给定一个字符串s由“X”，“Y”和“Z”只，任务是与S转换为通过选择字符和删除子仅由单一的不同字符的字符串，不包含该字符的最小次数。
注意：一旦选择了一个字符，其他字符就不能用于进一步的操作。
例子：
Input: S = “XXX”
Output: 0
Explanation: Since the given string already consists of a single distinct character, i.e. X, no removal is required. Therefore, the required count is 0.
Input: S = “XYZXYZX”
Output: 2
Explanation:
Selecting the character ‘X’ and removing the substrings “YZ” in two consecutive operations reduces the string to “XXX”, which consists of a single distinct character only.
方法：想法是使用 unordered_map 计算每个字符的出现次数，并计算每个字符所需的删除次数并打印最小值。请按照以下步骤解决问题：
- 初始化 unordered_map 并存储每个字符的出现索引。
- 迭代字符串S 的所有字符并更新映射中出现的字符“X”、“Y”和“Z” 。
- 遍历 Map 并为每个字符计算每个字符所需的移除次数。
- 计算每个字符，打印为任何字符获得的最小计数。
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 removals
// required to convert given string
// to single distinct characters only
int minimumOperations(string s, int n)
{
// Unordered map to store positions
// of characters X, Y and Z
unordered_map > mp;
// Update indices of X, Y, Z;
for (int i = 0; i < n; i++) {
mp[s[i]].push_back(i);
}
// Stores the count of
// minimum removals
int ans = INT_MAX;
// Traverse the Map
for (auto x : mp) {
int curr = 0;
int prev = 0;
bool first = true;
// Count the number of removals
// required for current character
for (int index : (x.second)) {
if (first) {
if (index > 0) {
curr++;
}
prev = index;
first = false;
}
else {
if (index != prev + 1) {
curr++;
}
prev = index;
}
}
if (prev != n - 1) {
curr++;
}
// Update the answer
ans = min(ans, curr);
}
// Print the answer
cout << ans;
}
// Driver Code
int main()
{
// Given string
string s = "YYXYZYXYZXY";
// Size of string
int N = s.length();
// Function call
minimumOperations(s, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to find minimum removals
// required to convert given string
// to single distinct characters only
static void minimumOperations(String s, int n)
{
// Unordered map to store positions
// of characters X, Y and Z
HashMap> mp = new HashMap<>();
// Update indices of X, Y, Z;
for(int i = 0; i < n; i++)
{
if (mp.containsKey(s.charAt(i)))
{
mp.get(s.charAt(i)).add(i);
}
else
{
mp.put(s.charAt(i), new ArrayList(Arrays.asList(i)));
}
}
// Stores the count of
// minimum removals
int ans = Integer.MAX_VALUE;
// Traverse the Map
for (Map.Entry> x : mp.entrySet())
{
int curr = 0;
int prev = 0;
boolean first = true;
// Count the number of removals
// required for current character
for(Integer index : (x.getValue()))
{
if (first)
{
if (index > 0)
{
curr++;
}
prev = index;
first = false;
}
else
{
if (index != prev + 1)
{
curr++;
}
prev = index;
}
}
if (prev != n - 1)
{
curr++;
}
// Update the answer
ans = Math.min(ans, curr);
}
// Print the answer
System.out.print(ans);
}
// Driver code
public static void main(String[] args)
{
// Given string
String s = "YYXYZYXYZXY";
// Size of string
int N = s.length();
// Function call
minimumOperations(s, N);
}
}
// This code is contributed by divyeshrabadiya07 Python3
# Python3 program for the above approach
import sys;
INT_MAX = sys.maxsize;
# Function to find minimum removals
# required to convert given string
# to single distinct characters only
def minimumOperations(s, n) :
# Unordered map to store positions
# of characters X, Y and Z
mp = {};
# Update indices of X, Y, Z;
for i in range(n) :
if s[i] in mp :
mp[s[i]].append(i);
else :
mp[s[i]] = [i];
# Stores the count of
# minimum removals
ans = INT_MAX;
# Traverse the Map
for x in mp :
curr = 0;
prev = 0;
first = True;
# Count the number of removals
# required for current character
for index in mp[x] :
if (first) :
if (index > 0) :
curr += 1;
prev = index;
first = False;
else :
if (index != prev + 1) :
curr += 1;
prev = index;
if (prev != n - 1) :
curr += 1;
# Update the answer
ans = min(ans, curr);
# Print the answer
print(ans);
# Driver Code
if __name__ == "__main__" :
# Given string
s = "YYXYZYXYZXY";
# Size of string
N = len(s);
# Function call
minimumOperations(s, N);
# This code is contributed by AnkThonC#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find minimum removals
// required to convert given string
// to single distinct characters only
static void minimumOperations(string s, int n)
{
// Unordered map to store positions
// of characters X, Y and Z
Dictionary> mp = new Dictionary>();
// Update indices of X, Y, Z;
for(int i = 0; i < n; i++)
{
if (mp.ContainsKey(s[i]))
{
mp[s[i]].Add(i);
}
else
{
mp[s[i]] = new List();
mp[s[i]].Add(i);
}
}
// Stores the count of
// minimum removals
int ans = Int32.MaxValue;
// Traverse the Map
foreach(KeyValuePair> x in mp)
{
int curr = 0;
int prev = 0;
bool first = true;
// Count the number of removals
// required for current character
foreach(int index in (x.Value))
{
if (first)
{
if (index > 0)
{
curr++;
}
prev = index;
first = false;
}
else
{
if (index != prev + 1)
{
curr++;
}
prev = index;
}
}
if (prev != n - 1)
{
curr++;
}
// Update the answer
ans = Math.Min(ans, curr);
}
// Print the answer
Console.Write(ans);
}
// Driver Code
static void Main()
{
// Given string
string s = "YYXYZYXYZXY";
// Size of string
int N = s.Length;
// Function call
minimumOperations(s, N);
}
}
// This code is contributed by divyesh072019 输出：
3时间复杂度： O(N)
辅助空间： O(N)
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