不同的子字符串中的开始和结束与给定的字符串

给定一个字符串s和其他两个字符串的开始和结束，找到字符串中不同子其开始与给定的开始和结束字符串号和结束。

例子：

Input : s = "geeksforgeeks"
        begin = "geeks"
        end = "for"
Output : 1

Input : s = "vishakha"
        begin = "h"
        end = "a"
Output : 2
Two different sub-strings are "ha" and "hakha".

方法：查找所有出现的字符串开始和字符串结束。将每个字符串的索引存储在两个不同的数组中。之后遍历整个字符串并在每次迭代中向已经看到的子字符串添加一个符号，并将新字符串映射到一些非负整数。由于字符串的结尾和开头以及不同长度的字符串映射到不同的数字（并且相等的字符串被相等地映射），因此只需计算一定长度的必要子字符串的数量。

C++

// Cpp program to find number of
// different sub strings
#include 
using namespace std;
 
// function to return number of different
// sub-strings
int numberOfDifferentSubstrings(string s, string a,
                                          string b)
{
    // initially our answer is zero.
    int ans = 0;
 
    // find the length of given strings
    int ls = s.size(), la = a.size(), lb = b.size();
 
    // currently make array and initially put zero.
    int x[ls] = { 0 }, y[ls] = { 0 };
 
    // find occurrence of "a" and "b" in string "s"
    for (int i = 0; i < ls; i++) {
        if (s.substr(i, la) == a)
            x[i] = 1;
        if (s.substr(i, lb) == b)
            y[i] = 1;
    }
 
    // We use a hash to make sure that same
    // substring is not counted twice.
    unordered_set hash;   
 
    // go through all the positions to find
    // occurrence of "a" first.
    string curr_substr = "";
    for (int i = 0; i < ls; i++) {
     
        // if we found occurrence of "a".
        if (x[i]) {
         
            // then go through all the positions
            // to find occurrence of "b".
            for (int j = i; j < ls; j++) {
             
                // if we do found "b" at index
                // j then add it to already
                // existed substring.
                if (!y[j])
                    curr_substr += s[j];
 
                // if we found occurrence of "b".
                if (y[j]) {
                 
                    // now add string "b" to
                    // already existed substring.
                    curr_substr += s.substr(j, lb);
                     
                    // If current substring is not
                    // included already.
                    if (hash.find(curr_substr) == hash.end())
                        ans++;
 
                    // put any non negative
                    // integer to make this
                    // string as already
                    // existed.
                    hash.insert(curr_substr);
                }
            }
 
            // make substring null.
            curr_substr = "";
        }
    }
 
    // return answer.
    return ans;
}
 
// Driver program for above function.
int main()
{
    string s = "codecppforfood";
    string begin = "c";
    string end = "d";
    cout << numberOfDifferentSubstrings(s, begin, end)
        << endl;
    return 0;
}

Java

// Java program to find number of
// different sub strings
import java.util.HashSet;
 
class GFG
{
 
// function to return number of
// different sub-strings
static int numberOfDifferentSubstrings(String s,
                                       char a, char b)
{
 
    // initially our answer is zero.
    int ans = 0;
 
    // find the length of given strings
    int ls = s.length();
 
    // currently make array and
    // initially put zero.
    int[] x = new int[ls];
    int[] y = new int[ls];
 
    // find occurrence of "a" and "b"
    // in string "s"
    for (int i = 0; i < ls; i++)
    {
        if (s.charAt(i) == a)
            x[i] = 1;
        if (s.charAt(i) == b)
            y[i] = 1;
    }
 
    // We use a hash to make sure that same
    // substring is not counted twice.
    HashSet hash = new HashSet<>();
 
    // go through all the positions to find
    // occurrence of "a" first.
    String curr_substr = "";
    for (int i = 0; i < ls; i++)
    {
 
        // if we found occurrence of "a".
        if (x[i] != 0)
        {
 
            // then go through all the positions
            // to find occurrence of "b".
            for (int j = i; j < ls; j++)
            {
 
                // if we do found "b" at index
                // j then add it to already
                // existed substring.
                if (y[j] == 0)
                    curr_substr += s.charAt(i);
 
                // if we found occurrence of "b".
                if (y[j] != 0)
                {
 
                    // now add string "b" to
                    // already existed substing.
                    curr_substr += s.charAt(j);
 
                    // If current substring is not
                    // included already.
                    if (!hash.contains(curr_substr))
                        ans++;
 
                    // put any non negative
                    // integer to make this
                    // string as already
                    // existed.
                    hash.add(curr_substr);
                }
            }
 
            // make substring null.
            curr_substr = "";
        }
    }
 
    // return answer.
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    String s = "codecppforfood";
    char begin = 'c';
    char end = 'd';
    System.out.println(
           numberOfDifferentSubstrings(s, begin, end));
}
}
 
// This code is contributed by
// sanjeev2552

Python3

# Python 3 program to find number of
# different sub strings
 
# function to return number of different
# sub-strings
def numberOfDifferentSubstrings(s, a, b):
 
    # initially our answer is zero.
    ans = 0
 
    # find the length of given strings
    ls = len(s)
    la = len(a)
    lb = len(b)
 
    # currently make array and initially
    # put zero.
    x = [0] * ls
    y = [0] * ls
 
    # find occurrence of "a" and "b" in string "s"
    for i in range(ls):
         
        if (s[i: la + i] == a):
            x[i] = 1
        if (s[i: lb + i] == b):
            y[i] = 1
 
