给定一个字符串S ，任务是通过将一个重复字符块与另一个不同重复字符块交换来计算按频率递增顺序对字符串进行排序的最小成本。每个操作的成本是两个块的绝对差值。
例子：
Input: S = “aabbcccdeffffggghhhhhii”
Output: 5
Explanation:
- Swap ‘d’ with ‘aa’. The Cost of this Operation is 1
- Swap ‘e’ with ‘bb’. The Cost of this Operation is 1
- Swap ‘ii’ with ‘ccc’. The Cost of this Operation is cost is 1
- Swap ‘ccc’ with ‘ffff’. The Cost of this Operation is 1
- Swap ‘ffff’ with ‘hhhhh’. The Cost of this Operation is 1
Input : S = “aaaa”
Output : 0
方法：
请按照以下步骤解决问题：
- 存储字符串中每个字符出现的频率。
- 对频率进行排序。
- 计算排序后的序列和原始序列中每个频率之间的差异。
- 所有差异的总和的一半是必需的答案。这是因为，对于每次交换，差异在上述步骤中计算了两次。
下面是上述方法的实现：
C++
// C++ program to implement
// above approach
#include
using namespace std;
int sortString(string S)
{
vector sorted, original;
bool insert = false;
// For a single character
if (S.length() == 1)
{
cout << 0 << endl;
}
// Stores count of repetitions
// of a character
int curr = 1;
for(int i = 0; i < (S.length() - 1); i++)
{
// If repeating character
if (S[i] == S[i + 1])
{
curr += 1;
insert = false;
}
// Otherwise
else
{
// Store frequency
sorted.push_back(curr);
original.push_back(curr);
// Reset count
curr = 1;
insert = true;
}
}
// Insert the last character block
if ((S[(S.length() - 1)] !=
S[(S.length() - 2)]) || insert == false)
{
sorted.push_back(curr);
original.push_back(curr);
}
// Sort the frequencies
sort(sorted.begin(), sorted.end());
// Stores the minimum cost of all operations
int t_cost = 0;
for(int i = 0; i < sorted.size(); i++)
{
// Store the absolute difference of
// i-th frequencies of ordered and
// unordered sequences
t_cost += abs(sorted[i] -
original[i]);
}
// Return the minimum cost
return (t_cost / 2);
}
// Driver Code
int main()
{
string S = "aabbcccdeffffggghhhhhii";
cout << sortString(S);
return 0;
} Java
// Java program to implement
// above approach
import java.util.*;
class GFG{
public static int sortString(String S)
{
Vector sorted = new Vector();
Vector original = new Vector();
boolean insert = false;
// For a single character
if (S.length() == 1)
{
System.out.println(0);
}
// Stores count of repetitions
// of a character
int curr = 1;
for(int i = 0; i < (S.length() - 1); i++)
{
// If repeating character
if (S.charAt(i) == S.charAt(i + 1))
{
curr += 1;
insert = false;
}
// Otherwise
else
{
// Store frequency
sorted.add(curr);
original.add(curr);
// Reset count
curr = 1;
insert = true;
}
}
// Insert the last character block
if ((S.charAt(S.length() - 1) !=
S.charAt(S.length() - 2)) ||
insert == false)
{
sorted.add(curr);
original.add(curr);
}
// Sort the frequencies
Collections.sort(sorted);
// Stores the minimum cost of all operations
int t_cost = 0;
for(int i = 0; i < sorted.size(); i++)
{
// Store the absolute difference of
// i-th frequencies of ordered and
// unordered sequences
t_cost += Math.abs(sorted.get(i) -
original.get(i));
}
// Return the minimum cost
return (t_cost / 2);
}
// Driver code
public static void main(String[] args)
{
String S = "aabbcccdeffffggghhhhhii";
System.out.print(sortString(S));
}
}
// This code is contributed by divyeshrabadiya07 Python3
# Python3 program to implement
# the above approach
def sortString(S):
sorted1 = []
original = []
insert = False
# For a single character
if (len(S) == 1):
print(0)
# Stores count of repetitions
# of a character
curr = 1
for i in range(len(S) - 1):
# If repeating character
if (S[i] == S[i + 1]):
curr += 1
insert = False
# Otherwise
else:
# Store frequency
sorted1.append(curr)
original.append(curr)
# Reset count
curr = 1
insert = True
# Insert the last character block
if ((S[(len(S) - 1)] != S[(len(S) - 2)]) or
insert == False):
sorted1.append(curr)
original.append(curr)
# Sort the frequencies
sorted1.sort()
# Stores the minimum cost of all operations
t_cost = 0
for i in range(len(sorted1)):
# Store the absolute difference of
# i-th frequencies of ordered and
# unordered sequences
t_cost += abs(sorted1[i] - original[i])
# Return the minimum cost
return (t_cost // 2)
# Driver Code
if __name__ == "__main__":
S = "aabbcccdeffffggghhhhhii"
print(sortString(S))
# This code is contributed by ChitranayalC#
// C# program to implement
// above approach
using System;
using System.Collections.Generic;
class GFG{
public static int sortString(string S)
{
List sorted = new List();
List original = new List();
bool insert = false;
// For a single character
if (S.Length == 1)
{
Console.WriteLine(0);
}
// Stores count of repetitions
// of a character
int curr = 1;
for(int i = 0; i < (S.Length - 1); i++)
{
// If repeating character
if (S[i] == S[i + 1])
{
curr += 1;
insert = false;
}
// Otherwise
else
{
// Store frequency
sorted.Add(curr);
original.Add(curr);
// Reset count
curr = 1;
insert = true;
}
}
// Insert the last character block
if ((S[S.Length - 1] !=
S[S.Length - 2]) || insert == false)
{
sorted.Add(curr);
original.Add(curr);
}
// Sort the frequencies
sorted.Sort();
// Stores the minimum cost of all operations
int t_cost = 0;
for(int i = 0; i < sorted.Count; i++)
{
// Store the absolute difference of
// i-th frequencies of ordered and
// unordered sequences
t_cost += Math.Abs(sorted[i] -
original[i]);
}
// Return the minimum cost
return (t_cost / 2);
}
// Driver Code
static void Main()
{
string S = "aabbcccdeffffggghhhhhii";
Console.Write(sortString(S));
}
}
// This code is contributed by divyesh072019 Javascript
输出：
5时间复杂度： O(NlogN)
辅助空间： O(N)
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