# Minimum number of characters required to be added to a String such that all lowercase alphabets occurs as a subsequence in increasing order

Last Updated : 09 Sep, 2021

Given a string S consisting of N characters, the task is to find the minimum number of characters that must be added to S such that all lowercase alphabets occur as a subsequence in increasing order in S.

Examples:

Input: S = “axyabzaxd”
Output: 22
Explanation:
The characters from b to w (bcdefghijklmnopqrstuvw) whose count is 22, is missing in the given string S.
So, to make all the lower case english characters as a subsequence in it we have to add all these 22 characters.
Therefore, the minimum characters to be added is 22.

Input: S = “abcdefghixyabzjklmnoaxpqrstd”
Output: 6

Naive Approach: The simplest approach to solve the given problem is to generate all possible subsequences of the given string S and check in which subsequence appending the minimum number of characters in the string gives the subsequence of all lowercase alphabets in increasing order. After checking for all the subsequences, print the minimum number of characters appended.
Time Complexity: O(2N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized by the following observation:

• Since all the lowercase characters must be present after appending characters i.e., string S must contain string T as “abcdefghijklmnopqrstuvwxyz” as a subsequence.
• The minimum number of characters that must be appended in any order is to first find the Longest Common Subsequence(say L) of string S and string T as this gives the maximum number of characters in increasing order that we don’t need to add
• And then print the value of (26 – L) as the resultant minimum characters required.

From the above observations, find the value of LCS of the string S and T and print the value of (26 – LCS) as the result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include "bits/stdc++.h"` `using` `namespace` `std;`   `// Function to find the LCS` `// of strings S and string T` `int` `LCS(string& S, ``int` `N, string& T, ``int` `M,` `        ``vector >& dp)` `{`   `    ``// Base Case` `    ``if` `(N == 0 or M == 0)` `        ``return` `0;`   `    ``// Already Calculated State` `    ``if` `(dp[N][M] != -1)` `        ``return` `dp[N][M];`   `    ``// If the characters are the same` `    ``if` `(S[N - 1] == T[M - 1]) {` `        ``return` `dp[N][M] = 1 + LCS(S, N - 1, T, M - 1, dp);` `    ``}`   `    ``// Otherwise` `    ``return` `dp[N][M] = max(LCS(S, N - 1, T, M, dp),` `                          ``LCS(S, N, T, M - 1, dp));` `}`   `// Function to find the minimum number of` `// characters that needs to be appended` `// in the string to get all lowercase` `// alphabets as a subsequences` `int` `minimumCharacter(string& S)` `{`   `    ``// String containing all the characters` `    ``string T = ``"abcdefghijklmnopqrstuvwxyz"``;`   `    ``int` `N = S.length(), M = T.length();`   `    ``// Stores the result of overlapping` `    ``// subproblems` `    ``vector > dp(N + 1, vector<``int``>(M + 1, -1));`   `    ``// Return the minimum characters` `    ``// required` `    ``return` `(26 - LCS(S, N, T, M, dp));` `}`   `// Driver Code` `int` `main()` `{`   `    ``string S = ``"abcdadc"``;` `    ``cout << minimumCharacter(S);`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.io.*;`   `class` `GFG` `{`   `  ``// Function to find the LCS` `  ``// of strings S and string T` `  ``static` `int` `LCS(String S, ``int` `N, String T,` `                 ``int` `M, ``int` `dp[][])` `  ``{`   `    ``// Base Case` `    ``if` `(N == ``0` `|| M == ``0``)` `      ``return` `0``;`   `    ``// Already Calculated State` `    ``if` `(dp[N][M] != ``0``)` `      ``return` `dp[N][M];`   `    ``// If the characters are the same` `    ``if` `(S.charAt(N - ``1``)== T.charAt(M - ``1``)) {` `      ``return` `dp[N][M] = ``1` `+ LCS(S, N - ``1``,` `                                ``T, M - ``1``, dp);` `    ``}`   `    ``// Otherwise` `    ``return` `dp[N][M] = Math.max(` `      ``LCS(S, N - ``1``, T, M, dp),` `      ``LCS(S, N, T, M - ``1``, dp));` `  ``}`   `  ``// Function to find the minimum number of` `  ``// characters that needs to be appended` `  ``// in the string to get all lowercase` `  ``// alphabets as a subsequences` `  ``static` `int` `minimumCharacter(String S)` `  ``{`   `    ``// String containing all the characters` `    ``String T = ``"abcdefghijklmnopqrstuvwxyz"``;`   `    ``int` `N = S.length(), M = T.length();`   `    ``// Stores the result of overlapping` `    ``// subproblems` `    ``int` `dp[][]= ``new` `int``[N+``1``][M+``1``];` `    ``// Return the minimum characters` `    ``// required` `    ``return` `(``26` `- LCS(S, N, T, M, dp));` `  ``}`   `  ``// Driver Code` `  ``public` `static` `void` `main (String[] args) {` `    ``String S = ``"abcdadc"``;`   `    ``System.out.println(minimumCharacter(S));` `  ``}` `}`   `// This code is contributed by Potta Lokesh`

