分割字符串的方法,以便每个分区以不同的字符开头

2021年4月4日18:56:22 发表评论 656 次浏览

本文概述

给定一个字符串s。让ķ是给定字符串可能的最大分区数, 每个分区均以不同的字符开头。任务是找到可将字符串s拆分成多种方式的方法ķ分区(非空), 以便每个分区以不同的字符开头

例子:

Input : s = "abb"
Output : 2
"abb" can be maximum split into 2 
partitions {a, bb} with distinct 
starting character, so k = 2. And, number of ways to split "abb" into 
2 partition with distinct starting 
character is 2 that are {a, bb} and
{ab, b}.

Input : s = "acbbcc"
Output : 6

首先, 我们需要找到k的值。观察到k将等于字符串中不同字符的数量, 因为只有最大数量的分区可以使每个分区具有不同的起始元素。

现在, 找到在每个分区中将字符串分成k个部分的方式的方法数量以不同的字符开始。首先观察一下, 第一个分区将始终固定字符串的第一个字符, 无论它有多长。现在, 我们需要处理除第一个字符以外的所有其他字符。

让我们举一个例子, 假设s =" acbbcc", 我们已经在上面讨论了第一个字符" a"。现在, 要处理" b"和" c", 请观察" b"出现在字符串的两个位置, 而" c"出现在三个位置。如果我们为" b"选择两个位置中的任何位置, 为" c"选择三个位置中的任何一个位置, 则可以在这些位置上对字符串进行分区。请注意, 部分数将等于3(等于s中的不同字符数, 即k)。

所以概括观察, 让我们数一世是字符i在s中出现的次数。所以我们的答案将是计数的乘积一世所有的i都使我出现在字符串中, 并且我不等于字符串的第一个字符。

以下是此方法的实现:

C ++

// CPP Program to find number of way 
// to split string such that each partition
// starts with distinct character with 
// maximum number of partitions.
#include <bits/stdc++.h>
  
using namespace std;
  
// Returns the number of we can split
// the string
int countWays(string s)
{
     int count[26] = { 0 };
  
     // Finding the frequency of each
     // character.
     for ( char x : s)
         count[x - 'a' ]++;
  
     // making frequency of first character
     // of string equal to 1.
     count展开 - 'a' ] = 1;
  
     // Finding the product of frequency 
     // of occurrence of each character.
     int ans = 1;
     for ( int i = 0; i < 26; ++i)
         if (count[i] != 0)
             ans *= count[i];
  
     return ans;
}
  
// Driven Program
int main()
{
     string s = "acbbcc" ;
     cout << countWays(s) << endl;
     return 0;
}

Java

// Java Program to find number 
// of way to split string such 
// that each partition starts 
// with distinct character with 
// maximum number of partitions.
import java.util.*;
import java.lang.*;
import java.io.*;
  
class GFG
{
  
// Returns the number of we 
// can split the string
static int countWays(String s)
{
     int count[] = new int [ 26 ];
  
     // Finding the frequency of
     // each character.
     for ( int i = 0 ; i < s.length(); i++)
         count展开++;
  
     // making frequency of first 
     // character of string equal to 1.
     count展开 = 1 ;
  
     // Finding the product of frequency 
     // of occurrence of each character.
     int ans = 1 ;
     for ( int i = 0 ; i < 26 ; ++i)
         if (count[i] != 0 )
             ans *= count[i];
  
     return ans;
}
  
// Driver Code
public static void main(String ags[])
{
     String s = "acbbcc" ;
     System.out.println(countWays(s));
}
}
  
// This code is contributed
// by Subhadeep

Python3

# Python3 Program to find number of way
# to split string such that each partition
# starts with distinct character with
# maximum number of partitions.
  
# Returns the number of we can split
# the string
def countWays(s):
     count = [ 0 ] * 26 ;
  
     # Finding the frequency of each
     # character.
     for x in s:
         count[ ord (x) - 
               ord ( 'a' )] = (count[ ord (x) - 
                                  ord ( 'a' )]) + 1 ;
  
     # making frequency of first character
     # of string equal to 1.
     count[ ord (s[ 0 ]) - ord ( 'a' )] = 1 ;
  
     # Finding the product of frequency
     # of occurrence of each character.
     ans = 1 ;
     for i in range ( 26 ):
         if (count[i] ! = 0 ):
             ans * = count[i];
  
     return ans;
  
# Driver Code
if __name__ = = '__main__' :
     s = "acbbcc" ;
     print (countWays(s));
  
# This code is contributed by Rajput-Ji

C#

// C# Program to find number 
// of way to split string such 
// that each partition starts 
// with distinct character with 
// maximum number of partitions.
  
using System;
   
class GFG
{
   
// Returns the number of we 
// can split the string
static int countWays( string s)
{
     int [] count = new int [26];
   
     // Finding the frequency of
     // each character.
     for ( int i = 0; i < s.Length; i++)
         count展开 - 'a' ]++;
   
     // making frequency of first 
     // character of string equal to 1.
     count展开 - 'a' ] = 1;
   
     // Finding the product of frequency 
     // of occurrence of each character.
     int ans = 1;
     for ( int i = 0; i < 26; ++i)
         if (count[i] != 0)
             ans *= count[i];
   
     return ans;
}
   
// Driver Code
public static void Main()
{
     string s = "acbbcc" ;
     Console.WriteLine(countWays(s));
}
}

输出如下:

6

木子山

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