Given a string S of length N consisting of lowercase letters and an integer K, where N % K = 0, the task is to find the minimum cost to convert the given string into a concatenated string of same K-length substrings by performing the following operations:
- A character can be replaced with another character.
- Cost of each operation is the absolute difference between the replacced and the replacing character. For example, if ‘a’ is replaced with ‘z’, then cost of the operation is |”a”-“z”| = 25.
Examples:
Input: S = “abcdef”, K = 2
Output: 8
Explanation:
One possible answer is “cdcdcd” and the repeatingK length substring is “cd”. The minimum cost to required to convert the string is calcualted by the following steps:
Step 1: Replace S[0] with “c”. Therefore, cost = |”a”-“c”| = 2.
Step 2: Replace S[1] with “d”. Therefore, cost = |”b”-“d”| = 2.
Step 3: Replace S[2] with “c”. Therefore, cost = |”c”-“c”| = 0.
Step 4: Replace S[3] with “d”. Therefore, cost = |”d”-“d”| = 0.
Step 5: Replace S[4] with “c”. Therefore, cost = |”e”-“c”| = 2.
Step 6: Replace S[5] with “d”. Therefore, cost = |”f”-“d”| = 2.
Therefore, the minimum cost required = 2 + 2 + 0 + 0 + 2 + 2 = 8.Input: S = “abcabc”, K = 3
Output: 0
Explanation:
The given string already consists a repeating substring “abc” of length K
Naive Approach: The simplest approach is to generate all possible permutations of length K and find the cost to convert the given string such that it has a repeating pattern of length K. Then, print the minimum cost among them.
Time Complexity: O(N*K26), where N is the length of the given string and K is the given integer.
Auxiliary Space: O(N)
Efficient Approach: The idea is to use a Greedy Technique and observe that for any position i from 0 to K – 1, characters at position i + j * K must be the same where 0 ≤ j < N/K. For example, if S = “abcbbc” and K = 3 then, characters at positions 0 and 3 must be equal, characters at positions 1 and 4 must the same, and characters at positions 2 and 5 must be equal. Therefore, the minimum cost for characters at positions i + j * K can be calcualted individually. Follow the steps below to solve the problem:
- Initialize a variable ans to store the minimum cost required.
- Traverse the string over the range [0, K – 1].
- For every position i, find the cost to place a character at positions i + j * K, for every character, where 0 ≤ j < N/K. Calcualte the minimum cost among them and update ans.
- After completing the above steps, print the value of ans as the required minimum cost.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find minimum cost// to convert given String into// String of K length same subStringvoid minCost(string s, int k){ // Stores length of String int n = s.size(); // Stores the minimum cost int ans = 0; // Traverse left subString // of k length for(int i = 0; i < k; i++) { // Stores the frequency int a[26]; for(int p = 0; p < 26; p++) { a[p] = 0; } for(int j = i; j < n; j += k) { a[s[j] - 'a']++; } // Stores minimum cost for // sequence of S[i]%k indices int min_cost = INT_MAX; // Check for optimal charcter for(int ch = 0; ch < 26; ch++) { int cost = 0; // Find sum of distance 'a'+ ch // from charcter S[i]%k indices for(int tr = 0; tr < 26; tr++) cost += abs(ch - tr) * a[tr]; // Choose minimum cost for // each index i min_cost = min(min_cost, cost); } // Increment ans ans += min_cost; } // Print minimum cost to // convert String cout << (ans);}// Driver Codeint main(){ // Given String S string S = "abcdefabc"; int K = 3; // Function Call minCost(S, K);}// This code is contributed by gauravrajput1 |
Java
// Java program for the above approachimport java.util.*;import java.lang.*;class GFG { // Function to find minimum cost // to convert given string into // string of K length same substring static void minCost(String s, int k) { // Stores length of string int n = s.length(); // Stores the minimum cost int ans = 0; // Traverse left substring // of k length for (int i = 0; i < k; i++) { // Stores the frequency int[] a = new int[26]; for (int j = i; j < n; j += k) { a[s.charAt(j) - 'a']++; } // Stores minimum cost for // sequence of S[i]%k indices int min_cost = Integer.MAX_VALUE; // Check for optimal charcter for (int ch = 0; ch < 26; ch++) { int cost = 0; // Find sum of distance 'a'+ ch // from charcter S[i]%k indices for (int tr = 0; tr < 26; tr++) cost += Math.abs(ch - tr) * a[tr]; // Choose minimum cost for // each index i min_cost = Math.min(min_cost, cost); } // Increment ans ans += min_cost; } // Print minimum cost to // convert string System.out.println(ans); } // Driver Code public static void main(String[] args) { // Given string S String S = "abcdefabc"; int K = 3; // Function Call minCost(S, K); }} |
Python3
# Python3 program for the # above approachimport sys# Function to find minimum cost# to convert given string into# string of K length same substringdef minCost(s, k): # Stores length of string n = len(s) # Stores the minimum cost ans = 0 # Traverse left substring # of k length for i in range(k): # Stores the frequency a = [0] * 26 for j in range(i, n, k): a[ord(s[j]) - ord('a')] += 1 # Stores minimum cost for # sequence of S[i]%k indices min_cost = sys.maxsize - 1 # Check for optimal charcter for ch in range(26): cost = 0 # Find sum of distance 'a'+ ch # from charcter S[i]%k indices for tr in range(26): cost += abs(ch - tr) * a[tr] # Choose minimum cost for # each index i min_cost = min(min_cost, cost) # Increment ans ans += min_cost # Print minimum cost to # convert string print(ans)# Driver Code # Given string SS = "abcdefabc" K = 3 # Function callminCost(S, K)# This code is contributed by code_hunt |
C#
// C# program for the // above approachusing System;class GFG{ // Function to find minimum cost// to convert given string into// string of K length same substringstatic void minCost(string s, int k){ // Stores length of string int n = s.Length; // Stores the minimum cost int ans = 0; // Traverse left substring // of k length for(int i = 0; i < k; i++) { // Stores the frequency int[] a = new int[26]; for(int j = i; j < n; j += k) { a[s[j] - 'a']++; } // Stores minimum cost for // sequence of S[i]%k indices int min_cost = Int32.MaxValue; // Check for optimal charcter for(int ch = 0; ch < 26; ch++) { int cost = 0; // Find sum of distance 'a'+ ch // from charcter S[i]%k indices for(int tr = 0; tr < 26; tr++) cost += Math.Abs(ch - tr) * a[tr]; // Choose minimum cost for // each index i min_cost = Math.Min(min_cost, cost); } // Increment ans ans += min_cost; } // Print minimum cost to // convert string Console.WriteLine(ans);}// Driver Codepublic static void Main(){ // Given string S string S = "abcdefabc"; int K = 3; // Function call minCost(S, K);}}// This code is contributed by sanjoy_62 |
9
Time Complexity: O(N + K)
Auxiliary Space: O(N)
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