给定大小为N的字符串S和正整数K (其中N％K = 0) ，任务是找到需要替换的最小字符数，以使该字符串为K-周期且K长度为周期字符串必须是回文。
例子：
Input: S = “abaaba”, K = 3
Output: 0
Explanation: The given string is already K-periodic and the periodic string “aba” is palindromic.
Input: S = “abaaba”, K = 2
Output: 2
Explanation: By changing the characters at index 1 and 4 to ‘a’, the updated sting “aaaaaa” is K-periodic and the periodic string “aa” is palindromic. Therefore, minimum changes required is 2.
方法：想法是从给定的字符串索引创建一个Graph并执行DFS遍历以查找所需的更改数量。请按照以下步骤解决此问题：
- 将变量total初始化为0，以存储所需的更改计数。
- 根据给定的条件，从字符串创建图形，并在最终字符串中的位置i，K − i + 1，K + i，2K − i + 1、2K + i，3K − i + 1，…处创建所有字符。对于所有1≤i≤K都应相等。
- 在[0，N]范围内进行迭代，并在索引i与(N – i – 1)之间添加一个无向边。
- 在[0，N – M]范围内进行迭代，并在索引i与(i + K)之间添加一个无向边。
- 为了最大程度地减少所需的操作次数，请使所有字母最多等于出现在这些位置的字母，通过对字符串执行DFS遍历可以很容易地找到它们。
- 对所有未访问的节点在创建的图上执行DFS遍历：
- 在该遍历中访问的字符中找到频率最高的最大元素(例如maxFrequency )。
- 通过DFS遍历中所有已访问字符的计数与上述步骤中的最大频率之差，更新字符的更改总数。
- 完成上述步骤后，打印总计值作为结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include "bits/stdc++.h"
using namespace std;
// Function to add an edge to graph
void addEdge(vector adj[], int u,
int v)
{
adj[u].push_back(v);
adj[v].push_back(u);
}
// Function to perform DFS traversal on the
// graph recursively from a given vertex u
void DFS(int u, vector adj[],
int& cnt, vector& visited,
int fre[], string S)
{
// Visit the current vertex
visited[u] = true;
// Total number of nodes
// in this component
cnt++;
// Increment the frequency of u
fre[S[u] - 'a']++;
for (int i = 0;
i < adj[u].size(); i++) {
if (!visited[adj[u][i]]) {
DFS(adj[u][i], adj, cnt,
visited, fre, S);
}
}
}
// Function for finding the minimum
// number changes required in given string
int minimumOperations(string& S, int m)
{
int V = 100;
vector adj[V];
int total = 0, N = S.length();
// Form the edges according to the
// given conditions
for (int i = 0; i < N; i++) {
addEdge(adj, i, N - i - 1);
addEdge(adj, N - i - 1, i);
}
for (int i = 0; i < N - m; i++) {
addEdge(adj, i, i + m);
addEdge(adj, i + m, i);
}
// Find minimum number of operations
vector visited(V, 0);
for (int i = 0; i < N; i++) {
// Frequency array for finding
// the most frequent character
if (!visited[i]) {
// Frequency array for finding
// the most frequent character
int fre[26] = { 0 };
int cnt = 0, maxx = -1;
DFS(i, adj, cnt, visited, fre, S);
// Finding most frequent character
for (int j = 0; j < 26; j++)
maxx = max(maxx, fre[j]);
// Change rest of the characters
// to most frequent one
total += cnt - maxx;
}
}
// Print total number of changes
cout << total;
}
// Driver Code
int main()
{
string S = "abaaba";
int K = 2;
// Function Call
minimumOperations(S, K);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to add an edge to graph
static void addEdge(Vector adj[], int u,
int v)
{
adj[u].add(v);
adj[v].add(u);
}
static int cnt = 0;
static boolean[] visited;
// Function to perform DFS traversal on the
// graph recursively from a given vertex u
static void DFS(int u, Vector adj[],
int fre[], String S)
{
// Visit the current vertex
visited[u] = true;
// Total number of nodes
// in this component
cnt++;
// Increment the frequency of u
fre[S.charAt(u) - 'a']++;
for(int i = 0; i < adj[u].size(); i++)
{
if (!visited[adj[u].get(i)])
{
DFS(adj[u].get(i), adj, fre, S);
}
}
}
// Function for finding the minimum
// number changes required in given String
static void minimumOperations(String S, int m)
{
int V = 100;
@SuppressWarnings("unchecked")
Vector []adj = new Vector[V];
int total = 0, N = S.length();
for(int i = 0; i < adj.length; i++)
adj[i] = new Vector();
// Form the edges according to the
// given conditions
for(int i = 0; i < N; i++)
{
addEdge(adj, i, N - i - 1);
addEdge(adj, N - i - 1, i);
}
for(int i = 0; i < N - m; i++)
{
addEdge(adj, i, i + m);
addEdge(adj, i + m, i);
}
// Find minimum number of operations
visited = new boolean[V];
for(int i = 0; i < N; i++)
{
// Frequency array for finding
