在字符串的不同子字符串中查找不同的字符

给定一个字符串str ，任务是找出给定字符串的所有不同子字符串中不同字符的计数。
例子：

Input: str = “ABCA” 
Output: 18 
 

Distinct sub-strings	Distinct characters
A	1
AB	2
ABC	3
ABCA	3
B	1
BC	2
BCA	3
C	1
CA	2

Hence, 1 + 2 + 3 + 3 + 1 + 2 + 3 + 1 + 2 = 18
Input: str = “AAAB” 
Output: 10 
 

方法：取给定字符串的所有可能的子字符串，并使用一个集合来检查当前子字符串之前是否已经被处理过。现在，对于每个不同的子字符串，计算其中的不同字符（同样可以使用 set 来计算）。所有不同子字符串的计数总和是最终答案。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the count of distinct
// characters in all the distinct
// sub-strings of the given string
int countTotalDistinct(string str)
{
    int cnt = 0;
 
    // To store all the sub-strings
    set items;
 
    for (int i = 0; i < str.length(); ++i) {
 
        // To store the current sub-string
        string temp = "";
 
        // To store the characters of the
        // current sub-string
        set ans;
        for (int j = i; j < str.length(); ++j) {
            temp = temp + str[j];
            ans.insert(str[j]);
 
            // If current sub-string hasn't
            // been stored before
            if (items.find(temp) == items.end()) {
 
                // Insert it into the set
                items.insert(temp);
 
                // Update the count of
                // distinct characters
                cnt += ans.size();
            }
        }
    }
 
    return cnt;
}
 
// Driver code
int main()
{
    string str = "ABCA";
 
    cout << countTotalDistinct(str);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.HashSet;
 
class geeks
{
 
    // Function to return the count of distinct
    // characters in all the distinct
    // sub-strings of the given string
    public static int countTotalDistinct(String str)
    {
        int cnt = 0;
 
        // To store all the sub-strings
        HashSet items = new HashSet<>();
 
        for (int i = 0; i < str.length(); ++i)
        {
 
            // To store the current sub-string
            String temp = "";
 
            // To store the characters of the
            // current sub-string
            HashSet ans = new HashSet<>();
            for (int j = i; j < str.length(); ++j)
            {
                temp = temp + str.charAt(j);
                ans.add(str.charAt(j));
 
                // If current sub-string hasn't
                // been stored before
                if (!items.contains(temp))
                {
 
                    // Insert it into the set
                    items.add(temp);
 
                    // Update the count of
                    // distinct characters
                    cnt += ans.size();
                }
            }
        }
 
        return cnt;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String str = "ABCA";
        System.out.println(countTotalDistinct(str));
    }
}
 
// This code is contributed by
// sanjeev2552

Python3

# Python3 implementation of the approach
 
# Function to return the count of distinct
# characters in all the distinct
# sub-strings of the given string
def countTotalDistinct(string) :
 
    cnt = 0;
 
    # To store all the sub-strings
    items = set();
 
    for i in range(len(string)) :
 
        # To store the current sub-string
        temp = "";
 
        # To store the characters of the
        # current sub-string
        ans = set();
        for j in range(i, len(string)) :
            temp = temp + string[j];
            ans.add(string[j]);
 
            # If current sub-string hasn't
            # been stored before
            if temp not in items :
 
                # Insert it into the set
                items.add(temp);
 
                # Update the count of
                # distinct characters
                cnt += len(ans);
 
    return cnt;
 
 
# Driver code
if __name__ == "__main__" :
 
    string = "ABCA";
 
    print(countTotalDistinct(string));
     
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class geeks
{
 
    // Function to return the count of distinct
    // characters in all the distinct
    // sub-strings of the given string
    public static int countTotalDistinct(String str)
    {
        int cnt = 0;
 
        // To store all the sub-strings
        HashSet items = new HashSet();
 
        for (int i = 0; i < str.Length; ++i)
        {
 
            // To store the current sub-string
            String temp = "";
 
            // To store the characters of the
            // current sub-string
            HashSet ans = new HashSet();
            for (int j = i; j < str.Length; ++j)
            {
                temp = temp + str[j];
                ans.Add(str[j]);
 
                // If current sub-string hasn't
                // been stored before
                if (!items.Contains(temp))
                {
 
                    // Insert it into the set
                    items.Add(temp);
 
                    // Update the count of
                    // distinct characters
                    cnt += ans.Count;
                }
            }
        }
 
        return cnt;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        String str = "ABCA";
        Console.WriteLine(countTotalDistinct(str));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// js implementation of the approach
 
// Function to return the count of distinct
// characters in all the distinct
// sub-strings of the given string
function countTotalDistinct(str)
{
    let cnt = 0;
 
    // To store all the sub-strings
    let items = new Set();
 
    for (let i = 0; i < str.length; ++i) {
 
        // To store the current sub-string
        let temp = "";
 
        // To store the characters of the
        // current sub-string
        let ans = new Set();
        for (let j = i; j < str.length; ++j) {
            temp = temp + str[j];
            ans.add(str[j]);
            // If current sub-string hasn't
            // been stored before
            if (!items.has(temp)) {
 
                // Insert it into the set
                items.add(temp);
 
                // Update the count of
                // distinct characters
                cnt += ans.size;
            }
        }
    }
 
    return cnt;
}
 
// Driver code
let str = "ABCA";
document.write(countTotalDistinct(str));
 
 

输出：

18

时间复杂度： O(n^2)

由于使用了嵌套循环，因此如果 n^2 复杂度是有序的

空间复杂度： O(n)

使用了两组大小 n，因此复杂度将是 O(2n)，只不过是 O(n)。