Given a string S, the task is to check whether the string can be typed using only a single row of the qwerty keyboard.
Examples:
Input: S = “Dad”
Output: Yes
Explanation:
Characters “D” and “a” are present in the same row of qwerty keypad. That is second Row.Input: S = “Mom”
Output: No
Explanation:
Characters “M” and “o” are not present in the same row of qwerty keypad.
Approach: The idea is to store the characters of the same row of the qwerty keypad into different hash-maps to check that all the characters of the string are from the same row.
Below is the implementation of the above approach:
C++
// C++ Program to check whether// the string can be printed// using same row of qwerty keypad#include <bits/stdc++.h>using namespace std;// Function to find the row of the// character in the qwerty keypadint checkQwertyRow(char x){ // Sets to include the characters // from the same row of the qwerty keypad set<char> first_row = { '1', '2', '3', '4', '5', '6', '7', '8', '9', '0', '-', '=' }; set<char> second_row = { 'Q', 'W', 'E', 'R', 'T', 'Y', 'U', 'I', 'O', 'P', '[', ']', 'q', 'w', 'e', 'r', 't', 'y', 'u', 'i', 'o', 'p' }; set<char> third_row = { 'A', 'S', 'D', 'F', 'G', 'H', 'J', 'K', 'L', ';', ':', 'a', 's', 'd', 'f', 'g', 'h', 'j', 'k', 'l' }; set<char> fourth_row = { 'Z', 'X', 'C', 'V', 'B', 'N', 'M', ',', '.', '/', 'z', 'x', 'c', 'v', 'b', 'n', 'm' }; // Condition to check the row of the // current character of the string if (first_row.count(x) > 0) { return 1; } else if (second_row.count(x) > 0) { return 2; } else if (third_row.count(x) > 0) { return 3; } else if (fourth_row.count(x) > 0) { return 4; } return 0;}// Function to check the characters are// from the same row of the qwerty keypadbool checkValidity(string str){ char x = str[0]; int row = checkQwertyRow(x); for (int i = 0; i < str.length(); i++) { x = str[i]; if (row != checkQwertyRow(x)) { return false; } } return true;}// Driver Codeint main(){ string str = "neveropen"; if (checkValidity(str)) cout << "Yes"; else cout << "No"; return (0);} |
Java
// Java program to check whether// the string can be printed// using same row of qwerty keypadimport java.util.*;class GFG{// Function to find the row of the// character in the qwerty keypadstatic int checkQwertyRow(char x){ // Sets to include the characters // from the same row of the qwerty keypad Character[] first_row1 = { '1', '2', '3', '4', '5', '6', '7', '8', '9', '0', '-', '=' }; Set<Character> first_row = new HashSet<>( Arrays.asList(first_row1)); Character[] second_row1 = { 'Q', 'W', 'E', 'R', 'T', 'Y', 'U', 'I', 'O', 'P', '[', ']', 'q', 'w', 'e', 'r', 't', 'y', 'u', 'i', 'o', 'p' }; Set<Character> second_row = new HashSet<>( Arrays.asList(second_row1)); Character[] third_row1 = { 'A', 'S', 'D', 'F', 'G', 'H', 'J', 'K', 'L', ';', ':', 'a', 's', 'd', 'f', 'g', 'h', 'j', 'k', 'l' }; Set<Character> third_row = new HashSet<>( Arrays.asList(third_row1)); Character[] fourth_row1 = { 'Z', 'X', 'C', 'V', 'B', 'N', 'M', ',', '.', '/', 'z', 'x', 'c', 'v', 'b', 'n', 'm' }; Set<Character> fourth_row = new HashSet<>( Arrays.asList(fourth_row1)); // Condition to check the row of the // current character of the string if (first_row.contains(x)) { return 1; } else if (second_row.contains(x)) { return 2; } else if (third_row.contains(x)) { return 3; } else if (fourth_row.contains(x)) { return 4; } return 0;}// Function to check the characters are// from the same row of the qwerty keypadstatic boolean checkValidity(String str){ char x = str.charAt(0); int row = checkQwertyRow(x); for(int i = 0; i < str.length(); i++) { x = str.charAt(i); if (row != checkQwertyRow(x)) { return false; } } return true;}// Driver codepublic static void main(String[] args){ String str = "neveropen"; if (checkValidity(str)) System.out.println("Yes"); else System.out.println("No");}}// This code is contributed by offbeat |
Python3