// the most frequent character
if (!visited[i])
{
// Frequency array for finding
// the most frequent character
int fre[] = new int[26];
cnt = 0;
int maxx = -1;
DFS(i, adj, fre, S);
// Finding most frequent character
for(int j = 0; j < 26; j++)
maxx = Math.max(maxx, fre[j]);
// Change rest of the characters
// to most frequent one
total += cnt - maxx;
}
}
// Print total number of changes
System.out.print(total);
}
// Driver Code
public static void main(String[] args)
{
String S = "abaaba";
int K = 2;
// Function Call
minimumOperations(S, K);
}
}
// This code is contributed by aashish1995 Python3
# Python3 program for the above approach
import sys
sys.setrecursionlimit(1500)
# Function to add an edge to graph
def addEdge(u, v):
global adj
adj[u].append(v)
adj[v].append(u)
# Function to perform DFS traversal on the
# graph recursively from a given vertex u
def DFS(u, fre, S):
global visited, adj, cnt
# Visit the current vertex
visited[u] = 1
# Total number of nodes
# in this component
cnt += 1
# Increment the frequency of u
fre[ord(S[u]) - ord('a')] += 1
for i in adj[u]:
if (visited[i] == 0):
DFS(i, fre, S)
# Function for finding the minimum
# number changes required in given string
def minimumOperations(S, m):
global adj, visited, cnt
total, N = 0, len(S)
# Form the edges according to the
# given conditions
for i in range(N):
addEdge(i, N - i - 1)
addEdge(N - i - 1, i)
for i in range(N-m):
addEdge(i, i + m)
addEdge(i + m, i)
for i in range(N):
# Frequency array for finding
# the most frequent character
if (not visited[i]):
# Frequency array for finding
# the most frequent character
fre = [0] * 26
cnt, maxx = 0, -1
DFS(i, fre, S)
# Finding most frequent character
for j in range(26):
maxx = max(maxx, fre[j])
# Change rest of the characters
# to most frequent one
total += cnt - maxx
# Print total number of changes
print (total)
# Driver Code
if __name__ == '__main__':
adj = [[] for i in range(101)]
visited, cnt = [0 for i in range(101)], 0
S = "abaaba"
K = 2
# Function Call
minimumOperations(S, K)
# This code is contributed by mohit kumar 29C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to add an edge to graph
static void addEdge(List []adj, int u,
int v)
{
adj[u].Add(v);
adj[v].Add(u);
}
static int cnt = 0;
static bool[] visited;
// Function to perform DFS traversal on the
// graph recursively from a given vertex u
static void DFS(int u, List []adj,
int []fre, String S)
{
// Visit the current vertex
visited[u] = true;
// Total number of nodes
// in this component
cnt++;
// Increment the frequency of u
fre[S[u] - 'a']++;
for(int i = 0; i < adj[u].Count; i++)
{
if (!visited[adj[u][i]])
{
DFS(adj[u][i], adj, fre, S);
}
}
}
// Function for finding the minimum
// number changes required in given String
static void minimumOperations(String S, int m)
{
int V = 100;
List []adj = new List[V];
int total = 0, N = S.Length;
for(int i = 0; i < adj.Length; i++)
adj[i] = new List();
// Form the edges according to the
// given conditions
for(int i = 0; i < N; i++)
{
addEdge(adj, i, N - i - 1);
addEdge(adj, N - i - 1, i);
}
for(int i = 0; i < N - m; i++)
{
addEdge(adj, i, i + m);
addEdge(adj, i + m, i);
}
// Find minimum number of operations
visited = new bool[V];
for(int i = 0; i < N; i++)
{
// Frequency array for finding
// the most frequent character
if (!visited[i])
{
// Frequency array for finding
// the most frequent character
int []fre = new int[26];
cnt = 0;
int maxx = -1;
DFS(i, adj, fre, S);
// Finding most frequent character
for(int j = 0; j < 26; j++)
maxx = Math.Max(maxx, fre[j]);
// Change rest of the characters
// to most frequent one
total += cnt - maxx;
}
}
// Print total number of changes
Console.Write(total);
}
// Driver Code
public static void Main(String[] args)
{
String S = "abaaba";
int K = 2;
// Function Call
minimumOperations(S, K);
}
}
// This code is contributed by aashish1995 输出：
2时间复杂度： O(N)
辅助空间： O(N)