最小化将给定字符串转换为长度为 K 的相等子字符串串联的成本

给定长度的字符串S N包括小写字母和整数K，其中N％K = 0，任务是找到通过执行给定的字符串转换成相同的K -length子串的串联的字符串的最小成本以下操作：

• 一个字符可以替换为另一个字符。
• 每个操作的成本是被替换字符和被替换字符之间的绝对差值。例如，如果将‘a’替换为‘z’ ，则操作成本为|”a”-“z”| = 25 。

例子：

Input: S = “abcdef”, K = 2
Output: 8
Explanation:
One possible answer is “cdcdcd” and the repeated k length substring is “cd”. The minimum cost required to convert the string is calculated 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

朴素的方法：最简单的方法是生成长度为K 的所有可能的排列，并找到转换给定字符串的成本，使其具有长度为K的重复模式。然后，打印其中的最小成本。

时间复杂度： O(N*K ²⁶ )，其中 N 是给定字符串的长度，K 是给定整数。
辅助空间： O(N)

有效的方法：想法是使用贪婪技术并观察对于从0到K – 1 的任何位置i ，位置i + j * K处的字符必须相同，其中0 ≤ j < N/K 。例如，如果S =“abcbbc”和K = 3，则在位置0和3的字符必须是相等的，在在2和5位1和4位必须在相同的字符，以及字符必须是相等的。因此，可以单独计算位置i + j * K处的字符的最小成本。请按照以下步骤解决问题：

• 初始化变量an以存储所需的最低成本。
• 在[0, K – 1]范围内遍历字符串。
• 对于每个位置i，找到放置一个字符的位置处的成本I + J * K，为每一个字符，其中0≤Ĵ计算其中的最小成本并更新ans 。
• 完成上述步骤后，打印ans的值作为所需的最小成本。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find minimum cost
// to convert given String into
// String of K length same subString
void 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 character
        for(int ch = 0; ch < 26; ch++)
        {
            int cost = 0;
 
            // Find sum of distance 'a'+ ch
            // from character 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 Code
int 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 approach
 
import 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 character
            for (int ch = 0; ch < 26; ch++) {
 
                int cost = 0;
 
                // Find sum of distance 'a'+ ch
                // from character 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 approach
import sys
 
# Function to find minimum cost
# to convert given string into
# string of K length same substring
def 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 character
        for ch in range(26):
            cost = 0
              
            # Find sum of distance 'a'+ ch
            # from character 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 S
S = "abcdefabc"
  
K = 3
  
# Function call
minCost(S, K)
 
# This code is contributed by code_hunt

C#

// C# program for the
// above approach
using System;
 
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[j] - 'a']++;
        }
 
        // Stores minimum cost for
        // sequence of S[i]%k indices
        int min_cost = Int32.MaxValue;
 
        // Check for optimal character
        for(int ch = 0; ch < 26; ch++)
        {
            int cost = 0;
             
            // Find sum of distance 'a'+ ch
            // from character 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 Code
public static void Main()
{
     
    // Given string S
    string S = "abcdefabc";
 
    int K = 3;
 
    // Function call
    minCost(S, K);
}
}
 
// This code is contributed by sanjoy_62

Javascript

// Javascript program for the above approach
 
// Function to find minimum cost
// to convert given String into
// String of K length same subString
function minCost(s, k)
{
     
    // Stores length of String
    var n = s.length;
     
    // Stores the minimum cost
    var ans = 0;
 
    // Traverse left subString
    // of k length
    for(var i = 0; i < k; i++)
    {
         
        // Stores the frequency
        var a = new Array(26).fill(0);
 
        for(var j = i; j < n; j += k)
        {
            a[s[j].charCodeAt(0) - 'a'.charCodeAt(0)]++;
        }
 
        // Stores minimum cost for
        // sequence of S[i]%k indices
        var min_cost = 1000000000;
 
        // Check for optimal character
        for(var ch = 0; ch < 26; ch++)
        {
            var cost = 0;
 
            // Find sum of distance 'a'+ ch
            // from character S[i]%k indices
            for(var 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
    document.write(ans);
}
 
// Driver Code
// Given String S
var S = "abcdefabc";
var K = 3;
// Function Call
minCost(S, K);
 

输出：

9

时间复杂度： O(N + K)
辅助空间： O(N)

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