Given two strings X and Y both consisting of N uppercase alphabets and an integer K, the task is to find the minimum count of cyclic increments of any character in the string X by 1 or K required to convert the string X into string Y.
Cyclic Increments: Incrementing character āZā by 1 or K starts from the character āAā.
Examples:
Input: X = āABCTā, Y = āPBDIā, K = 13
Output: 7
Explanation:Ā
Convert the character A to P. Increments required are 13 (A to N), 1 (N to O), 1 (O to P). Therefore, the total count is 3.
Character B remains unchanged. Therefore, the total count remains 3.
Convert the character C to D. Increments required are 1 (C to D). Therefore, the total count is 4.
Convert the character T to I. Increments required is of 13 (T to G), 1 (G to H), 1 (H to I). Therefore, the total count is 7.Input: X = āABCDā, Y = āABCDā, K = 6
Output: 0
Approach: Follow the below steps to solve this problem:
- Initialize a variable, say count, to store the minimum increment required to convert string X into string Y.
- Traverse the string X and for every character in string X, there are three following cases:
- When characters in both the strings X and Y are the same then continue.
- When characters in the string Y is greater than X string then add the value (Y[i] ā X[i])/K and (Y[i] ā X[i]) modulo K to the count.
- When characters in string X is greater than string Y then subtract the character X[i] from 90 and subtract 64 from Y[i] to calculate the cyclic difference and then add the difference modulo K and difference/K to the count.
- After completing the above steps, print count as the minimum number of increments required.
Below is the implementation of the above approach:
C++
// CPP program for the above approach#include<bits/stdc++.h>using namespace std;Ā
// Function to count minimum increments// by 1 or K required to convert X to Yvoid countOperations(Ā Ā string X, string Y, int K){Ā
Ā Ā int count = 0;Ā
Ā Ā // Traverse the string XĀ Ā for (int i = 0; i < X.length(); i++) Ā Ā {Ā
Ā Ā Ā Ā int c = 0;Ā
Ā Ā Ā Ā // Case 1Ā Ā Ā Ā if (X[i] == Y[i])Ā Ā Ā Ā Ā Ā continue;Ā
Ā Ā Ā Ā // Case 2Ā Ā Ā Ā else if (X[i] < Y[i])Ā Ā Ā Ā {Ā
Ā Ā Ā Ā Ā Ā // Add the difference/K toĀ Ā Ā Ā Ā Ā // the countĀ Ā Ā Ā Ā Ā if ((Y[i] - X[i]) >= K) {Ā Ā Ā Ā Ā Ā Ā Ā c = (Y[i] - X[i]) / K; Ā Ā Ā Ā Ā Ā }Ā
Ā Ā Ā Ā Ā Ā // Add the difference%K toĀ Ā Ā Ā Ā Ā // the countĀ Ā Ā Ā Ā Ā c += (Y[i] - X[i]) % K;Ā Ā Ā Ā }Ā
Ā Ā Ā Ā // Case 3Ā Ā Ā Ā else {Ā Ā Ā Ā Ā Ā int t = 90 - X[i];Ā Ā Ā Ā Ā Ā t += Y[i] - 65 + 1;Ā
Ā Ā Ā Ā Ā Ā // Add the difference/K toĀ Ā Ā Ā Ā Ā // the countĀ Ā Ā Ā Ā Ā if (t >= K)Ā Ā Ā Ā Ā Ā Ā Ā c = t / K;Ā
Ā Ā Ā Ā Ā Ā // Add the difference%K toĀ Ā Ā Ā Ā Ā // the countĀ Ā Ā Ā Ā Ā c += (t % K);Ā Ā Ā Ā }Ā Ā Ā Ā count += c;Ā Ā }Ā
Ā Ā // Print the answerĀ Ā cout << count << endl;}Ā
// Driver Codeint main(){Ā Ā string X = "ABCT", Y = "PBDI";Ā Ā int K = 6;Ā Ā countOperations(X, Y, K);}Ā
// This code is contributed by ipg2016107. |
Java
// Java program for the above approachĀ
import java.io.*;Ā
class GFG {Ā
Ā Ā Ā Ā // Function to count minimum incrementsĀ Ā Ā Ā // by 1 or K required to convert X to YĀ Ā Ā Ā public static void countOperations(Ā Ā Ā Ā Ā Ā Ā Ā String X, String Y, int K)Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā // Convert String to character arrayĀ Ā Ā Ā Ā Ā Ā Ā char ch1[] = X.toCharArray();Ā Ā Ā Ā Ā Ā Ā Ā char ch2[] = Y.toCharArray();Ā Ā Ā Ā Ā Ā Ā Ā int count = 0;Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Traverse the string XĀ Ā Ā Ā Ā Ā Ā Ā for (int i = 0; i < X.length(); i++) {Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā int c = 0;Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Case 1Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (ch1[i] == ch2[i])Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā continue;Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Case 2Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā else if (ch1[i] < ch2[i]) {Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Add the difference/K toĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // the countĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (((int)ch2[i]Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā - (int)ch1[i])Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā >= K) {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā