Given string str of length N and an integer K, the task is to return the largest string in Dictionary Order by erasing K characters from that string.
A largest string Dictionary order is the last string when strings are arranged in alphabetical order.
Examples:
Input: str = “ritz” K = 2
Output: tz
Explanation:
There are 6 possible ways of deleting two characters from s: “ri”, “rt”, “rz”, “it”, “iz”, “tz”.
Among these strings “tz” is the largest in dictionary order.
Thus “tz” is the desired output.Input: str = “jackie” K = 2
Output: jkie
Explanation:
The characters “a” and “c” are deleted to get the largest possible string.
Naive Approach: The idea is to find all the subsequence of the given string length N – K. Store those subsequences in a list. There will be nCm Such sequences. After the above steps print the largest string in alphabetical order stored in the list.
Time Complexity: O(2N-K)
Efficient Approach: The idea is to use a Deque. Below are the steps:
- Store all the characters of the string in the deque.
- Traverse the given string and for each character in the string keep popping the characters from deque if it is less than the last character stored in the deque. Perform this operation until K is non-zero.
- Now after the above operations, insert the current character in the deque.
- After the above operations, the string formed by the characters stored in the deque is the resultant string.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the largest // string after deleting k characters string largestString( int n, int k, string s)
{ // Deque dq used to find the
// largest string in dictionary
// after deleting k characters
deque< char > deq;
// Iterate till the length
// of the string
for ( int i = 0; i < n; ++i)
{
// Condition for popping
// characters from deque
while (deq.size() > 0 &&
deq.back() < s[i] &&
k > 0)
{
deq.pop_front();
k--;
}
deq.push_back(s[i]);
}
// To store the resultant string
string st = "" ;
// To form resultant string
for ( char c : deq)
st = st + c;
// Return the resultant string
return st;
} // Driver code int main()
{ int n = 4;
int k = 2;
// Given String
string sc = "ritz" ;
// Function call
string result = largestString(n, k, sc);
// Print the answer
cout << result << endl;
return 0;
} // This code is contributed by divyeshrabadiya07 |
// Java program for the above approach import java.util.ArrayDeque;
import java.util.Deque;
import java.util.Scanner;
import java.io.IOException;
public class GFG {
// Function to find the largest
// string after deleting k characters
public static String
largestString( int n, int k, String sc)
{
char [] s = sc.toCharArray();
// Deque dq used to find the
// largest string in dictionary
// after deleting k characters
Deque<Character> deq
= new ArrayDeque<>();
// Iterate till the length
// of the string
for ( int i = 0 ; i < n; ++i) {
// Condition for popping
// characters from deque
while (deq.size() > 0
&& deq.getLast() < s[i]
&& k > 0 ) {
deq.pollLast();
k--;
}
deq.add(s[i]);
}
// To store the resultant string
String st = "" ;
// To form resultant string
for ( char c : deq)
st = st + Character.toString(c);
// Return the resultant string
return st;
}
// Driver Code
public static void main(String[] args)
throws IOException
{
int n = 4 ;
int k = 2 ;
// Given String
String sc = "ritz" ;
// Function call
String result = largestString(n, k, sc);
// Print the answer
System.out.println(result);
}
} |
# Python3 program for the above approach from collections import deque
# Function to find the largest # string after deleting k characters def largestString(n, k, sc):
s = [i for i in sc]
# Deque dq used to find the
# largest string in dictionary
# after deleting k characters
deq = deque()
# Iterate till the length
# of the string
for i in range (n):
# Condition for popping
# characters from deque
while ( len (deq) > 0 and
deq[ - 1 ] < s[i] and
k > 0 ):
deq.popleft()
k - = 1
deq.append(s[i])
# To store the resultant string
st = ""
# To form resultant string
for c in deq:
st = st + c
# Return the resultant string
return st
# Driver Code if __name__ = = '__main__' :
n = 4
k = 2
# Given String
sc = "ritz"
# Function call
result = largestString(n, k, sc)
# Print the answer
print (result)
# This code is contributed by mohit kumar 29 |
using System;
using System.Linq;
using System.Collections.Generic;
class GFG
{ // Function to find the largest
// string after deleting k characters
public static string LargestString( int n, int k, string sc)
{
char [] s = sc.ToCharArray();
// Deque dq used to find the
// largest string in dictionary
// after deleting k characters
List< char > deq = new List< char >();
// Iterate till the length
// of the string
for ( int i = 0; i < n; ++i)
{
// Condition for popping
// characters from deque
while (deq.Count > 0
&& deq[deq.Count - 1] < s[i]
&& k > 0)
{
deq.RemoveAt(deq.Count - 1);
k--;
}
deq.Add(s[i]);
}
// To store the resultant string
string st = "" ;
// To form resultant string
foreach ( char c in deq)
st = st + c.ToString();
// Return the resultant string
return st;
}
// Driver Code
public static void Main( string [] args)
{
int n = 4;
int k = 2;
// Given String
string sc = "ritz" ;
// Function call
string result = LargestString(n, k, sc);
// Print the answer
Console.WriteLine(result);
}
} |
<script> // Javascript program for the above approach // Function to find the largest // string after deleting k characters function largestString(n, k, s)
{ // Deque dq used to find the
// largest string in dictionary
// after deleting k characters
var deq = [];
// Iterate till the length
// of the string
for ( var i = 0; i < n; ++i)
{
// Condition for popping
// characters from deque
while (deq.length > 0 &&
deq[deq.length - 1] < s[i] &&
k > 0)
{
deq.shift();
k--;
}
deq.push(s[i]);
}
// To store the resultant string
var st = "" ;
// To form resultant string
deq.forEach(c => {
st = st + c;
});
// Return the resultant string
return st;
} // Driver code var n = 4;
var k = 2;
// Given String var sc = "ritz" ;
// Function call var result = largestString(n, k, sc);
// Print the answer document.write(result); // This code is contributed by rutvik_56 </script> |
tz
Time Complexity: O(N)
Auxiliary Space: O(N)