# Minimize cost to empty a given string by removing characters alphabetically

• Last Updated : 19 Oct, 2021

Given string str, the task is to minimize the total cost to remove all the characters from the string in alphabetical order.

The cost of removing any character at i th index from the string will be i. The indexing is 1-based.

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Examples:

Input: str = “abcab”
Output:
Explanation:
First char ‘a’ at index 1 is removed, str[] becomes “bcab”,
Then char ‘a’ with index 3 is removed, str[] becomes “bcb”
After that char ‘b’ with index 1 is removed, str[] becomes “cb”,
Then char ‘b’ with index 2 is removed, str[] becomes “c”,
Finally, char ‘c’ is removed.
Total points = 1+3 + 1 + 2 + 1 = 8.

Input: str = “def”
Output: 3

Naive Approach: The simplest approach is to remove the smallest character with a smaller index in the string in each step and keep on adding the cost to the total cost. Print the final cost after this operation.

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by precomputing for each character, the number of smaller characters preceding it in the given string. Below are the steps:

1. Initialize the total cost to 0.
2. Transverse the given string and for each character, count the number of characters that are less than the current character and have occurred before it.
3. If this count is 0, this means that the current character will be removed at the present index, so add the index of the character to the resultant cost.
4. Else subtract the count from its current index and then add it to the total cost.
5. Print the total cost after all the above steps.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``#include ``using` `namespace` `std;`` ` `// Function to find the minimum cost``// required to remove each character``// of the string in alphabetical order``int` `minSteps(string str, ``int` `N)``{``    ``int` `smaller, cost = 0;`` ` `    ``// Stores the frequency of``    ``// characters of the string``    ``int` `f = { 0 };`` ` `    ``// Iterate through the string``    ``for` `(``int` `i = 0; i < N; i++) {``        ``int` `curr_ele = str[i] - ``'a'``;``        ``smaller = 0;`` ` `        ``// Count the number of characters``        ``// smaller than the present character``        ``for` `(``int` `j = 0; j <= curr_ele; j++) {``            ``if` `(f[j])``                ``smaller += f[j];``        ``}`` ` `        ``// If no smaller character``        ``// precedes current character``        ``if` `(smaller == 0)``            ``cost += (i + 1);``        ``else``            ``cost += (i - smaller + 1);`` ` `        ``// Increase the frequency of``        ``// the current character``        ``f[str[i] - ``'a'``]++;``    ``}`` ` `    ``// Return the``    ``// total cost``    ``return` `cost;``}`` ` `// Driver Code``int` `main()``{``    ``// Given string str``    ``string str = ``"abcab"``;``    ``int` `N = str.size();`` ` `    ``// Function call``    ``cout << minSteps(str, N);``    ``return` `0;``}`

## Java

 `// Java program for ``// the above approach``import` `java.io.*;``class` `GFG{`` ` `    ``// Function to find the minimum cost``    ``// required to remove each character``    ``// of the string in alphabetical order``    ``static` `int` `minSteps(String str, ``int` `N)``    ``{``        ``int` `smaller, cost = ``0``;`` ` `        ``// Stores the frequency of``        ``// characters of the string``        ``int` `f[] = ``new` `int``[``26``];`` ` `        ``// Iterate through the string``        ``for` `(``int` `i = ``0``; i < N; i++) ``        ``{``            ``int` `curr_ele = str.charAt(i) - ``'a'``;``            ``smaller = ``0``;`` ` `            ``// Count the number of characters``            ``// smaller than the present character``            ``for` `(``int` `j = ``0``; j <= curr_ele; j++) ``            ``{``                ``if` `(f[j] != ``0``)``                    ``smaller += f[j];``            ``}`` ` `            ``// If no smaller character``            ``// precedes current character``            ``if` `(smaller == ``0``)``                ``cost += (i + ``1``);``            ``else``                ``cost += (i - smaller + ``1``);`` ` `            ``// Increase the frequency of``            ``// the current character``            ``f[str.charAt(i) - ``'a'``]++;``        ``}`` ` `        ``// Return the``        ``// total cost``        ``return` `cost;``    ``}`` ` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``// Given string str``        ``String str = ``"abcab"``;`` ` `        ``int` `N = str.length();`` ` `        ``// Function call``        ``System.out.println(minSteps(str, N));``    ``}``}`` ` `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 program for the above approach `` ` `# Function to find the minimum cost``# required to remove each character``# of the string in alphabetical order ``def` `minSteps(``str``, N):`` ` `    ``cost ``=` `0`` ` `    ``# Stores the frequency of``    ``# characters of the string``    ``f ``=` `[``0``] ``*` `26`` ` `    ``# Iterate through the string``    ``for` `i ``in` `range``(N):``        ``curr_ele ``=` `ord``(``str``[i]) ``-` `ord``(``'a'``)``        ``smaller ``=` `0`` ` `        ``# Count the number of characters``        ``# smaller than the present character``        ``for` `j ``in` `range``(curr_ele ``+` `1``):``            ``if` `(f[j]):``                ``smaller ``+``=` `f[j]`` ` `        ``# If no smaller character``        ``# precedes current character``        ``if` `(smaller ``=``=` `0``):``            ``cost ``+``=` `(i ``+` `1``)``        ``else``:``            ``cost ``+``=` `(i ``-` `smaller ``+` `1``)`` ` `        ``# Increase the frequency of``        ``# the current character``        ``f[``ord``(``str``[i]) ``-` `ord``(``'a'``)] ``+``=` `1`` ` `    ``# Return the total cost``    ``return` `cost`` ` `# Driver Code`` ` `# Given string str``str` `=` `"abcab"`` ` `N ``=` `len``(``str``)`` ` `# Function call``print``(minSteps(``str``, N))`` ` `# This code is contributed by Shivam Singh`

## C#

 `// C# program for ``// the above approach``using` `System;``class` `GFG{`` ` `// Function to find the minimum cost``// required to remove each character``// of the string in alphabetical order``static` `int` `minSteps(``string` `str, ``int` `N)``{``    ``int` `smaller, cost = 0;`` ` `    ``// Stores the frequency of``    ``// characters of the string``    ``int``[] f = ``new` `int``;`` ` `    ``// Iterate through the string``    ``for` `(``int` `i = 0; i < N; i++) ``    ``{``        ``int` `curr_ele = str[i] - ``'a'``;``        ``smaller = 0;`` ` `        ``// Count the number of characters``        ``// smaller than the present character``        ``for` `(``int` `j = 0; j <= curr_ele; j++) ``        ``{``            ``if` `(f[j] != 0)``            ``smaller += f[j];``        ``}``         ` `        ``// If no smaller character``        ``// precedes current character``        ``if` `(smaller == 0)``            ``cost += (i + 1);``        ``else``            ``cost += (i - smaller + 1);`` ` `        ``// Increase the frequency of``        ``// the current character``        ``f[str[i] - ``'a'``]++;``    ``}`` ` `    ``// Return the``    ``// total cost``    ``return` `cost;``}`` ` `// Driver Code``public` `static` `void` `Main()``{`` ` `    ``// Given string str``    ``string` `str = ``"abcab"``;`` ` `    ``int` `N = str.Length;`` ` `    ``// Function call``    ``Console.Write(minSteps(str, N));``}``}`` ` `// This code is contributed by Chitranayal`

## Javascript

 ``
Output:
`8`

Time Complexity: O(N)
Auxiliary Space: O(26)

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