# Python3 program to check whether # the string can be printed # using same row of qwerty keypad# Function to find the row of the # character in the qwerty keypad def checkQwertyRow(x): # Sets to include the # characters from the # same row of the qwerty keypad first_row = ['1', '2', '3', '4', '5', '6', '7', '8', '9', '0', '-', '='] second_row = ['Q', 'W', 'E', 'R', 'T', 'Y', 'U', 'I', 'O', 'P', '[', ']', 'q', 'w', 'e', 'r', 't', 'y', 'u', 'i', 'o', 'p'] third_row = ['A', 'S', 'D', 'F', 'G', 'H', 'J', 'K', 'L', ';', ':', 'a', 's', 'd', 'f', 'g', 'h', 'j', 'k', 'l'] fourth_row = ['Z', 'X', 'C', 'V', 'B', 'N', 'M', ',', '.', '/', 'z', 'x', 'c', 'v', 'b', 'n', 'm'] # Condition to check the # row of the current character # of the string if(first_row.count(x) > 0): return 1 elif(second_row.count(x) > 0): return 2 elif(third_row.count(x) > 0): return 3 elif(fourth_row.count(x) > 0): return 4 return 0# Function to check the # characters are from the # same row of the qwerty keypaddef checkValidity(str): x = str[0] row = checkQwertyRow(x) for i in range(len(str)): x = str[i] if(row != checkQwertyRow(x)): return False return True# Driver Code str = "neveropen"if(checkValidity(str)): print("Yes")else: print("No")# This code is contributed by avanitrachhadiya2155 |
C#
// C# program to check whether// the string can be printed// using same row of qwerty keypadusing System;using System.Collections.Generic;class GFG{// Function to find the row of the// character in the qwerty keypadstatic int checkQwertyRow(char x){ // Sets to include the characters // from the same row of the qwerty keypad char[] first_row1 = { '1', '2', '3', '4', '5', '6', '7', '8', '9', '0', '-', '=' }; HashSet<char> first_row = new HashSet<char>( first_row1); char[] second_row1 = { 'Q', 'W', 'E', 'R', 'T', 'Y', 'U', 'I', 'O', 'P', '[', ']', 'q', 'w', 'e', 'r', 't', 'y', 'u', 'i', 'o', 'p' }; HashSet<char> second_row = new HashSet<char>( second_row1); char[] third_row1 = { 'A', 'S', 'D', 'F', 'G', 'H', 'J', 'K', 'L', ';', ':', 'a', 's', 'd', 'f', 'g', 'h', 'j', 'k', 'l' }; HashSet<char> third_row = new HashSet<char>( third_row1); char[] fourth_row1 = { 'Z', 'X', 'C', 'V', 'B', 'N', 'M', ',', '.', '/', 'z', 'x', 'c', 'v', 'b', 'n', 'm' }; HashSet<char> fourth_row = new HashSet<char>( fourth_row1); // Condition to check the row of the // current character of the string if (first_row.Contains(x)) { return 1; } else if (second_row.Contains(x)) { return 2; } else if (third_row.Contains(x)) { return 3; } else if (fourth_row.Contains(x)) { return 4; } return 0;}// Function to check the characters are// from the same row of the qwerty keypadstatic bool checkValidity(String str){ char x = str[0]; int row = checkQwertyRow(x); for(int i = 0; i < str.Length; i++) { x = str[i]; if (row != checkQwertyRow(x)) { return false; } } return true;}// Driver codepublic static void Main(String[] args){ String str = "neveropen"; if (checkValidity(str)) Console.WriteLine("Yes"); else Console.WriteLine("No");}}// This code is contributed by Amit Katiyar |
Javascript
<script> // JavaScript program to check whether // the string can be printed // using same row of qwerty keypad // Function to find the row of the // character in the qwerty keypad function checkQwertyRow(x) { // Sets to include the characters // from the same row of the qwerty keypad var first_row = [ "1", "2", "3", "4", "5", "6", "7", "8", "9", "0", "-", "=", ]; var second_row = [ "Q", "W", "E", "R", "T", "Y", "U", "I", "O", "P", "[", "]", "q", "w", "e", "r", "t", "y", "u", "i", "o", "p", ]; var third_row = [ "A", "S", "D", "F", "G", "H", "J", "K", "L", ";", ":", "a", "s", "d", "f", "g", "h", "j", "k", "l", ]; var fourth_row = [ "Z", "X", "C", "V", "B", "N", "M", ",", ".", "/", "z", "x", "c", "v", "b", "n", "m", ]; // Condition to check the row of the // current character of the string if (first_row.includes(x)) { return 1; } else if (second_row.includes(x)) { return 2; } else if (third_row.includes(x)) { return 3; } else if (fourth_row.includes(x)) { return 4; } return 0; } // Function to check the characters are // from the same row of the qwerty keypad function checkValidity(str) { var x = str[0]; var row = checkQwertyRow(x); for (var i = 0; i < str.length; i++) { x = str[i]; if (row !== checkQwertyRow(x)) { return false; } } return true; } // Driver code var str = "neveropen"; if (checkValidity(str)) document.write("Yes"); else document.write("No"); </script> |
Output:
No
Time Complexity: O(N), where N is the length of the given string.|
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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