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

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

Time Complexity: O(n), for using indexof() function which takes a linear time.
Auxiliary Space: O(n), where n is the length of the given string.