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 ‘Z’ by 1 or K starts from the character ‘A’.
Examples:
Input: X = “ABCT”, Y = “PBDI”, K = 13
Output: 7
Explanation:
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 = “ABCD”, Y = “ABCD”, 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] – X[i])/K and (Y[i] – 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;// Function to count minimum increments// by 1 or K required to convert X to Yvoid countOperations( string X, string Y, int K){ int count = 0; // Traverse the string X for (int i = 0; i < X.length(); i++) { int c = 0; // Case 1 if (X[i] == Y[i]) continue; // Case 2 else if (X[i] < Y[i]) { // Add the difference/K to // the count if ((Y[i] - X[i]) >= K) { c = (Y[i] - X[i]) / K; } // Add the difference%K to // the count c += (Y[i] - X[i]) % K; } // Case 3 else { int t = 90 - X[i]; t += Y[i] - 65 + 1; // Add the difference/K to // the count if (t >= K) c = t / K; // Add the difference%K to // the count c += (t % K); } count += c; } // Print the answer cout << count << endl;}// Driver Codeint main(){ string X = "ABCT", Y = "PBDI"; int K = 6; countOperations(X, Y, K);}// This code is contributed by ipg2016107. |
Java
// Java program for the above approachimport java.io.*;class GFG { // Function to count minimum increments // by 1 or K required to convert X to Y public static void countOperations( String X, String Y, int K) { // Convert String to character array char ch1[] = X.toCharArray(); char ch2[] = Y.toCharArray(); int count = 0; // Traverse the string X for (int i = 0; i < X.length(); i++) { int c = 0; // Case 1 if (ch1[i] == ch2[i]) continue; // Case 2 else if (ch1[i] < ch2[i]) { // Add the difference/K to // the count if (((int)ch2[i] - (int)ch1[i]) >= K) { c = ((int)ch2[i] - (int)ch1[i]) / K; } // Add the difference%K to // the count c += ((int)ch2[i] - (int)ch1[i]) % K; } // Case 3 else { int t = 90 - (int)ch1[i]; t += (int)ch2[i] - 65 + 1; // Add the difference/K to // the count if (t >= K) c = t / K; // Add the difference%K to // the count c += (t % K); } count += c; } // Print the answer System.out.print(count); } // Driver Code public static void main(String[] args) { String X = "ABCT", Y = "PBDI"; int K = 6; countOperations(X, Y, K); }} |
Python3
# Python program for the above approach# Function to count minimum increments# by 1 or K required to convert X to Ydef countOperations(X, Y, K): count = 0 # Traverse the string X for i in range(len(X)): c = 0 # Case 1 if (X[i] == Y[i]): continue # Case 2 elif (X[i] < Y[i]): # Add the difference/K to # the count if ((ord(Y[i]) - ord(X[i])) >= K): c = (ord(Y[i]) - ord(X[i])) // K # Add the difference%K to # the count c += (ord(Y[i]) - ord(X[i])) % K # Case 3 else: t = 90 - ord(X[i]) t += ord(Y[i]) - 65 + 1 # Add the difference/K to # the count if (t >= K): c = t // K # Add the difference%K to # the count c += (t % K) count += c # Print the answer print(count)# Driver CodeX = "ABCT"Y = "PBDI"K = 6countOperations(X, Y, K)# This code is contributed by aditya7409. |
C#
// C# program for the above approach using System;class GFG { // Function to count minimum increments // by 1 or K required to convert X to Y static void countOperations( string X, string Y, int K) { // Convert String to character array char[] ch1 = X.ToCharArray(); char[] ch2 = Y.ToCharArray(); int count = 0; // Traverse the string X for (int i = 0; i < X.Length; i++) { int c = 0; // Case 1 if (ch1[i] == ch2[i]) continue; // Case 2 else if (ch1[i] < ch2[i]) { // Add the difference/K to // the count if (((int)ch2[i] - (int)ch1[i]) >= K) { c = ((int)ch2[i] - (int)ch1[i]) / K; } // Add the difference%K to // the count c += ((int)ch2[i] - (int)ch1[i]) % K; } // Case 3 else { int t = 90 - (int)ch1[i]; t += (int)ch2[i] - 65 + 1; // Add the difference/K to // the count if (t >= K) c = t / K; // Add the difference%K to // the count c += (t % K); } count += c; } // Print the answer Console.WriteLine(count); } // Driver code static void Main() { string X = "ABCT", Y = "PBDI"; int K = 6; countOperations(X, Y, K); }}// This code is contributed by susmitakundugoaldanga. |
Javascript
<script> // JavaScript program for the above approach // Function to count minimum increments // by 1 or K required to convert X to Y function countOperations(X, Y, K) { var count = 0; // Traverse the string X for (var i = 0; i < X.length; i++) { var c = 0; // Case 1 if (X[i] === Y[i]) continue; // Case 2 else if (X[i].charCodeAt(0) < Y[i].charCodeAt(0)) { // Add the difference/K to // the count if (Y[i].charCodeAt(0) - X[i].charCodeAt(0) >= K) { c = (Y[i].charCodeAt(0) - X[i].charCodeAt(0)) / K; } // Add the difference%K to // the count c += (Y[i].charCodeAt(0) - X[i].charCodeAt(0)) % K; } // Case 3 else { var t = 90 - X[i].charCodeAt(0); t += Y[i].charCodeAt(0) - 65 + 1; // Add the difference/K to // the count if (t >= K) { c = parseInt(t / K); } // Add the difference%K to // the count c += parseInt(t % K); } count = parseInt(count + c); } // Print the answer document.write(count + "<br>"); } // Driver Code var X = "ABCT", Y = "PBDI"; var K = 6; countOperations(X, Y, K); // This code is contributed by rdtank. </script> |
Output:
11
Time Complexity: O(N)
Auxiliary Space: O(1) because it is using constant space
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