Check if a string can be split into 3 substrings such that one of them is a substring of the other two

Given a string S consisting of N lowercase alphabets, the task is to check if it is possible to split the string S into three non-empty substrings such that Y is the substring of the strings X and Z. If it is possible to split the string S then, print “Yes”. Otherwise, print “No”.

Examples:

Input: S = “geekseekforgeeks”
Output: Yes
Explanation:
The given string S = “geeksforgeeks” can be splitted as “geeks”, “eek”, and Z = “forgeeks”.
The string “eeks” is a substring of both the strings “geeks” and “forgeeks”.

Input: S = “naturalxws”
Output: No

Approach: The given problem can be solved by storing the frequency of unique characters and observe the fact that if there exists any character having frequency at least 3, then the string can be split into 3 substrings satisfying the given conditions. Follow the steps below to solve the problem:

• Initialize an array, say hash[], to store the frequency of characters in the string S.
• Iterate over the characters of the string and store the frequency of each character present in the string S in the array hash[].
• Traverse the array hash[] and if the frequency of any character is greater than 2, then print “Yes” and break out of the loop.
• After completing the above steps, if there exists no such character having a frequency at least 3, then print “No”.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to check if string S contains
// any character with frequency >= 3 or not
string freqCheck(string S, int N)
{
    // Stores frequency of characters
    int hash[26] = { 0 };

    // Iterate over the string
    for (int i = 0; i < N; i++) {

        // Update the frequency
        // of current character
        hash[S[i] - 'a']++;
    }

    // Iterate over the hash array
    for (int i = 0; i < 26; i++) {

        // If any character has
        // frequency >= 3
        if (hash[i] > 2) {
            return "Yes";
        }
    }

    // Otherwise
    return "No";
}

// Driver Code
int main()
{
    string S = "geekseekforgeeks";
    int N = S.length();
    cout << freqCheck(S, N);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG{

// Function to check if string S contains
// any character with frequency >= 3 or not
static String freqCheck(String S, int N)
{

    // Stores frequency of characters
    int hash[] = new int[26];

    // Iterate over the string
    for(int i = 0; i < N; i++)
    {

        // Update the frequency
        // of current character
        hash[S.charAt(i) - 'a']++;
    }

    // Iterate over the hash array
    for(int i = 0; i < 26; i++)
    {

        // If any character has
        // frequency >= 3
        if (hash[i] > 2)
        {
            return "Yes";
        }
    }

    // Otherwise
    return "No";
}

// Driver Code
public static void main(String[] args)
{
    String S = "geekseekforgeeks";
    int N = S.length();

    System.out.println(freqCheck(S, N));
}
}

// This code is contributed by Kingash

Python3

# Python3 program for the above approach

# Function to check if string S contains
# any character with frequency >= 3 or not
def freqCheck(S, N):

    # Stores frequency of characters
    hash = [0] * 26

    # Iterate over the string
    for i in range(N):

        # Update the frequency
        # of current character
        hash[ord(S[i]) - ord('a')] += 1

    # Iterate over the hash array
    for i in range(26):

        # If any character has
        # frequency >= 3
        if (hash[i] > 2):
            return "Yes"

    # Otherwise
    return "No"

# Driver Code
if __name__ == "__main__":

    S = "geekseekforgeeks"
    N = len(S)

    print(freqCheck(S, N))

# This code is contributed by ukasp

C#

// C# program for the above approach
using System;

class GFG{

// Function to check if string S contains
// any character with frequency >= 3 or not
static string freqCheck(string S, int N)
{

    // Stores frequency of characters
    int[] hash = new int[26];

    // Iterate over the string
    for(int i = 0; i < N; i++)
    {

        // Update the frequency
        // of current character
        hash[S[i] - 'a']++;
    }

    // Iterate over the hash array
    for(int i = 0; i < 26; i++)
    {

        // If any character has
        // frequency >= 3
        if (hash[i] > 2)
        {
            return "Yes";
        }
    }

    // Otherwise
    return "No";
}

// Driver code
static public void Main()
{
    string S = "geekseekforgeeks";
    int N = S.Length;

    Console.WriteLine(freqCheck(S, N));
}
}

// This code is contributed by offbeat

Javascript

<script>
    // Javascript program for the above approach

// Function to check if string S contains
// any character with frequency >= 3 or not
function freqCheck(S, N){

    // Stores frequency of characters
    let hash = new Array(26).fill(0)

    // Iterate over the string
    for(let i = 0; i < N; i++){

        // Update the frequency
        // of current character
        hash[S.charCodeAt(i) - 'a'.charCodeAt(0)] += 1
    }
    // Iterate over the hash array
    for(let i = 0; i < 26; i++){

        // If any character has
        // frequency >= 3
        if (hash[i] > 2){
            return "Yes"
        }
    }

    // Otherwise
    return "No"
}
// Driver Code

    let S = "geekseekforgeeks"
    let N = S.length

    document.write(freqCheck(S, N))

// This code is contributed by gfgking
</script>

Output:

Yes

Time Complexity: O(N)
Auxiliary Space: O(1)