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 conditionsstring 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 Codeint 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 approachimport java.util.*;class GFG{// Function to check whether a pair of// Strings exists satisfying the conditionsstatic 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 Codepublic 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 conditionsdef 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 Codeif __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 conditionsfunction 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 Inputlet 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 approachusing System;using System.Collections.Generic;public class GFG{// Function to check whether a pair of// Strings exists satisfying the conditionsstatic 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 Codepublic 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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