伯爵具有字符串作为第一个字符中最常见的字符子的

给定一个字符串s组成的大小为N小写字母的，任务是要算其中包含字符串的第一个字符中最常见的字符的所有字符串。

注意：如果超过一个字符具有最大频率，则考虑其中字典序最小的字符。

例子：

Input: S = “abcab”
Output: 7
Explanation:
There are two characters a and b occurring maximum times i.e., 2 times.
Selecting the lexicographically smaller character i.e. ‘a’.
Substrings starts with ‘a’ are: “a”, “ab”, “abc”, “abca”, “abcab”, “a”, “ab”.
Therefore the count is 7.

Input: S= “cccc”
Output: 10

做法：思路是先找出出现次数最多的字符，然后统计字符串中以该字符开头的子字符串。请按照以下步骤解决问题：

• 将计数初始化为0 ，这将存储字符串的总数。
• 找出字符串S 中出现次数最多的字符。令该字符为ch 。
• 使用变量i遍历字符串，如果第i^个索引处的字符与ch相同，则将计数增加(N – i) 。
• 完成以上步骤后，打印count的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include
using namespace std;
 
// Function to find all substrings
// whose first character occurs
// maximum number of times
int substringCount(string s)
{
     
    // Stores freqency of characters
    vector freq(26, 0);
 
    // Stores character that appears
    // maximum number of times
    char max_char = '#';
 
    // Stores max freqency of character
    int maxfreq = INT_MIN;
 
    // Updates frequecy of characters
    for(int i = 0; i < s.size(); i++)
    {
        freq[s[i] - 'a']++;
 
        // Update maxfreq
        if (maxfreq < freq[s[i] - 'a'])
            maxfreq = freq[s[i] - 'a'];
    }
 
    // Character that occures
    // maximum number of times
    for(int i = 0; i < 26; i++)
    {
         
        // Update the maximum frequency
        // character
        if (maxfreq == freq[i])
        {
            max_char = (char)(i + 'a');
            break;
        }
    }
 
    // Stores all count of substrings
    int ans = 0;
 
    // Traverse over string
    for(int i = 0; i < s.size(); i++)
    {
         
        // Get the current character
        char ch = s[i];
         
        // Update count of substrings
        if (max_char == ch)
        {
            ans += (s.size() - i);
        }
    }
     
    // Return the count of all
    // valid substrings
    return ans;
}
 
// Driver Code
int main()
{
    string S = "abcab";
 
    // Function Call
    cout << (substringCount(S));
}
 
// This code is contributed by mohit kumar 29

Java

// Java program for the above approach
import java.util.*;
 
class GFG {
 
    // Function to find all substrings
    // whose first character occurs
    // maximum number of times
    static int substringCount(String s)
    {
 
        // Stores freqency of characters
        int[] freq = new int[26];
 
        // Stores character that appears
        // maximum number of times
        char max_char = '#';
 
        // Stores max freqency of character
        int maxfreq = Integer.MIN_VALUE;
 
        // Updates frequecy of characters
        for (int i = 0;
             i < s.length(); i++) {
            freq[s.charAt(i) - 'a']++;
 
            // Update maxfreq
            if (maxfreq
                < freq[s.charAt(i) - 'a'])
                maxfreq
                    = freq[s.charAt(i) - 'a'];
        }
 
        // Character that occures
        // maximum number of times
        for (int i = 0; i < 26; i++) {
 
            // Update the maximum frequency
            // character
            if (maxfreq == freq[i]) {
                max_char = (char)(i + 'a');
                break;
            }
        }
 
        // Stores all count of substrings
        int ans = 0;
 
        // Traverse over string
        for (int i = 0;
             i < s.length(); i++) {
 
            // Get the current character
            char ch = s.charAt(i);
 
            // Update count of substrings
            if (max_char == ch) {
                ans += (s.length() - i);
            }
        }
 
        // Return the count of all
        // valid substrings
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        String S = "abcab";
 