c = ((int)ch2[i]Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā - (int)ch1[i])Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā / K;Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Add the difference%K toĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // the countĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā c += ((int)ch2[i]Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā - (int)ch1[i])Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā % K;Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Case 3Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā else {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā int t = 90 - (int)ch1[i];Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā t += (int)ch2[i] - 65 + 1;Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Add the difference/K toĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // the countĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (t >= K)Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā c = t / K;Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Add the difference%K toĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // the countĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā c += (t % K);Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā count += c;Ā Ā Ā Ā Ā Ā Ā Ā }Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Print the answerĀ Ā Ā Ā Ā Ā Ā Ā System.out.print(count);Ā Ā Ā Ā }Ā
Ā Ā Ā Ā // Driver CodeĀ Ā Ā Ā public static void main(String[] args)Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā String X = "ABCT", Y = "PBDI";Ā Ā Ā Ā Ā Ā Ā Ā int K = 6;Ā Ā Ā Ā Ā Ā Ā Ā countOperations(X, Y, K);Ā Ā Ā Ā }} |
Python3
# Python program for the above approachĀ
# Function to count minimum increments# by 1 or K required to convert X to Ydef countOperations(X, Y, K):Ā Ā Ā Ā count = 0Ā Ā Ā Ā Ā Ā Ā Ā Ā # Traverse the string XĀ Ā Ā Ā for i in range(len(X)):Ā Ā Ā Ā Ā Ā Ā Ā c = 0Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # Case 1Ā Ā Ā Ā Ā Ā Ā Ā if (X[i] == Y[i]):Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā continueĀ
Ā Ā Ā Ā Ā Ā Ā Ā # Case 2Ā Ā Ā Ā Ā Ā Ā Ā elif (X[i] < Y[i]):Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # Add the difference/K toĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # the countĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if ((ord(Y[i]) - ord(X[i])) >= K):Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā c = (ord(Y[i]) - ord(X[i])) // KĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # Add the difference%K toĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # the countĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā c += (ord(Y[i]) - ord(X[i])) % KĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # Case 3Ā Ā Ā Ā Ā Ā Ā Ā else:Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā t = 90 - ord(X[i])Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā t += ord(Y[i]) - 65 + 1Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # Add the difference/K toĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # the countĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (t >= K):Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā c = t // KĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # Add the difference%K toĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # the countĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā c += (t % K)Ā Ā Ā Ā Ā Ā Ā Ā count += cĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # Print the answerĀ Ā Ā Ā print(count)Ā
# Driver CodeX = "ABCT"Y = "PBDI"K = 6countOperations(X, Y, K)Ā
# This code is contributed by aditya7409. |
C#
// C# program for the above approach using System;class GFG {Ā
Ā Ā // Function to count minimum incrementsĀ Ā // by 1 or K required to convert X to YĀ Ā static void countOperations(Ā Ā Ā Ā string X, string Y, int K)Ā Ā {Ā Ā Ā Ā // Convert String to character arrayĀ Ā Ā Ā char[] ch1 = X.ToCharArray();Ā Ā Ā Ā char[] ch2 = Y.ToCharArray();Ā Ā Ā Ā int count = 0;Ā
Ā Ā Ā Ā // Traverse the string XĀ Ā Ā Ā for (int i = 0; i < X.Length; i++) {Ā
Ā Ā Ā Ā Ā Ā int c = 0;Ā
Ā Ā Ā Ā Ā Ā // Case 1Ā Ā Ā Ā Ā Ā if (ch1[i] == ch2[i])Ā Ā Ā Ā Ā Ā Ā Ā continue;Ā
