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Remove minimum characters from string to split it into three substrings under given constraints

  • Difficulty Level : Medium
  • Last Updated : 05 May, 2021
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Given a string str of lowercase alphabets, the task is to remove minimum characters from the given string so that string can be break into 3 substrings str1, str2, and str3 such that each substring can be empty or can contains only characters ‘a’, ‘b’, and ‘c’ respectively.
Example: 
 

Input: str = “aaaabaaxccac” 
Output:
Explanation: 
String after removing b, x, and a then, string str become “aaaaaaccc” 
Now str1 = “aaaaaa”, str2 = “”, str3 = “ccc”. 
The minimum character removed is 3.
Input: str = “baabcbbdcca” 
Output:
Explanation: 
String after removing b, c, d, and a then, string str become “aabbbcc” 
Now str1 = “aa”, str2 = “bbb”, str3 = “cc”. 
The minimum character removed is 4. 
 

Approach: This problem can be solved using Greedy Approach. We will use three prefix array to make prefix array of characters ‘a’, ‘b’, and ‘c’. Each prefix array will store the count of letter ‘a’, ‘b’, and ‘c’ at any index i respectively. Below are the steps:
 

  1. Create three prefix array as: 
    • prefa[i] represents the count of letter “a” in prefix of length i.
    • prefb[i] represents the count of letter “b” in prefix of length i.
    • prefc[i] represents the count of letter “c” in prefix of length i.
  2. In order to delete minimum number of character, the resultant string should be of maximum size.
  3. The idea is to fix two positions i and j in string, 0 ? i ? j ≤ N, in order to split string into three parts of all possible length and do the following: 
    • Remove all character except ‘a’ from the prefix, which ends in i, this will be the string str1.
    • Remove all character except ‘c’ from the suffix, which starts in j, this will be the string str3.
    • Remove all character except ‘b’ which is between positions i and j, this will be the string str2.
  4. Therefore, the total length of the resultant string is given by: 
     

total length of (str1 + str2 + str3) = (prefa[i]) + (prefb[j] – prefb[i]) + (prefc[n] – prefc[j]) 
 

  1. Subtract the length of the resultant string from the length of the given string str to get the minimum characters to removed.

Below is the implementation of the above approach:
 



C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that counts minimum
// character that must be removed
void min_remove(string str)
{
    // Length of string
    int N = str.length();
 
    // Create prefix array
    int prefix_a[N + 1];
    int prefix_b[N + 1];
    int prefix_c[N + 1];
 
    // Initialize first position
    prefix_a[0] = 0;
    prefix_b[0] = 0;
    prefix_c[0] = 0;
 
    // Fill prefix array
    for (int i = 1; i <= N; i++) {
        prefix_a[i]
            = prefix_a[i - 1]
              + (str[i - 1] == 'a');
 
        prefix_b[i]
            = prefix_b[i - 1]
              + (str[i - 1] == 'b');
 
        prefix_c[i]
            = prefix_c[i - 1]
              + (str[i - 1] == 'c');
    }
 
    // Initialise maxi
    int maxi = INT_MIN;
 
    // Check all the possibilities by
    // putting i and j at different
    // position & find maximum among them
    for (int i = 0; i <= N; i++) {
 
        for (int j = i; j <= N; j++) {
 
            maxi = max(maxi,
                       (prefix_a[i]
                        + (prefix_b[j]
                           - prefix_b[i])
                        + (prefix_c[N]
                           - prefix_c[j])));
        }
    }
 
    // Print the characters to be removed
    cout << (N - maxi) << endl;
}
 
// Driver Code
int main()
{
    // Given String
    string str = "aaaabaaxccac";
 
    // Function Call
    min_remove(str);
 
    return 0;
}

Java




// Java program for the above approach
class GFG{
     
// Function that counts minimum
// character that must be removed
static void min_remove(String str)
{
     
    // Length of string
    int N = str.length();
 
    // Create prefix array
    int []prefix_a = new int[N + 1];
    int []prefix_b = new int[N + 1];
    int []prefix_c = new int[N + 1];
 
    // Initialize first position
    prefix_a[0] = 0;
    prefix_b[0] = 0;
    prefix_c[0] = 0;
 
    // Fill prefix array
    for(int i = 1; i <= N; i++)
    {
        prefix_a[i] = prefix_a[i - 1] +
                     (int)((str.charAt(
                            i - 1) == 'a') ? 1 : 0);
 
        prefix_b[i] = prefix_b[i - 1] +
                      (int)((str.charAt(i - 1) ==
                                     'b') ? 1 : 0);
 
        prefix_c[i] = prefix_c[i - 1] +
                      (int)((str.charAt(i - 1) ==
                                     'c') ? 1 : 0);
    }
 
    // Initialise maxi
    int maxi = Integer.MIN_VALUE;
 
    // Check all the possibilities by
    // putting i and j at different
    // position & find maximum among them
    for(int i = 0; i <= N; i++)
    {
        for(int j = i; j <= N; j++)
        {
            maxi = Math.max(maxi, (prefix_a[i] +
                                  (prefix_b[j] -
                                   prefix_b[i]) +
                                  (prefix_c[N] -
                                   prefix_c[j])));
        }
    }
 
    // Print the characters to be removed
    System.out.println((N - maxi));
}
 
// Driver Code
public static void main(String []args)
{
     
    // Given String
    String str = "aaaabaaxccac";
 