## Python

 `# Python3 program for the above approach` `import` `numpy as np`   `# Function to find the LCS` `# of strings S and string T` `def` `LCS(S, N, T, M, dp) :`   `    ``# Base Case` `    ``if` `(N ``=``=` `0` `or` `M ``=``=` `0``) :` `        ``return` `0``;`   `    ``# Already Calculated State` `    ``if` `(dp[N][M] !``=` `0``) :` `        ``return` `dp[N][M];`   `    ``# If the characters are the same` `    ``if` `(S[N ``-` `1``] ``=``=` `T[M ``-` `1``]) :` `        ``dp[N][M] ``=` `1` `+` `LCS(S, N ``-` `1``, T, M ``-` `1``, dp);` `        ``return` `dp[N][M]` `    `  `    ``# Otherwise` `    ``dp[N][M] ``=` `max``( LCS(S, N ``-` `1``, T, M, dp), LCS(S, N, T, M ``-` `1``, dp));` `    `  `    ``return` `dp[N][M]`     `# Function to find the minimum number of` `# characters that needs to be appended` `# in the string to get all lowercase` `# alphabets as a subsequences` `def` `minimumCharacter(S) :`   `    ``# String containing all the characters` `    ``T ``=` `"abcdefghijklmnopqrstuvwxyz"``;`   `    ``N ``=` `len``(S); M ``=` `len``(T);`   `    ``# Stores the result of overlapping` `    ``# subproblems` `    ``dp ``=` `np.zeros((N ``+` `1``, M ``+` `1``));`   `    ``# Return the minimum characters` `    ``# required` `    ``return` `(``26` `-` `LCS(S, N, T, M, dp));`     `# Driver Code` `if` `__name__ ``=``=` `"__main__"` `:`   `    ``S ``=` `"abcdadc"``;` `    ``print``(minimumCharacter(S));`   `    ``# This code is contributed by AnkThon`

## C#

 `// C# program for the above approach`     `using` `System;`   `public` `class` `GFG` `{`   `  ``// Function to find the LCS` `  ``// of strings S and string T` `  ``static` `int` `LCS(String S, ``int` `N, String T,` `                 ``int` `M, ``int` `[,]dp)` `  ``{`   `    ``// Base Case` `    ``if` `(N == 0 || M == 0)` `      ``return` `0;`   `    ``// Already Calculated State` `    ``if` `(dp[N,M] != 0)` `      ``return` `dp[N,M];`   `    ``// If the characters are the same` `    ``if` `(S[N - 1]== T[M - 1]) {` `      ``return` `dp[N,M] = 1 + LCS(S, N - 1,` `                                ``T, M - 1, dp);` `    ``}`   `    ``// Otherwise` `    ``return` `dp[N,M] = Math.Max(` `      ``LCS(S, N - 1, T, M, dp),` `      ``LCS(S, N, T, M - 1, dp));` `  ``}`   `  ``// Function to find the minimum number of` `  ``// characters that needs to be appended` `  ``// in the string to get all lowercase` `  ``// alphabets as a subsequences` `  ``static` `int` `minimumchar(String S)` `  ``{`   `    ``// String containing all the characters` `    ``String T = ``"abcdefghijklmnopqrstuvwxyz"``;`   `    ``int` `N = S.Length, M = T.Length;`   `    ``// Stores the result of overlapping` `    ``// subproblems` `    ``int` `[,]dp= ``new` `int``[N+1,M+1];` `    ``// Return the minimum characters` `    ``// required` `    ``return` `(26 - LCS(S, N, T, M, dp));` `  ``}`   `  ``// Driver Code` `  ``public` `static` `void` `Main(String[] args) {` `    ``String S = ``"abcdadc"``;`   `    ``Console.WriteLine(minimumchar(S));` `  ``}` `}`       `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:

`22`

Time Complexity: O(26*N)
Auxiliary Space: O(1)

Previous
Next