长度为K的子字符串包含相同字符

给定一个字符串’str’和一个整数’k’，任务是计算长度为’k’的包含相同字符的子字符串的数量。给定的字符串仅包含小写字母。
例子：

Input: str = "aaaabbbccdddd", k=4
Output: 2
The sub-strings of length 4 which contain identical 
characters are 'aaaa' and 'dddd'. So, the count is 2.

Input: str = "aaaaaa", k=4
Output: 3

一种简单的方法：

• 查找所有的字符串是长的“K”子串。
• 检查这些子字符串是否仅由’1’字符。
• 如果满足以上条件，则增加计数。
• 显示计数。

高效的方法：我们将使用“窗口滑动”技术来解决此问题。

• 取一个长度为’k’的子字符串(这是第一个’k’字符)。
• 然后，将下一个字符添加到子字符串。
• 如果子字符串的长度大于’k’，则从字符串的开头删除字符。
• 并且，借助映射来计数子字符串中字符的频率。
• 如果子字符串的一个字符的频率等于子字符串本身的长度，即所有字符都相同。
• 然后，增加计数，否则重复上述步骤，直到最后没有其他字符可添加为止。
• 最后显示总数。

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
 
// Function that counts all the
// sub-strings of length 'k'
// which have all identical characters
void solve(string s, int k)
{
    // count of sub-strings, length,
    // initial position of sliding window
    int count = 0, length = 0, pos = 0;
 
    // map to store the frequency of
    // the characters of sub-string
    map m;
 
    for (int i = 0; i < s.length(); i++) {
 
        // increase the frequency of the character
        // and length of the sub-string
        m[s[i]]++;
        length++;
 
        // if the length of the sub-string
        // is greater than K
        if (length > k) {
 
            // remove the character from
            // the beginning of sub-string
            m[s[pos++]]--;
            length--;
        }
 
        // if the length of the sub string
        // is equal to k and frequency of one
        // of its characters is equal to the
        // length of the sub-string
        // i.e. all the characters are same
        // increase the count
        if (length == k && m[s[i]] == length)
            count++;
    }
 
    // display the number
    // of valid sub-strings
    cout << count << endl;
}
 
// Driver code
int main()
{
    string s = "aaaabbbccdddd";
    int k = 4;
 
    solve(s, k);
 
    return 0;
}

Java

// Java implementation of above approach
import java.util.*;
 
class GFG
{
    // Function that counts all the
    // sub-strings of length 'k'
    // which have all identical characters
    static void solve(String s, int k)
    {
        // count of sub-strings, length,
        // initial position of sliding window
        int count = 0, length = 0, pos = 0;
     
        // map to store the frequency of
        // the characters of sub-string
        HashMap m =
                            new HashMap();
     
        for (int i = 0; i < s.length(); i++)
        {
     
            // increase the frequency of the character
            // and length of the sub-string
            if(m.containsKey(s.charAt(i)))
                m.put(s.charAt(i), m.get(s.charAt(i))+1);
            else
                m.put(s.charAt(i), 1);
                 
            length++;
     
            // if the length of the sub-string
            // is greater than K
            if (length > k)
            {
     
                // remove the character from
                // the beginning of sub-string
                m.put(s.charAt(pos), m.get(s.charAt(pos)) -1 );
                pos++;
                length--;
            }
     
            // if the length of the sub string
            // is equal to k and frequency of one
            // of its characters is equal to the
            // length of the sub-string
            // i.e. all the characters are same
            // increase the count
            if (length == k && m.get(s.charAt(i)) == length)
                count++;
        }
     
        // display the number
        // of valid sub-strings
        System.out.println( count);
    }
     
    // Driver code
    public static void main (String[] args)
    {
         
        String s = "aaaabbbccdddd";
        int k = 4;
     
        solve(s, k);
    }
}
 
// This code is contributed by ihritik

Python 3

# Python3 implementation of above
# approach
 
# Function that counts all the
# sub-strings of length 'k'
# which have all identical characters
def solve(s, k) :
 
    # count of sub-strings, length,
    # initial position of sliding window
    count, length, pos = 0, 0, 0
 
    # dictionary to store the frequency of
    # the characters of sub-string
    m = dict.fromkeys(s,0)
 
    for i in range(len(s)) :
         
