可以使用超出范围的字符来形成范围为[L，R]的子字符串的方式的数目

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📜 可以使用超出范围的字符来形成范围为[L，R]的子字符串的方式的数目

📅 最后修改于: 2021-04-27 19:57:10 🧑 作者: Mango

给定字符串S和范围[L，R] 。任务是找到使用存在于字符串但不在S [L，R]范围内的字符构造S [L，R]范围内的子字符串的方法数量。

例子：

Input: s = “cabcaab”, l = 1, r = 3
Output: 2
The substring is “abc”
s[4] + s[6] + s[0] = ‘a’ + ‘b’ + ‘c’ = “abc”
s[5] + s[6] + s[0] = ‘a’ + ‘b’ + ‘c’ = “abc”

Input: s = “aaaa”, l = 1, r = 2
Output: 2

为什么编程需要懂一点英语

方法：可以使用哈希表和组合器解决问题。可以按照以下步骤解决上述问题：

计算哈希表中每个不在L和R范围内的字符的频率(例如freq)。
从L到R分别进行迭代，并计算路数。
对于范围L和R中的每个字符，路数乘以freq [s [i]-‘a’]，并将freq [s [i]-‘a’]的值减少1。
如果freq [s [i]-‘a’]的值为0，则我们没有任何字符可以填充该位置，因此路数将为0。
最后，整体乘法将是我们的答案。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the number of
// ways to form the sub-string
int calculateWays(string s, int n, int l, int r)
{
  
    // Initialize a hash-table
    // with 0
    int freq[26];
    memset(freq, 0, sizeof freq);
  
    // Iterate in the string and count
    // the frequency of characters that
    // do not lie in the range L and R
    for (int i = 0; i < n; i++) {
  
        // Out of range characters
        if (i < l || i > r)
            freq[s[i] - 'a']++;
    }
  
    // Stores the final number of ways
    int ways = 1;
  
    // Iterate for the sub-string in the range
    // L and R
    for (int i = l; i <= r; i++) {
  
        // If exists then mulitply
        // the number of ways and
        // decrement the frequency
        if (freq[s[i] - 'a']) {
            ways = ways * freq[s[i] - 'a'];
            freq[s[i] - 'a']--;
        }
  
        // If does not exist
        // the sub-string cannot be formed
        else {
            ways = 0;
            break;
        }
    }
  
    // Return the answer
    return ways;
}
  
// Driver code
int main()
{
    string s = "cabcaab";
    int n = s.length();
  
    int l = 1, r = 3;
    cout << calculateWays(s, n, l, r);
  
    return 0;
}

Java

// Java implementation of the approach 
class GfG {
  
// Function to return the number of 
// ways to form the sub-string 
static int calculateWays(String s, int n, int l, int r) 
{ 
  
    // Initialize a hash-table 
    // with 0 
    int freq[] = new int[26]; 
  
    // Iterate in the string and count 
    // the frequency of characters that 
    // do not lie in the range L and R 
    for (int i = 0; i < n; i++) { 
  
        // Out of range characters 
        if (i < l || i > r) 
            freq[s.charAt(i)-'a']++; 
    } 
  
    // Stores the final number of ways 
    int ways = 1; 
  
    // Iterate for the sub-string in the range 
    // L and R 
    for (int i = l; i <= r; i++) { 
  
        // If exists then mulitply 
        // the number of ways and 
        // decrement the frequency 
        if (freq[s.charAt(i) - 'a'] != 0) { 
            ways = ways * freq[s.charAt(i) - 'a']; 
            freq[s.charAt(i) - 'a']--; 
        } 
  
        // If does not exist 
        // the sub-string cannot be formed 
        else { 
            ways = 0; 
            break; 
        } 
    } 
  
    // Return the answer 
    return ways; 
} 
  
// Driver code 
public static void main(String[] args) 
{ 
    String s = "cabcaab"; 
    int n = s.length(); 
  
    int l = 1, r = 3; 
    System.out.println(calculateWays(s, n, l, r)); 
  
} 
}

Python3

# Python 3 implementation of the approach
  
# Function to return the number of
# ways to form the sub-string
def calculateWays(s, n, l, r):
      
    # Initialize a hash-table
    # with 0
    freq = [0 for i in range(26)]
  
    # Iterate in the string and count
    # the frequency of characters that
    # do not lie in the range L and R
    for i in range(n):
          
        # Out of range characters
        if (i < l or i > r):
            freq[ord(s[i]) - ord('a')] += 1
  
    # Stores the final number of ways
    ways = 1
  
    # Iterate for the sub-string in the range
    # L and R
    for i in range(l, r + 1, 1):
          
        # If exists then mulitply
        # the number of ways and
        # decrement the frequency
        if (freq[ord(s[i]) - ord('a')]):
            ways = ways * freq[ord(s[i]) - ord('a')]
            freq[ord(s[i]) - ord('a')] -= 1
  
        # If does not exist
        # the sub-string cannot be formed
        else:
            ways = 0
            break
  
    # Return the answer
    return ways
  
# Driver code
if __name__ == '__main__':
    s = "cabcaab"
    n = len(s)
  
    l = 1
    r = 3
    print(calculateWays(s, n, l, r))
  
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach 
using System;
  
class GfG
{
  
// Function to return the number of 
// ways to form the sub-string 
static int calculateWays(String s, int n, int l, int r) 
{ 
  
    // Initialize a hash-table 
    // with 0 
    int []freq = new int[26]; 
  
    // Iterate in the string and count 
    // the frequency of characters that 
    // do not lie in the range L and R 
    for (int i = 0; i < n; i++) 
    { 
  
        // Out of range characters 
        if (i < l || i > r) 
            freq[s[i]-'a']++; 
    } 
  
    // Stores the final number of ways 
    int ways = 1; 
  
    // Iterate for the sub-string in the range 
    // L and R 
    for (int i = l; i <= r; i++) 
    { 
  
        // If exists then mulitply 
        // the number of ways and 
        // decrement the frequency 
        if (freq[s[i] - 'a'] != 0) { 
            ways = ways * freq[s[i] - 'a']; 
            freq[s[i] - 'a']--; 
        } 
  
        // If does not exist 
        // the sub-string cannot be formed 
        else { 
            ways = 0; 
            break; 
        } 
    } 
  
    // Return the answer 
    return ways; 
} 
  
// Driver code 
public static void Main() 
{ 
    String s = "cabcaab"; 
    int n = s.Length; 
  
    int l = 1, r = 3; 
    Console.WriteLine(calculateWays(s, n, l, r)); 
  
} 
} 
  
/* This code contributed by PrinciRaj1992 */

PHP

 $r)
            $freq[ord($s[$i]) - 97]++;
    }
  
    // Stores the final number of ways
    $ways = 1;
  
    // Iterate for the sub-string in the range
    // L and R
    for ($i = $l; $i <= $r; $i++) 
    {
  
        // If exists then mulitply
        // the number of ways and
        // decrement the frequency
        if ($freq[ord($s[$i]) - 97]) 
        {
            $ways = $ways * $freq[ord($s[$i]) - 97];
            $freq[ord($s[$i]) - 97]--;
        }
  
        // If does not exist
        // the sub-string cannot be formed
        else 
        {
            $ways = 0;
            break;
        }
    }
  
    // Return the answer
    return $ways;
}
  
// Driver code
$s = "cabcaab";
$n = strlen($s);
  
$l = 1;
$r = 3;
echo calculateWays($s, $n, $l, $r);
  
// This code is contributed by ihritik
?>

输出：

时间复杂度： O(N)，其中N是字符串的长度。
辅助空间： O(1)