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: 3
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: 4
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:
- 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.
- In order to delete minimum number of character, the resultant string should be of maximum size.
- 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.
- 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])
- 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++ 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 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 |
# 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# 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 |
<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])));
}
}
// Print 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> |
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.