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Maximize cost of deletions to obtain string having no pair of similar adjacent characters
• Last Updated : 11 Sep, 2020

Given string str of lowercase alphabets and an array cost[], where cost[i] represents the cost of deletion of ith character in the given string. The task is to find the maximum possible cost to obtain a string having no adjacent pair of same characters, by deleting the same pair of consecutive characters.

Examples:

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:
Explanation:
No consecutive characters are found.

Naive Approach: The simplest approach is to traverse the given string to check for each character from a to z if there are substrings having only that character. Follow the steps below to solve the problem:

1. Traverse from a to z to check substrings having only that character.
2. If any substring is found, calculate the sum for removal of each character and subtract the minimum cost from it.
3. Add the above sum in the total count.
4. Repeat the above steps for every substring.
5. Print the total count.

Time Complexity: O(N*26), where N is the length of the given string
Auxiliary Space: O(N)

Efficient Approach: Follow the below steps to optimize the above approach:

1. Traverse the given string over the range of indices [0, N – 2] and initialize a variable, say maxCost, to store the maximum possible cost.
2. Check if characters S[i] and S[i+1] are equal.
3. If adjacent characters are equal, then add the maximum cost from cost[i] and cost[i + 1] to maxCost and shift the minimum cost to right by setting cost[i + 1] = min(cost[i], cost[i + 1]).
4. After traversing the given string, print the value of maxCost.

Below is the implementation of the above approach:

## 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`
Output:
```4
```

Time Complexity: O(N)
Auxiliary Space: O(N)

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