    # We use a hash to make sure that same
    # substring is not counted twice.
    hash = []
 
    # go through all the positions to find
    # occurrence of "a" first.
    curr_substr = ""
    for i in range(ls):
     
        # if we found occurrence of "a".
        if (x[i]):
         
            # then go through all the positions
            # to find occurrence of "b".
            for j in range( i, ls):
             
                # if we do found "b" at index
                # j then add it to already
                # existed substring.
                if (not y[j]):
                    curr_substr += s[j]
 
                # if we found occurrence of "b".
                if (y[j]):
                 
                    # now add string "b" to
                    # already existed substring.
                 
                    curr_substr += s[j: lb + j]
                     
                    # If current substring is not
                    # included already.
                    if curr_substr not in hash:
                        ans += 1
 
                    # put any non negative integer
                    # to make this string as already
                    # existed.
                    hash.append(curr_substr)
 
            # make substring null.
            curr_substr = ""
 
    # return answer.
    return ans
 
# Driver Code
if __name__ == "__main__":
     
    s = "codecppforfood"
    begin = "c"
    end = "d"
    print(numberOfDifferentSubstrings(s, begin, end))
 
# This code is contributed by ita_c

C#

// C# program to find number of
// different sub strings
using System;
using System.Collections.Generic;
 
class GFG
{
 
// function to return number of
// different sub-strings
static int numberOfDifferentSubstrings(String s,
                                    char a, char b)
{
 
    // initially our answer is zero.
    int ans = 0;
 
    // find the length of given strings
    int ls = s.Length;
 
    // currently make array and
    // initially put zero.
    int[] x = new int[ls];
    int[] y = new int[ls];
 
    // find occurrence of "a" and "b"
    // in string "s"
    for (int i = 0; i < ls; i++)
    {
        if (s[i] == a)
            x[i] = 1;
        if (s[i] == b)
            y[i] = 1;
    }
 
    // We use a hash to make sure that same
    // substring is not counted twice.
    HashSet hash = new HashSet();
 
    // go through all the positions to find
    // occurrence of "a" first.
    String curr_substr = "";
    for (int i = 0; i < ls; i++)
    {
 
        // if we found occurrence of "a".
        if (x[i] != 0)
        {
 
            // then go through all the positions
            // to find occurrence of "b".
            for (int j = i; j < ls; j++)
            {
 
                // if we do found "b" at index
                // j then add it to already
                // existed substring.
                if (y[j] == 0)
                    curr_substr += s[i];
 
                // if we found occurrence of "b".
                if (y[j] != 0)
                {
 
                    // now add string "b" to
                    // already existed substring.
                    curr_substr += s[j];
 
                    // If current substring is not
                    // included already.
                    if (!hash.Contains(curr_substr))
                        ans++;
 
                    // put any non negative
                    // integer to make this
                    // string as already
                    // existed.
                    hash.Add(curr_substr);
                }
            }
 
            // make substring null.
            curr_substr = "";
        }
    }
 
    // return answer.
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
    String s = "codecppforfood";
    char begin = 'c';
    char end = 'd';
    Console.WriteLine(
        numberOfDifferentSubstrings(s, begin, end));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// Javascript program to find number of
// different sub strings
     
// function to return number of
// different sub-strings
function numberOfDifferentSubstrings(s,a,b)
{
    // initially our answer is zero.
    let ans = 0;
   
    // find the length of given strings
    let ls = s.length;
   
    // currently make array and
    // initially put zero.
    let x = new Array(ls);
    let y = new Array(ls);
     
    for(let i=0;i<ls;i++)
    {
        x[i]=0;
        y[i]=0;
    }
   
    // find occurrence of "a" and "b"
    // in string "s"
    for (let i = 0; i < ls; i++)
    {
        if (s[i] == a)
            x[i] = 1;
        if (s[i] == b)
            y[i] = 1;
    }
   
    // We use a hash to make sure that same
    // substring is not counted twice.
    let hash = new Set();
   
    // go through all the positions to find
    // occurrence of "a" first.
    let curr_substr = "";
    for (let i = 0; i < ls; i++)
    {
   
        // if we found occurrence of "a".
        if (x[i] != 0)
        {
   
            // then go through all the positions
            // to find occurrence of "b".
            for (let j = i; j < ls; j++)
            {
   
                // if we do found "b" at index
                // j then add it to already
                // existed substring.
                if (y[j] == 0)
                    curr_substr += s[i];
   
                // if we found occurrence of "b".
                if (y[j] != 0)
                {
   
                    // now add string "b" to
                    // already existed substring.
                    curr_substr += s[j];
   
                    // If current substring is not
                    // included already.
                    if (!hash.has(curr_substr))
                        ans++;
   
                    // put any non negative
                    // integer to make this
                    // string as already
                    // existed.
                    hash.add(curr_substr);
                }
            }
   
            // make substring null.
            curr_substr = "";
        }
    }
   
    // return answer.
    return ans;
    }
     
    // Driver Code
    let s = "codecppforfood";
    let begin = 'c';
    let end = 'd';
    document.write(numberOfDifferentSubstrings(s, begin, end));
     
    // This code is contributed by rag2127
     

输出：

3

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