Lexicographically largest subsequence such that every character occurs at least k times
Given a string S and an integer K. The task is to find lexicographically largest subsequence of S, say T, such that every character in T must occur at least K times.
Examples:
Input : S = "banana", K = 2.
Output : nn
Possible subsequence where each character exists at least 2 times are:
From the above subsequences, "nn" is the lexicographically largest.
The idea is to solve greedily the above problem. If we want to make the subsequence lexicographically largest, we must give priority to lexicographically larger characters. ‘z’ is the largest character, let suppose z occurs fz times in S. If fz >= K, append ‘z’z k times in the string T and keep removing characters from the left of S until all the z’s are removed. Apply the strategy with ‘y’, ‘w’, ….., ‘a’. In the end, you will find the answer.
Let see an example. Suppose S = “zzwzawa” and K = 2. Start with the largest character ‘z’. Here fz = 3 >= K. So T will become “zzz” and we will remove letters from the left of S until all the z’s are removed. So now S will become “awa”. Next largest is ‘y’ but that occurs 0 times in k so we will skip it. We will skip ‘w’, ‘v’ etc also until we go to ‘a’ which occurs 2 times. Now T will become “zzzaa” and S will become a empty string. Our answer is “zzzaa”.
Below is implementation of this approach:
C++
Java
import java.util.Arrays;
class GFG {
static void subsequence( char s[], char t[], int n, int k)
{
int last = 0 , cnt = 0 , new_last = 0 , size = 0 ;
for ( char ch = 'z' ; ch >= 'a' ; ch--) {
cnt = 0 ;
for ( int i = last; i < n; i++) {
if (s[i] == ch)
cnt++;
}
if (cnt >= k) {
for ( int i = last; i < n; i++) {
if (s[i] == ch) {
t[size++] = ch;
new_last = i;
}
}
last = new_last;
}
}
t[size] = '\0' ;
}
public static void main(String[] args) {
char s[] = { 'b' , 'a' , 'n' , 'a' , 'n' , 'a' };
int n = s.length;
int k = 2 ;
char t[] = new char [n];
subsequence(s, t, n - 1 , k);
for ( int i = 0 ;i<t.length;i++)
if (t[i]!= 0 )
System.out.print(t[i]);
}
}
|
Python3
C#
Javascript
Time Complexity: O(n)
Auxiliary Space: O(n)
Last Updated :
20 Jul, 2022
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