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Check if a pair of strings exists that starts with and without the character K or not

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  • Last Updated : 25 Jan, 2022
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Given an array arr[] consisting of N strings of lowercase characters and a character K such that any string may start with the character K, the task is to check if there exists any pair of strings that are starting and not starting (‘!’) with the character K. If found to be true, then print “Yes“. Otherwise, print “No“.

Examples:

Input: arr[] = {“a”, “!a”, “b”, “!c”, “d”, “!d”}, K = ‘!’
Output: Yes
Explanation:
There exists valid pairs of the strings are {(“a”, “!a”), (“!d”, “d”)}.

Input: arr[] = {“red”, “red”, “red”, “!orange”, “yellow”, “!blue”, “cyan”, “!green”, “brown”, “!gray”}, K = ‘!’
Output: No

Naive Approach: The simplest approach to solve the given problem is to find all possible pairs from the array and check if the strings pair satisfy the given condition or not.

Time Complexity: O(N2*M), where M is the maximum length of the string in the given array arr[].
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be solved by using dictionary. Follow the steps below to solve the problem:

  • Initialize a dictionary, say, visited to store the previously visited strings.
  • Iterate over the list arr[] and in each iteration, if the starting character of the current string is the character K then check for string without the character K in visited otherwise, check for the string with the character K in visited. If the string is found then return “Yes“.
  • In each iteration, add the string S into the map visited.
  • After completing the above steps, print “No” if the above conditions are not satisfied.

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 whether a pair of
// strings exists satisfying the conditions
string checkhappy(vector<string> arr, char K, int N)
{
   
    // Stores the visited strings
    set<string> visited;
 
    // Iterate over the array arr[]
    for (string s : arr) {
 
        // If first character of current
           // string is K
        if(s[0] == K)
            if (visited.find(s.substr(1)) != visited.end())
                return "Yes";
               
        // Otherwise
        else
            if (visited.find((K + s)) != visited.end())
                return "Yes";
         
        // Adding to the visited
        visited.insert(s);
      }
 
    return "No";
}
 
// Driver Code
int main() {
 
    // Given Input
    vector<string> arr = {"a", "! a", "b", "! c", "d", "! d"};
    char K = '!';
    int N = arr.size();
     
    cout << checkhappy(arr, K, N) << endl;
                   
    return 0;
}
 
// This code is contributed Dharanendra L V.


Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to check whether a pair of
// Strings exists satisfying the conditions
static String checkhappy(String[] arr, char K, int N)
{
   
    // Stores the visited Strings
    HashSet<String> visited = new HashSet<String> ();
 
    // Iterate over the array arr[]
    for (String s : arr) {
 
        // If first character of current
           // String is K
        if(s.charAt(0) == K)
            if (visited.contains(s.substring(1)))
                return "Yes";
               
        // Otherwise
        else
            if (visited.contains((K + s)))
                return "Yes";
         
        // Adding to the visited
        visited.add(s);
      }
 
    return "No";
}
 
// Driver Code
public static void main(String[] args) {
 
    // Given Input
    String[] arr = {"a", "! a", "b", "! c", "d", "! d"};
    char K = '!';
    int N = arr.length;
     
    System.out.print(checkhappy(arr, K, N) +"\n");
                   
}
}
 
// This code is contributed by shikhasingrajput


Python3




# Python program for the above approach
 
# Function to check whether a pair of
# strings exists satisfying the conditions
def checkhappy(arr, K, N):
   
    # Stores the visited strings
    visited = set()
 
    # Iterate over the array arr[]
    for s in arr:
 
        # If first character of current
        # string is K
        if(s[0] == K):
            if s[1:] in visited:
                return 'Yes'
               
        # Otherwise
        else:
            if (K + s) in visited:
                return 'Yes'
         
        # Adding to the visited
        visited.add(s)
 
    return "No"
 
 
# Driver Code
if __name__ == '__main__':
     
    # Given Input
    arr = ['a', '! a', 'b', '! c', 'd', '! d']
    K = '!'
    N = len(arr)
     
    print(checkhappy(arr, K, N))


Javascript




<script>
 
// Javascript program for the above approach
 
// Function to check whether a pair of
// strings exists satisfying the conditions
function checkhappy(arr, K, N)
{
     
    // Stores the visited strings
    let visited = new Set();
     
    // Iterate over the array arr[]
    for(let s of arr)
    {
         
        // If first character of current
        // string is K
        if (s[0] == K)
        {
            if (visited.has(s.slice(1)))
                return "Yes";
        }
         
        // Otherwise
        else
        {
            if (visited.has(K + s))
                return "Yes";
        }
         
        // Adding to the visited
        visited.add(s);
    }
    return "No";
}
 
// Driver Code
 
// Given Input
let arr = [ "a", "! a", "b", "! c",
            "d", "! d" ];
let K = "!";
let N = arr.length;
 
document.write(checkhappy(arr, K, N));
 
// This code is contributed by gfgking
 
</script>


C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
public class GFG{
 
// Function to check whether a pair of
// Strings exists satisfying the conditions
static String checkhappy(String[] arr, char K, int N)
{
   
    // Stores the visited Strings
    HashSet<String> visited = new HashSet<String> ();
 
    // Iterate over the array []arr
    foreach (String s in arr) {
 
        // If first character of current
           // String is K
        if(s[0] == K)
            if (visited.Contains(s.Substring(1)))
                return "Yes";
               
        // Otherwise
        else
            if (visited.Contains((K + s)))
                return "Yes";
         
        // Adding to the visited
        visited.Add(s);
      }
 
    return "No";
}
 
// Driver Code
public static void Main(String[] args) {
 
    // Given Input
    String[] arr = {"a", "! a", "b", "! c", "d", "! d"};
    char K = '!';
    int N = arr.Length;
     
    Console.Write(checkhappy(arr, K, N) +"\n");
                   
}
}
 
  
 
// This code contributed by shikhasingrajput


Output: 

No

 

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


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