给定一个包含小写英文字符的字符串str ,我们可以对给定的字符串执行以下两个操作:
- 删除整个字符串。
- 仅在字符串str [0…i]等于子字符串str [(i + 1)…(2 * i + 1)]时删除前缀。
任务是找到删除整个字符串所需的最大操作数。
例子:
Input: str = “abababab”
Output: 4
Operation 1: Delete prefix “ab” and the string becomes “ababab”.
Operation 2: Delete prefix “ab” and the string becomes “abab”.
Operation 3: Delete prefix “ab”, str = “ab”.
Operation 4: Delete the entire string.
Input: s = “abc”
Output: 1
方法:为了最大程度地增加操作数,要删除的前缀必须具有最小长度,即从单个字符开始,然后逐个追加连续的字符,以找到满足给定条件的最小长度前缀。将需要执行一个操作来删除该前缀,例如str [0…i],然后递归调用同一函数以找到子字符串str [i + 1…2 * i + 1]的结果。如果没有这样的前缀可以删除,则返回1,因为唯一可能的操作是删除整个字符串。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the maximum number
// of given operations required to
// remove the given string entirely
int find(string s)
{
// If length of the string is zero
if (s.length() == 0)
return 0;
// Single operation can delete the entire string
int c = 1;
// To store the prefix of the string
// which is to be deleted
string d = "";
for (int i = 0; i < s.length(); i++) {
// Prefix s[0..i]
d += s[i];
// To store the substring s[i+1...2*i+1]
string s2 = s.substr(i + 1, d.length());
// If the prefix s[0...i] can be deleted
if (s2 == d) {
// 1 operation to remove the current prefix
// and then recursively find the count of
// operations for the substring s[i+1...n-1]
c = 1 + find(s.substr(i + 1));
break;
}
}
// Entire string has to be deleted
return c;
}
// Driver code
int main()
{
string s = "abababab";
cout << find(s);
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG{
// Function to return the maximum number
// of given operations required to
// remove the given string entirely
static int find(String s)
{
// If length of the string is zero
if (s.length() == 0)
return 0;
// Single operation can delete
// the entire string
int c = 1;
// To store the prefix of the string
// which is to be deleted
String d = "";
for(int i = 0; i < s.length(); i++)
{
// Prefix s[0..i]
d += s.charAt(i);
// To store the substring s[i+1...2*i+1]
String s2 = "";
if (i + d.length() < s.length())
{
s2 = s.substring(i + 1,
d.length() + (i + 1));
}
// If the prefix s[0...i] can be deleted
if (s2.equals(d))
{
// 1 operation to remove the current prefix
// and then recursively find the count of
// operations for the substring s[i+1...n-1]
c = 1 + find(s.substring(i + 1));
break;
}
}
// Entire string has to be deleted
return c;
}
// Driver code
public static void main(String[] args)
{
String s = "abababab";
System.out.print(find(s));
}
}
// This code is contributed by pratham76
Python3
# Python3 implementation of the approach
# Function to return the maximum number
# of given operations required to
# remove the given entirely
def find(s):
# If length of the is zero
if (len(s) == 0):
return 0
# Single operation can delete
# the entire string
c = 1
# To store the prefix of the string
# which is to be deleted
d = ""
for i in range(len(s)):
# Prefix s[0..i]
d += s[i]
# To store the subs[i+1...2*i+1]
s2 = s[i + 1:i + 1 + len(d)]
# If the prefix s[0...i] can be deleted
if (s2 == d):
# 1 operation to remove the current prefix
# and then recursively find the count of
# operations for the subs[i+1...n-1]
c = 1 + find(s[i + 1:])
break
# Entire has to be deleted
return c
# Driver code
s = "abababab"
print(find(s))
# This code is contributed by Mohit Kumar
C#
// C# implementation of the approach
using System;
class GFG{
// Function to return the maximum number
// of given operations required to
// remove the given string entirely
static int find(string s)
{
// If length of the string is zero
if (s.Length == 0)
return 0;
// Single operation can delete the entire string
int c = 1;
// To store the prefix of the string
// which is to be deleted
string d = "";
for (int i = 0; i < s.Length; i++)
{
// Prefix s[0..i]
d += s[i];
// To store the substring s[i+1...2*i+1]
string s2 = "";
if(i + d.Length < s.Length)
{
s2 = s.Substring(i + 1, d.Length);
}
// If the prefix s[0...i] can be deleted
if (s2 == d)
{
// 1 operation to remove the current prefix
// and then recursively find the count of
// operations for the substring s[i+1...n-1]
c = 1 + find(s.Substring(i + 1));
break;
}
}
// Entire string has to be deleted
return c;
}
// Driver code
public static void Main(string[] args)
{
string s = "abababab";
Console.Write(find(s));
}
}
// This code is contributed by rutvik
输出:
4