计算具有不同首尾字符的子字符串

给定字符串S ，任务是打印给定字符串的子字符串计数，该字符串的第一个字符和最后一个字符不同。

例子：

Input: S = “abcab”
Output: 8
Explanation: 
There are 8 substrings having first and last characters different {ab, abc, abcab, bc, bca, ca, cab, ab}.

Input: S = “aba”
Output: 2
Explanation: 
There are 2 substrings having first and last characters different {ab, ba}.

天真的方法：想法是生成给定字符串的所有可能的子字符串，并为每个子字符串检查第一个和最后一个字符是否不同。如果发现为真，则将计数增加1并检查下一个子字符串。遍历所有子字符串后打印计数。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count the substrings
// having different first and last
// characters
int countSubstring(string s, int n)
{
    // Store the final count
    int ans = 0;
 
    // Loop to traverse the string
    for (int i = 0; i < n; i++) {
 
        // Counter for each iteration
        int cnt = 0;
 
        // Iterate over substrings
        for (int j = i + 1; j < n; j++) {
 
            // Compare the characters
            if (s[j] != s[i])
 
                // Increase count
                cnt++;
        }
 
        // Adding count of substrings
        // to the final count
        ans += cnt;
    }
 
    // Print the final count
    cout << ans;
}
 
// Driver Code
int main()
{
    // Given string
    string S = "abcab";
 
    // Length of the string
    int N = 5;
 
    // Function Call
    countSubstring(S, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG{
     
// Function to count the substrings
// having different first and last
// characters
static void countSubstring(String s, int n)
{
     
    // Store the final count
    int ans = 0;
 
    // Loop to traverse the string
    for(int i = 0; i < n; i++)
    {
         
        // Counter for each iteration
        int cnt = 0;
 
        // Iterate over substrings
        for(int j = i + 1; j < n; j++)
        {
             
            // Compare the characters
            if (s.charAt(j) != s.charAt(i))
 
                // Increase count
                cnt++;
        }
 
        // Adding count of substrings
        // to the final count
        ans += cnt;
    }
 
    // Print the final count
    System.out.print(ans);
}
     
// Driver Code
public static void main(String[] args)
{
     
    // Given string
    String S = "abcab";
 
    // Length of the string
    int N = 5;
 
    // Function call
    countSubstring(S, N);
}
}
 
// This code is contributed by code_hunt

Python3

# Python3 program for the above approach
 
# Function to count the substrings
# having different first and last
# characters
def countSubstring(s, n):
 
    # Store the final count
    ans = 0
 
    # Loop to traverse the string
    for i in range(n):
 
        # Counter for each iteration
        cnt = 0
 
        # Iterate over substrings
        for j in range(i + 1, n):
 
            # Compare the characters
            if (s[j] != s[i]):
 
                # Increase count
                cnt += 1
         
        # Adding count of substrings
        # to the final count
        ans += cnt
     
    # Print the final count
    print(ans)
 
# Driver Code
 
# Given string
S = "abcab"
 
# Length of the string
N = 5
 
# Function call
countSubstring(S, N)
 
# This code is contributed by code_hunt

C#

// C# program for the above approach
using System;
using System.Collections;
using System.Collections.Generic;
using System.Text;
 
class GFG{
     
// Function to count the substrings
// having different first and last
// characters
static void countSubstring(string s, int n)
{
     
    // Store the final count
    int ans = 0;
 
    // Loop to traverse the string
    for(int i = 0; i < n; i++)
    {
         
        // Counter for each iteration
        int cnt = 0;
 
        // Iterate over substrings
        for(int j = i + 1; j < n; j++)
        {
             
            // Compare the characters
            if (s[j] != s[i])
 
                // Increase count
                cnt++;
        }
 
        // Adding count of substrings
        // to the final count
        ans += cnt;
    }
 
    // Print the final count
    Console.Write(ans);
}
     
// Driver Code
public static void Main(string[] args)
{
     
    // Given string
    string S = "abcab";
 
    // Length of the string
    int N = 5;
 
    // Function call
    countSubstring(S, N);
}
}
 
// This code is contributed by rutvik_56

Javascript

 
// JavaScript program for
// the above approach
 
// Function to count the substrings
// having different first and last
// characters
function countSubstring(s, n)
{
      
    // Store the final count
    let ans = 0;
  
    // Loop to traverse the string
    for(let i = 0; i < n; i++)
    {
          
        // Counter for each iteration
        let cnt = 0;
  
        // Iterate over substrings
        for(let j = i + 1; j < n; j++)
        {
              
            // Compare the characters
            if (s[j] != s[i])
  
                // Increase count
                cnt++;
        }
  
        // Adding count of substrings
        // to the final count
        ans += cnt;
    }
  
    // Print the final count
    document.write(ans);
}
 
 
// Driver code
 
    // Given string
    let S = "abcab";
  
    // Length of the string
    let N = 5;
  
    // Function call
    countSubstring(S, N);
      
     // This code is contributed by splevel62.

