Minimum number of characters required to be added to a String such that all lowercase alphabets occurs as a subsequence in increasing order
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<vector< int > >& 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<vector< int > > 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
<script> // javascript program for the above approach // Function to find the LCS // of strings S and string T function LCS(S , N, T, M , 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 function minimumCharacter(S) { // String containing all the characters var T = "abcdefghijklmnopqrstuvwxyz" ; var N = S.length, M = T.length; // Stores the result of overlapping // subproblems var dp= Array(N+1).fill(0).map(x => Array(M+1).fill(0)); // Return the minimum characters // required return (26 - LCS(S, N, T, M, dp)); } // Driver Code var S = "abcdadc" ; document.write(minimumCharacter(S)); // This code is contributed by 29AjayKumar </script> |
22
Time Complexity: O(26*N)
Auxiliary Space: O(1)