Ā Ā Ā Ā Ā Ā // Case 2Ā Ā Ā Ā Ā Ā else if (ch1[i] < ch2[i]) {Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Add the difference/K toĀ Ā Ā Ā Ā Ā Ā Ā // the countĀ Ā Ā Ā Ā Ā Ā Ā if (((int)ch2[i]Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā - (int)ch1[i])Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā >= K) {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā c = ((int)ch2[i]Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā - (int)ch1[i])Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā / K;Ā Ā Ā Ā Ā Ā Ā Ā }Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Add the difference%K toĀ Ā Ā Ā Ā Ā Ā Ā // the countĀ Ā Ā Ā Ā Ā Ā Ā c += ((int)ch2[i]Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā - (int)ch1[i])Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā % K;Ā Ā Ā Ā Ā Ā }Ā
Ā Ā Ā Ā Ā Ā // Case 3Ā Ā Ā Ā Ā Ā else {Ā Ā Ā Ā Ā Ā Ā Ā int t = 90 - (int)ch1[i];Ā Ā Ā Ā Ā Ā Ā Ā t += (int)ch2[i] - 65 + 1;Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Add the difference/K toĀ Ā Ā Ā Ā Ā Ā Ā // the countĀ Ā Ā Ā Ā Ā Ā Ā if (t >= K)Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā c = t / K;Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Add the difference%K toĀ Ā Ā Ā Ā Ā Ā Ā // the countĀ Ā Ā Ā Ā Ā Ā Ā c += (t % K);Ā Ā Ā Ā Ā Ā }Ā
Ā Ā Ā Ā Ā Ā count += c;Ā Ā Ā Ā }Ā
Ā Ā Ā Ā // Print the answerĀ Ā Ā Ā Console.WriteLine(count);Ā Ā }Ā
Ā Ā // Driver codeĀ Ā static void Main()Ā Ā {Ā Ā Ā Ā string X = "ABCT", Y = "PBDI";Ā Ā Ā Ā int K = 6;Ā Ā Ā Ā countOperations(X, Y, K);Ā Ā }}Ā
// This code is contributed by susmitakundugoaldanga. |
Javascript
<script>Ā Ā Ā Ā Ā Ā // JavaScript program for the above approachĀ Ā Ā Ā Ā Ā // Function to count minimum incrementsĀ Ā Ā Ā Ā Ā // by 1 or K required to convert X to YĀ Ā Ā Ā Ā Ā function countOperations(X, Y, K) Ā Ā Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā var count = 0;Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Traverse the string XĀ Ā Ā Ā Ā Ā Ā Ā for (var i = 0; i < X.length; i++) Ā Ā Ā Ā Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā var c = 0;Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Case 1Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (X[i] === Y[i]) continue;Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Case 2Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā else if (X[i].charCodeAt(0) < Y[i].charCodeAt(0))Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Add the difference/K toĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // the countĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (Y[i].charCodeAt(0) - X[i].charCodeAt(0) >= K)Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā c = (Y[i].charCodeAt(0) - X[i].charCodeAt(0)) / K;Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Add the difference%K toĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // the countĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā c += (Y[i].charCodeAt(0) - X[i].charCodeAt(0)) % K;Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Case 3Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā else {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā var t = 90 - X[i].charCodeAt(0);Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā t += Y[i].charCodeAt(0) - 65 + 1;Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Add the difference/K toĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // the countĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (t >= K) {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā c = parseInt(t / K);Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Add the difference%K toĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // the countĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā c += parseInt(t % K);Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā count = parseInt(count + c);Ā Ā Ā Ā Ā Ā Ā Ā }Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Print the answerĀ Ā Ā Ā Ā Ā Ā Ā document.write(count + "<br>");Ā Ā Ā Ā Ā Ā }Ā
Ā Ā Ā Ā Ā Ā // Driver CodeĀ Ā Ā Ā Ā Ā var X = "ABCT",Ā Ā Ā Ā Ā Ā Ā Ā Y = "PBDI";Ā Ā Ā Ā Ā Ā var K = 6;Ā Ā Ā Ā Ā Ā countOperations(X, Y, K);Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // This code is contributed by rdtank.Ā Ā Ā Ā </script> |
Output:Ā
11
Ā
Time Complexity: O(N)
Auxiliary Space: O(1) because it is using constant space
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