鉴于长度为N的字符串str和一个整数K,任务是通过从字符串删除ķ要返回的字符在字典顺序最大的字符串。
A largest string Dictionary order is the last string when strings are arranged in alphabetical order.
例子:
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.
朴素的方法:这个想法是找到给定字符串长度N – K 的所有子序列。将这些子序列存储在列表中。将有n C m这样的序列。在上述步骤之后,按字母顺序打印存储在列表中的最大字符串。
时间复杂度: O(2 NK )
有效的方法:这个想法是使用一个 德克。以下是步骤:
- 将字符串的所有字符存储在双端队列中。
- 遍历定的字符串,在字符串保持从双端队列弹出的字符每一个字符,如果它小于存储在双端队列的最后一个字符。执行此操作直到K不为零。
- 现在完成上述操作后,在双端队列中插入当前字符。
- 上述操作后,由存储在双端队列中的字符形成的字符串是所得字符串。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
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 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
// 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 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
# 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
Javascript
tz
时间复杂度: O(N)
辅助空间: O(N)
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