Given a string S consisting of only lowercase English letters. The task is to find the minimum number of times of finger moves to type the given string. The move is considered when you press the key which is not in the row to currently pressed key on the keyboard.
QWERTY keyboard consisting of only English letters.
Examples:
Input: S = “neveropen”
Output: 7
Explanation:
move 1 >> g
move 2 >> e
move 2 >> e
move 3 >> k
move 3 >> s
move 3 >> f
move 4 >> o
move 4 >> r
move 5 >> g
move 6 >> e
move 6 >> e
move 7 >> k
move 7 >> sInput: S = “radhamohan”
Output: 6
Approach: This can be done by initially setting the row number of each character in the QWERTY keyboard. Below are the steps:
- Store the row of each character in an array row[].
- Initialize move = 1.
- Traverse the given string and check if the row of the current character in the string is equal to the previous character or not.
- If current character is not equal then increment the move as we need to change the current row while printing the character.
- Else check for the next pairs of characters.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function that calculates the moves// required to print the current stringint numberMoves(string s){ // row1 has qwertyuiop, row2 has // asdfghjkl, and row3 has zxcvbnm // Store the row number of // each character int row[] = { 2, 3, 3, 2, 1, 2, 2, 2, 1, 2, 2, 2, 3, 3, 1, 1, 1, 1, 2, 1, 1, 3, 1, 3, 1, 3 }; // String length int n = s.length(); // Initialise move to 1 int move = 1; // Traverse the string for (int i = 1; i < n; i++) { // If current row number is // not equal to previous row // then increment the moves if (row[s[i] - 'a'] != row[s[i - 1] - 'a']) { move++; } } // Return the moves return move;}// Driver Codeint main(){ // Given String str string str = "neveropen"; // Function Call cout << numberMoves(str); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function that calculates the moves// required to print the current Stringstatic int numberMoves(String s){ // row1 has qwertyuiop, row2 has // asdfghjkl, and row3 has zxcvbnm // Store the row number of // each character int row[] = { 2, 3, 3, 2, 1, 2, 2, 2, 1, 2, 2, 2, 3, 3, 1, 1, 1, 1, 2, 1, 1, 3, 1, 3, 1, 3 }; // String length int n = s.length(); // Initialise move to 1 int move = 1; // Traverse the String for (int i = 1; i < n; i++) { // If current row number is // not equal to previous row // then increment the moves if (row[s.charAt(i) - 'a'] != row[s.charAt(i-1) - 'a']) { move++; } } // Return the moves return move;}// Driver Codepublic static void main(String[] args){ // Given String str String str = "neveropen"; // Function Call System.out.print(numberMoves(str));}}// This code is contributed by sapnasingh4991 |
Python3
# Python3 program for the above approach# Function that calculates the moves# required to print current Stringdef numberMoves(s): # row1 has qwertyuiop, row2 has # asdfghjkl, and row3 has zxcvbnm # Store the row number of # each character row = [2, 3, 3, 2, 1, 2, 2, 2, 1, 2, 2, 2, 3, 3, 1, 1, 1, 1, 2, 1, 1, 3, 1, 3, 1, 3]; # String length n = len(s); # Initialise move to 1 move = 1; # Traverse the String for i in range(1, n): # If current row number is # not equal to previous row # then increment the moves if(row[ord(s[i]) - ord('a')] != row[ord(s[i - 1]) - ord('a')]): move += 1; # Return the moves return move;# Driver Codeif __name__ == '__main__': # Given String str str = "neveropen"; # Function Call print(numberMoves(str));# This code is contributed by Rajput-Ji Add |
C#
// C# program for the above approachusing System;class GFG{// Function that calculates the moves// required to print the current Stringstatic int numberMoves(String s){ // row1 has qwertyuiop, row2 has // asdfghjkl, and row3 has zxcvbnm // Store the row number of // each character int []row = { 2, 3, 3, 2, 1, 2, 2, 2, 1, 2, 2, 2, 3, 3, 1, 1, 1, 1, 2, 1, 1, 3, 1, 3, 1, 3 }; // String length int n = s.Length; // Initialise move to 1 int move = 1; // Traverse the String for (int i = 1; i < n; i++) { // If current row number is // not equal to previous row // then increment the moves if (row[s[i] - 'a'] != row[s[i - 1] - 'a']) { move++; } } // Return the moves return move;}// Driver Codepublic static void Main(String[] args){ // Given String str String str = "neveropen"; // Function Call Console.Write(numberMoves(str));}}// This code is contributed by sapnasingh4991 |
Javascript
// Javascript program for the above approach// Function that calculates the moves// required to print the current stringfunction numberMoves(s){ // row1 has qwertyuiop, row2 has // asdfghjkl, and row3 has zxcvbnm // Store the row number of // each character var row = [ 2, 3, 3, 2, 1, 2, 2, 2, 1, 2, 2, 2, 3, 3, 1, 1, 1, 1, 2, 1, 1, 3, 1, 3, 1, 3 ]; // String length var n = s.length; // Initialise move to 1 var move = 1; // Traverse the string for (var i = 1; i < n; i++) { // If current row number is // not equal to previous row // then increment the moves if (row[s[i].charCodeAt(0) - 'a'.charCodeAt(0)] != row[s[i - 1].charCodeAt(0) - 'a'.charCodeAt(0)]) { move++; } } // Return the moves return move;}// Driver Code // Given String str var str = "neveropen"; // Function Call console.log(numberMoves(str)); // This code is contributed by Abhijeet Kumar(abhijeet19403) |
Output:
7
Time Complexity: O(N), where N is the length of the string.
Space Complexity: O(1)
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