输出：

8

时间复杂度： O(N ² )
辅助空间： O(1)

高效的方法：可以使用Map by对上述方法进行优化，以存储字符串字符的频率。请按照以下步骤解决问题：

1. 初始化两个变量，一个变量用于每次迭代计算不同的字符(例如cur )，另一个变量用于存储子字符串的最终计数(例如ans )。
2. 初始化映射M来存储其中所有字符的频率。
3. 遍历给定的字符串，对于每个字符，请按照以下步骤操作：
   • 迭代地图M。
   • 如果第一个元素(即映射的键)与当前字符，则继续。
   • 否则，添加与当前字符相对应的值。
4. 遍历Map后，将cur添加到最终结果中，即ans + = cur 。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count the substrings
// having different first & last character
int countSubstring(string s, int n)
{
    // Stores frequency of each char
    map m;
 
    // Loop to store frequency of
    // the characters in a Map
    for (int i = 0; i < n; i++)
        m[s[i]]++;
 
    // To store final result
    int ans = 0;
 
    // Traversal of string
    for (int i = 0; i < n; i++) {
 
        // Store answer for every
        // iteration
        int cnt = 0;
        m[s[i]]--;
 
        // Map traversal
        for (auto value : m) {
 
            // Compare current char
            if (value.first == s[i]) {
                continue;
            }
            else {
                cnt += value.second;
            }
        }
 
        ans += cnt;
    }
 
    // Print the final count
    cout << ans;
}
 
// Driver Code
int main()
{
    // Given string
    string S = "abcab";
 
    // Length of the string
    int N = 5;
 
    // Function Call
    countSubstring(S, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to count the subStrings
// having different first & last character
static void countSubString(char []s, int n)
{
     
    // Stores frequency of each char
    HashMap mp = new HashMap();
 
    // Loop to store frequency of
    // the characters in a Map
    for(int i = 0; i < n; i++)
 
        if (mp.containsKey(s[i]))
        {
            mp.put(s[i], mp.get(s[i]) + 1);
        }
        else
        {
            mp.put(s[i], 1);
        }
 
    // To store final result
    int ans = 0;
 
    // Traversal of String
    for(int i = 0; i < n; i++)
    {
         
        // Store answer for every
        // iteration
        int cnt = 0;
        if (mp.containsKey(s[i]))
        {
            mp.put(s[i], mp.get(s[i]) - 1);
 
            // Map traversal
            for(Map.Entry value : mp.entrySet())
            {
                 
                // Compare current char
                if (value.getKey() == s[i])
                {
                    continue;
                }
                else
                {
                    cnt += value.getValue();
                }
            }
            ans += cnt;
        }
    }
     
    // Print the final count
    System.out.print(ans);
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given String
    String S = "abcab";
 
    // Length of the String
    int N = 5;
 
    // Function call
    countSubString(S.toCharArray(), N);
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach
 
# Function to count the substrings
# having different first & last character
def countSubstring(s, n):
  
    # Stores frequency of each char
    m = {}
 
    # Loop to store frequency of
    # the characters in a Map
    for i in range(n):
        if s[i] in m:
            m[s[i]] += 1
        else:
            m[s[i]] = 1
 
    # To store final result
    ans = 0
 
    # Traversal of string
    for i in range(n):
 
        # Store answer for every
        # iteration
        cnt = 0
         
        if s[i] in m:
            m[s[i]] -= 1
        else:
            m[s[i]] = -1
 
        # Map traversal
        for value in m:
 
            # Compare current char
            if (value == s[i]):
                continue
            else:
                cnt += m[value]
 
        ans += cnt
 
    # Print the final count
    print(ans)
 
# Driver code
 
# Given string
S = "abcab"
 
# Length of the string
N = 5
 
# Function Call
countSubstring(S, N)
 
# This code is contributed by divyeshrabadiya07

C＃

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to count the subStrings
// having different first & last character
static void countSubString(char []s, int n)
{
     
    // Stores frequency of each char
    Dictionary mp = new Dictionary();
 
    // Loop to store frequency of
    // the characters in a Map
    for(int i = 0; i < n; i++)
 
        if (mp.ContainsKey(s[i]))
        {
            mp[s[i]] = mp[s[i]] + 1;
        }
        else
        {
            mp.Add(s[i], 1);
        }
 
    // To store readonly result
    int ans = 0;
 
    // Traversal of String
    for(int i = 0; i < n; i++)
    {
         
        // Store answer for every
        // iteration
        int cnt = 0;
        if (mp.ContainsKey(s[i]))
        {
            mp[s[i]] = mp[s[i]] - 1;
 
            // Map traversal
            foreach(KeyValuePair value in mp)
            {
                 
                // Compare current char
                if (value.Key == s[i])
                {
                    continue;
                }
                else
                {
                    cnt += value.Value;
                }
            }
            ans += cnt;
        }
    }
     
    // Print the readonly count
    Console.Write(ans);
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given String
    String S = "abcab";
 
    // Length of the String
    int N = 5;
 
    // Function call
    countSubString(S.ToCharArray(), N);
}
}
 
// This code is contributed by Amit Katiyar

输出：

8

时间复杂度： O(N * 26)
辅助空间： O(N)