    // Function call
    min_remove(str);
}
}
 
// This code is contributed by grand_master

Python3




# Python 3 program for the above approach
import sys
 
# Function that counts minimum
# character that must be removed
def min_remove(st):
 
    # Length of string
    N = len(st)
 
    # Create prefix array
    prefix_a = [0]*(N + 1)
    prefix_b = [0]*(N + 1)
    prefix_c = [0]*(N + 1)
 
    # Initialize first position
    prefix_a[0] = 0
    prefix_b[0] = 0
    prefix_c[0] = 0
 
    # Fill prefix array
    for i in range(1, N + 1):
         
        if (st[i - 1] == 'a'):
            prefix_a[i] = (prefix_a[i - 1] + 1)
        else:
            prefix_a[i] = prefix_a[i - 1]
 
        if (st[i - 1] == 'b'):
            prefix_b[i] = (prefix_b[i - 1] + 1)
        else:
            prefix_b[i]= prefix_b[i - 1]
 
        if (st[i - 1] == 'c'):
            prefix_c[i] = (prefix_c[i - 1] + 1)
        else:
            prefix_c[i] = prefix_c[i - 1]
     
    # Initialise maxi
    maxi = -sys.maxsize -1;
 
    # Check all the possibilities by
    # putting i and j at different
    # position & find maximum among them
    for i in range( N + 1):
        for j in range(i, N + 1):
            maxi = max(maxi,
                      (prefix_a[i] +
                      (prefix_b[j] -
                       prefix_b[i]) +
                      (prefix_c[N] -
                       prefix_c[j])))
 
    # Print the characters to be removed
    print((N - maxi))
 
# Driver Code
if __name__ == "__main__":
     
    # Given String
    st = "aaaabaaxccac"
 
    # Function Call
    min_remove(st)
 
# This code is contributed by Chitranayal

C#




// C# program for the above approach
using System;
 
class GFG{
     
// Function that counts minimum
// character that must be removed
static void min_remove(string str)
{
     
    // Length of string
    int N = str.Length;
 
    // Create prefix array
    int []prefix_a = new int[N + 1];
    int []prefix_b = new int[N + 1];
    int []prefix_c = new int[N + 1];
 
    // Initialize first position
    prefix_a[0] = 0;
    prefix_b[0] = 0;
    prefix_c[0] = 0;
 
    // Fill prefix array
    for(int i = 1; i <= N; i++)
    {
        prefix_a[i] = prefix_a[i - 1] +
                    (int)((str[i - 1] == 'a') ?
                                   1 : 0);
 
        prefix_b[i] = prefix_b[i - 1] +
                    (int)((str[i - 1] == 'b') ?
                                   1 : 0);
 
        prefix_c[i] = prefix_c[i - 1] +
                    (int)((str[i - 1] == 'c') ?
                                   1 : 0);
    }
 
    // Initialise maxi
    int maxi = Int32.MinValue;
 
    // Check all the possibilities by
    // putting i and j at different
    // position & find maximum among them
    for(int i = 0; i <= N; i++)
    {
        for(int j = i; j <= N; j++)
        {
            maxi = Math.Max(maxi, (prefix_a[i] +
                                  (prefix_b[j] -
                                   prefix_b[i]) +
                                  (prefix_c[N] -
                                   prefix_c[j])));
        }
    }
 
    // Print the characters to be removed
    Console.WriteLine((N - maxi));
}
 
// Driver Code
public static void Main()
{
     
    // Given String
    string str = "aaaabaaxccac";
 
    // Function call
    min_remove(str);
}
}
 
// This code is contributed by sanjoy_62

Javascript




<script>
 
// JavaScript program to implement
// the above approach
 
// Function that counts minimum
// character that must be removed
function min_remove(str)
{
       
    // Length of string
    let N = str.length;
   
    // Create prefix array
    let prefix_a = Array.from({length: N + 1}, (_, i) => 0);
    let prefix_b = Array.from({length: N + 1}, (_, i) => 0);
    let prefix_c = Array.from({length: N + 1}, (_, i) => 0);
   
    // Initialize first position
    prefix_a[0] = 0;
    prefix_b[0] = 0;
    prefix_c[0] = 0;
   
    // Fill prefix array
    for(let i = 1; i <= N; i++)
    {
        prefix_a[i] = prefix_a[i - 1] +
                     ((str[
                            i - 1] == 'a') ? 1 : 0);
   
        prefix_b[i] = prefix_b[i - 1] +
                      ((str[i - 1] ==
                                     'b') ? 1 : 0);
   
        prefix_c[i] = prefix_c[i - 1] +
                      ((str[i - 1] ==
                                     'c') ? 1 : 0);
    }
   
    // Initialise maxi
    let maxi = Number.MIN_VALUE;
   
    // Check all the possibilities by
    // putting i and j at different
    // position & find maximum among them
    for(let i = 0; i <= N; i++)
    {
        for(let j = i; j <= N; j++)
        {
            maxi = Math.max(maxi, (prefix_a[i] +
                                  (prefix_b[j] -
                                   prefix_b[i]) +
                                  (prefix_c[N] -
                                   prefix_c[j])));
        }
    }
   
    // Prlet the characters to be removed
    document.write((N - maxi));
}
 
// Driver code
 
     // Given String
    let str = "aaaabaaxccac";
   
    // Function call
    min_remove(str);
 
// This code is contributed by code_hunt.
</script>
Output: 
3

 

Time Complexity: O(N2), where N is the length of the given string. 
Space Complexity: O(N), where N is the length of the given string.
 




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