        // Function Call
        System.out.println(substringCount(S));
    }
}

Python3

# Python3 program for the above approach
import sys
 
# Function to find all substrings
# whose first character occurs
# maximum number of times
def substringCount(s):
     
    # Stores freqency of characters
    freq = [0 for i in range(26)]
 
    # Stores character that appears
    # maximum number of times
    max_char = '#'
 
    # Stores max freqency of character
    maxfreq = -sys.maxsize - 1
 
    # Updates frequecy of characters
    for i in range(len(s)):
        freq[ord(s[i]) - ord('a')] += 1
 
        # Update maxfreq
        if (maxfreq < freq[ord(s[i]) - ord('a')]):
            maxfreq = freq[ord(s[i]) - ord('a')]
 
    # Character that occures
    # maximum number of times
    for i in range(26):
         
        # Update the maximum frequency
        # character
        if (maxfreq == freq[i]):
            max_char = chr(i + ord('a'))
            break
 
    # Stores all count of substrings
    ans = 0
 
    # Traverse over string
    for i in range(len(s)):
         
        # Get the current character
        ch = s[i]
         
        # Update count of substrings
        if (max_char == ch):
            ans += (len(s) - i)
     
    # Return the count of all
    # valid substrings
    return ans
 
# Driver Code
if __name__ == '__main__':
     
    S = "abcab"
     
    # Function Call
    print(substringCount(S))
 
# This code is contributed by ipg2016107

C#

// C# program for the above approach
using System;
  
class GFG{
     
// Function to find all substrings
// whose first character occurs
// maximum number of times
static int substringCount(string s)
{
     
    // Stores freqency of characters
    int[] freq = new int[26];
     
    // Stores character that appears
    // maximum number of times
    char max_char = '#';
 
    // Stores max freqency of character
    int maxfreq = Int32.MinValue;
 
    // Updates frequecy of characters
    for(int i = 0; i < s.Length; i++)
    {
        freq[s[i] - 'a']++;
 
        // Update maxfreq
        if (maxfreq < freq[s[i] - 'a'])
            maxfreq = freq[s[i] - 'a'];
    }
 
    // Character that occures
    // maximum number of times
    for(int i = 0; i < 26; i++)
    {
         
        // Update the maximum frequency
        // character
        if (maxfreq == freq[i])
        {
            max_char = (char)(i + 'a');
            break;
        }
    }
 
    // Stores all count of substrings
    int ans = 0;
 
    // Traverse over string
    for(int i = 0; i < s.Length; i++)
    {
         
        // Get the current character
        char ch = s[i];
 
        // Update count of substrings
        if (max_char == ch)
        {
            ans += (s.Length - i);
        }
    }
 
    // Return the count of all
    // valid substrings
    return ans;
}
 
// Driver Code
public static void Main()
{
    string S = "abcab";
 
    // Function Call
    Console.WriteLine(substringCount(S));
}
}
 
// This code is contributed by susmitakundugoaldanga

Javascript

 
// JavaScript program for the above approach
 
// Function to find all substrings
// whose first character occurs
// maximum number of times
function substringCount(s)
{
     
    // Stores freqency of characters
    var freq = new Array(26).fill(0);
     
    // Stores character that appears
    // maximum number of times
    var max_char = "#";
     
    // Stores max freqency of character
    var maxfreq = -21474836487;
     
    // Updates frequecy of characters
    for(var i = 0; i < s.length; i++)
    {
        freq[s[i].charCodeAt(0) -
              "a".charCodeAt(0)]++;
     
        // Update maxfreq
        if (maxfreq < freq[s[i].charCodeAt(0) -
                            "a".charCodeAt(0)])
            maxfreq = freq[s[i].charCodeAt(0) -
                            "a".charCodeAt(0)];
    }
     
    // Character that occures
    // maximum number of times
    for(var i = 0; i < 26; i++)
    {
         
        // Update the maximum frequency
        // character
        if (maxfreq === freq[i])
        {
            max_char = String.fromCharCode(
                i + "a".charCodeAt(0));
            break;
        }
    }
     
    // Stores all count of substrings
    var ans = 0;
     
    // Traverse over string
    for(var i = 0; i < s.length; i++)
    {
         
        // Get the current character
        var ch = s[i];
         
        // Update count of substrings
        if (max_char === ch)
        {
            ans += s.length - i;
        }
    }
     
    // Return the count of all
    // valid substrings
    return ans;
}
 
// Driver Code
var S = "abcab";
 
// Function Call
document.write(substringCount(S));
 
// This code is contributed by rdtank
 

输出：

7

时间复杂度： O(N)
辅助空间： O(1)

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