        # increase the frequency of the character
        # and length of the sub-string
        m[s[i]] += 1
        length += 1
 
        # if the length of the sub-string
        # is greater than K
        if length > k :
 
            # remove the character from
            # the beginning of sub-string
            m[s[pos]] -= 1
            pos += 1
            length -= 1
 
        # if the length of the sub string
        # is equal to k and frequency of one
        # of its characters is equal to the
        # length of the sub-string
        # i.e. all the characters are same
        # increase the count
        if length == k and m[s[i]] == length :
            count += 1
 
    # display the number
    # of valid sub-strings
    print(count)
             
 
 
# Driver code    
if __name__ == "__main__" :
 
    s = "aaaabbbccdddd"
    k = 4
 
    # Function call
    solve(s, k)
                 
# This code is contributed by
# ANKITRAI1

C#

// C# implementation of above approach
 
using System;
using System.Collections.Generic;
class GFG
{
    // Function that counts all the
    // sub-strings of length 'k'
    // which have all identical characters
    static void solve(string s, int k)
    {
        // count of sub-strings, length,
        // initial position of sliding window
        int count = 0, length = 0, pos = 0;
     
        // map to store the frequency of
        // the characters of sub-string
        Dictionary m =
                            new Dictionary();
     
        for (int i = 0; i < s.Length; i++) {
     
            // increase the frequency of the character
            // and length of the sub-string
            if(m.ContainsKey(s[i]))
                m[s[i]]++;
            else
                m[s[i]] = 1;
                 
            length++;
     
            // if the length of the sub-string
            // is greater than K
            if (length > k) {
     
                // remove the character from
                // the beginning of sub-string
                m[s[pos]]--;
                pos++;
                length--;
            }
     
            // if the length of the sub string
            // is equal to k and frequency of one
            // of its characters is equal to the
            // length of the sub-string
            // i.e. all the characters are same
            // increase the count
            if (length == k && m[s[i]] == length)
                count++;
        }
     
        // display the number
        // of valid sub-strings
        Console.WriteLine(count);
    }
     
    // Driver code
    public static void Main () {
         
        string s = "aaaabbbccdddd";
        int k = 4;
     
        solve(s, k);
     
    }
 
}
 
 
// This code is contributed by ihritik

PHP

 $k)
        {
 
            // remove the character from
            // the beginning of sub-string
            $m[$s[$pos++]]--;
            $length--;
        }
 
        // if the length of the sub string
        // is equal to k and frequency of one
        // of its characters is equal to the
        // length of the sub-string
        // i.e. all the characters are same
        // increase the count
        if ($length == $k && $m[$s[$i]] == $length)
            $count++;
    }
 
    // display the number
    // of valid sub-strings
    echo $count . "\n";
}
 
// Driver code
$s = "aaaabbbccdddd";
$k = 4;
 
solve($s, $k);
 
// This code is contributed
// by ChitraNayal
?>

Javascript

 
 
// Javascript implementation of above approach
 
// Function that counts all the
// sub-strings of length 'k'
// which have all identical characters
function solve( s,  k)
{
    // count of sub-strings, length,
    // initial position of sliding window
    var count = 0, length = 0, pos = 0;
 
    // map to store the frequency of
    // the characters of sub-string
    var m = new Map();
 
    for (var i = 0; i < s.length; i++) {
 
        // increase the frequency of the character
        // and length of the sub-string
        if(!m.has(s[i]))
        {
            m.set(s[i], 0);
        }
        m.set(s[i], m.get(s[i])+1);
        length++;
 
        // if the length of the sub-string
        // is greater than K
        if (length > k) {
 
            // remove the character from
            // the beginning of sub-string
            if(!m.has(s[pos]))
            {
                m.set(s[pos], 0);
            }
            m.set(s[pos], m[s[pos]]-1);
            pos+=1;
            length--;
        }
 
        // if the length of the sub string
        // is equal to k and frequency of one
        // of its characters is equal to the
        // length of the sub-string
        // i.e. all the characters are same
        // increase the count
        if (length == k && m.get(s[i]) == length)
            count++;
    }
 
    // display the number
    // of valid sub-strings
    document.write( count);
}
 
// Driver code
var s = "aaaabbbccdddd";
var k = 4;
solve(s, k);
 
 

输出：

2

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