缺失的最大限度地提高成本，以获得不具有对类似的相邻字符字符串

给定小写字母字符串str和数组cost[] ，其中cost[i]表示删除给定字符串^第i^个字符的成本。任务是通过删除相同的连续字符对，找到获得没有相邻的相同字符对的字符串的最大可能成本。

例子：

Input: str = “abaac”, cost = {1, 2, 3, 4, 5}
Output: 4
Explanation:
Remove character ‘a’ from position 4 in one based indexing

Input: str = “abc”, cost = {1, 2, 3}
Output: 0 
Explanation:
No consecutive characters are found.

朴素的方法：最简单的方法是遍历给定的字符串以检查从a到z 的每个字符是否存在仅包含该字符的子字符串。请按照以下步骤解决问题：

1. 从a到z遍历以检查仅包含该字符的子字符串。
2. 如果找到任何子串，计算每个字符的去除总和，并从中减去最小成本。
3. 将上述总和添加到总计数中。
4. 对每个子串重复上述步骤。
5. 打印总计数。

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

高效方法：按照以下步骤优化上述方法：

1. 在索引[0, N – 2]的范围内遍历给定的字符串并初始化一个变量，比如maxCost ，以存储最大可能的成本。
2. 检查字符S[i]和S[i+1]是否相等。
3. 如果相邻字符相等，则将cost[i]和cost[i + 1]的最大成本添加到maxCost并通过设置cost[i + 1] = min(cost[i], cost[我 + 1]) 。
4. 遍历给定的字符串，打印maxCost的值。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find maximum cost to
// remove consecutive characters
int Maxcost(string s, int cost[])
{
    // Initialize the count
    int count = 0;
 
    // Maximum cost
    int maxcost = 0, i = 0;
 
    // Traverse from 0 to len(s) - 2
    while (i < s.size() - 1)
    {
        // If characters are identical
        if (s[i] == s[i + 1])
        {
            // Add cost[i] if its maximum
            if (cost[i] > cost[i + 1])
                maxcost += cost[i];
 
            // Add cost[i + 1]
            // if its maximum
            else
            {
                maxcost += cost[i + 1];
                cost[i + 1] = cost[i];
            }
        }
     
        // Increment i
        i += 1;
    }
 
    // Return the final max count
    return maxcost;
}
 
// Driver Code
int main()
{
    // Given string s
    string s = "abaac";
 
    // Given cost of removal
    int cost[] = {1, 2, 3, 4, 5};
 
    // Function Call
    cout << Maxcost(s, cost);
    return 0;
}
// This is code contributed by gauravrajput1

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
     
// Function to find maximum cost to
// remove consecutive characters
static int Maxcost(String s, int []cost)
{
     
    // Maximum cost
    int maxcost = 0;
    int i = 0;
     
    // Traverse from 0 to len(s) - 2
    while (i < s.length() - 1)
    {
         
        // If characters are identical
        if (s.charAt(i) == s.charAt(i + 1))
        {
             
            // Add cost[i] if its maximum
            if (cost[i] > cost[i + 1])
                maxcost += cost[i];
 
            // Add cost[i + 1]
            // if its maximum
            else
            {
                maxcost += cost[i + 1];
                cost[i + 1] = cost[i];
            }
        }
 
        // Increment i
        i++;
    }
     
    // Return the final max count
    return maxcost;
}
 
// Driver code
public static void main (String[] args)
{
     
    // Given string s
    String s = "abaac";
 
    // Given cost of removal
    int []cost = { 1, 2, 3, 4, 5 };
 
    // Function call
    System.out.print(Maxcost(s, cost));
}
}
 
// This code is contributed by code_hunt

Python3

# Python3 program for the above approach
 
# Function to find maximum cost to
# remove consecutive characters
def Maxcost(s, cost):
 
    # Initialize the count
    count = 0
 
    # Maximum cost
    maxcost = 0
    i = 0
 
    # Traverse from 0 to len(s) - 2
    while i < len(s) - 1:
         
        # If characters are identical
        if s[i] == s[i + 1]:
 
            # Add cost[i] if its maximum
            if cost[i] > cost[i + 1]:
                maxcost += cost[i]
             
            # Add cost[i + 1] if its
            # maximum
            else:
                maxcost += cost[i + 1]
                cost[i + 1] = cost[i]
         
        # Increment i
        i += 1
 
    # Return the final max count
    return maxcost
 
# Driver Code
 
# Given string s
s = "abaac"
 
# Given cost of removal
cost = [1, 2, 3, 4, 5]
 
# Function Call
print(Maxcost(s, cost))

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to find maximum cost to
// remove consecutive characters
static int Maxcost(string s, int []cost)
{
     
    // Maximum cost
    int maxcost = 0;
    int i = 0;
     
    // Traverse from 0 to len(s) - 2
    while (i < s.Length - 1)
    {
         
        // If characters are identical
        if (s[i] == s[i + 1])
        {
             
            // Add cost[i] if its maximum
            if (cost[i] > cost[i + 1])
                maxcost += cost[i];
  
            // Add cost[i + 1]
            // if its maximum
            else
            {
                maxcost += cost[i + 1];
                cost[i + 1] = cost[i];
            }
        }
 
        // Increment i
        i++;
    }
     
    // Return the final max count
    return maxcost;
}
 
// Driver code
public static void Main (string[] args)
{
     
    // Given string s
    string s = "abaac";
 
    // Given cost of removal
    int []cost = {1, 2, 3, 4, 5};
 
    // Function call
    Console.Write(Maxcost(s, cost));
}
}
 
// This code is contributed by rutvik_56

Javascript

// javascript program for the
// above approach
 
// Function to find maximum cost to
// remove consecutive characters
function Maxcost(s, cost)
{
       
    // Maximum cost
    let maxcost = 0;
    let i = 0;
       
    // Traverse from 0 to len(s) - 2
    while (i < s.length - 1)
    {
           
        // If characters are identical
        if (s[i] == s[i + 1])
        {
               
            // Add cost[i] if its maximum
            if (cost[i] > cost[i + 1])
                maxcost += cost[i];
   
            // Add cost[i + 1]
            // if its maximum
            else
            {
                maxcost += cost[i + 1];
                cost[i + 1] = cost[i];
            }
        }
   
        // Increment i
        i++;
    }
       
    // Return the final max count
    return maxcost;
}
  
// Driver Code
 
     // Given string s
    let s = "abaac";
   
    // Given cost of removal
    let cost = [ 1, 2, 3, 4, 5 ];
   
    // Function call
    document.write(Maxcost(s, cost));
  
 // This code is contributed by avijitmondal1998.

输出：

4

时间复杂度： O(N)
辅助空间： O(N)

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