = "geekeekeeg";
 
    if (isValid(s))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by Arnab Kundu

Python3




# Python 3 implementation of the approach
 
# Function that returns true
# if s contains two non overlapping
# sub strings "geek" and "keeg"
def isValid(s):
    p=""
 
    # If "geek" and "keeg" are both present
    # in s without over-lapping and "keeg"
    # starts after "geek" ends
    p=s.find("geek")
    if (s.find("keeg",p+4)):
        return True
 
    return False
 
# Driver code
if __name__ == "__main__":
    s = "geekeekeeg"
 
    if (isValid(s)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by ChitraNayal

C#




// C# implementation of the approach
 
using System;
 
class GFG
{
 
// Function that returns true
// if s contains two non overlapping
// sub Strings "geek" and "keeg"
static bool isValid(string s)
{
    // If "geek" and "keeg" are both present
    // in s without over-lapping and "keeg"
    // starts after "geek" ends
    if ((s.IndexOf( "geek")!=-1) &&
        (s.IndexOf( "keeg",s.IndexOf( "geek") + 4)!=-1))
        return true;
 
    return false;
}
 
// Driver code
public static void Main()
{
    string s = "geekeekeeg";
 
    if (isValid(s))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by AnkitRai01

Javascript




<script>
 
// JavaScript implementation of the approach
 
// Function that returns true
// if s contains two non overlapping
// sub Strings "geek" and "keeg"
function isValid(s)
{
    // If "geek" and "keeg" are both present
    // in s without over-lapping and "keeg"
    // starts after "geek" ends
    if ((s.indexOf("geek") != -1) &&
        (s.indexOf("keeg", s.indexOf("geek") + 4) != -1))
        return true;
 
    return false;
}
 
// Driver Code
var s = "geekeekeeg";
 
if (isValid(s))
    document.write("Yes");
else
    document.write("No");
    
// This code is contributed by Khushboogoyal499
 
</script>

Output: 

Yes

 


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