给一根绳子 str 。任务是通过在给定字符串中用字母替换“$”来找到所有可能的字符串数。 笔记 ：字母表的位置应确保字符串的元音和辅音始终交替，并且字符串必须始终以辅音开头。假设这样的字符串总是可能的，即不需要关心“$”以外的字符。 例子 :
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Input: str = "y$s"Output: 5$ can be replaced with any of the 5 vowels.So, there can be 5 strings.Input: str = "s$$e$"Output: 2205
方法： 已知字符串将以辅音开头。所以，如果“$”位于偶数位置（考虑到基于0的索引），那么应该有一个辅音，否则应该有一个元音。此外，考虑到不需要关心“$”以外的字符，即“$”以外的字符被正确地放置在字符串中，以保持交替的辅音和元音序列。让我们用一个例子来理解这个问题。
str=“s$$e$” 在这里，我们必须找到使用给定约束形成字符串的多种方法。
- $第一次出现在第二个位置，即第一个索引，因此我们可以使用5个元音。
- $第二次出现在第三位，所以我们可以使用21个辅音。
- $第三次出现在第五位，所以我们可以使用21个辅音。
所以，形成上述字符串的方法总数是 = 5*21*21 = 2205
以下是上述方法的实施情况：
C++
// C++ implementation of above approach #include <bits/stdc++.h> using namespace std; // Function to find the count of strings int countStrings(string s) { // Variable to store the final result long sum = 1; // Loop iterating through string for ( int i = 0; i < s.size(); i++) { // If '$' is present at the even // position in the string if (i % 2 == 0 && s[i] == '$' ) //'sum' is multiplied by 21 sum *= 21; // If '$' is present at the odd // position in the string else if (s[i] == '$' ) //'sum' is multiplied by 5 sum *= 5; } return sum; } // Driver code int main() { // Let the string 'str' be s$$e$ string str = "s$$e$" ; // Print result cout << countStrings(str) << endl; return 0; } |
JAVA
// Java implementation of above approach import java.util.*; import java.lang.*; import java.io.*; class GFG{ // Function to find the count of strings static int countStrings(String s) { // Variable to store the final result int sum = 1 ; // Loop iterating through string for ( int i = 0 ; i < s.length(); i++) { // If '$' is present at the even // position in the string if (i % 2 == 0 && s.charAt(i) == '$' ) //'sum' is multiplied by 21 sum *= 21 ; // If '$' is present at the odd // position in the string else if (s.charAt(i) == '$' ) //'sum' is multiplied by 5 sum *= 5 ; } return sum; } // Driver code public static void main(String args[]) { // Let the string 'str' be s$$e$ String str = "s$$e$" ; // Print result System.out.println(countStrings(str)); } } |
Python 3
# Python 3 implementation of above approach # Function to find the count of strings def countStrings(s): # Variable to store the final result sum = 1 # Loop iterating through string for i in range ( len (s)): # If '$' is present at the even # position in the string if (i % 2 = = 0 and s[i] = = '$' ): #'sum' is multiplied by 21 sum * = 21 # If '$' is present at the odd # position in the string elif (s[i] = = '$' ): # 'sum' is multiplied by 5 sum * = 5 return sum # Driver code if __name__ = = "__main__" : # Let the string 'str' be s$$e$ str = "s$$e$" # Print result print (countStrings( str )) # this code is contributed by ChitraNayal |
C#
// C# implementation of above approach using System; class GFG{ // Function to find the count of strings static int countStrings(String s) { // Variable to store the final result int sum = 1; // Loop iterating through string for ( int i = 0; i < s.Length; i++) { // If '$' is present at the even // position in the string if (i % 2 == 0 && s[i] == '$' ) //'sum' is multiplied by 21 sum *= 21; // If '$' is present at the odd // position in the string else if (s[i] == '$' ) //'sum' is multiplied by 5 sum *= 5; } return sum; } // Driver code public static void Main() { // Let the string 'str' be s$$e$ String str = "s$$e$" ; // Print result Console.WriteLine(countStrings(str)); } } |
PHP
<?php // PHP implementation of above approach // Function to find the count of strings function countStrings( $s ) { // Variable to store the // final result $sum = 1; // Loop iterating through string for ( $i = 0; $i < strlen ( $s ); $i ++) { // If '$' is present at the even // position in the string if ( $i % 2 == 0 && $s [ $i ] == '$' ) //'sum' is multiplied by 21 $sum *= 21; // If '$' is present at the odd // position in the string else if ( $s [ $i ] == '$' ) //'sum' is multiplied by 5 $sum *= 5; } return $sum ; } // Driver code // Let the string 'str' be s$$e$ $str = "s$$e$" ; // Print result echo countStrings( $str ); // This code is contributed by Ryuga ?> |
Javascript
<script> // javascript implementation of above approach // Function to find the count of strings function countStrings( s) { // Variable to store the final result let sum = 1; // Loop iterating through string for (let i = 0; i < s.length; i++) { // If '$' is present at the even // position in the string if (i % 2 == 0 && s[i] == '$' ) //'sum' is multiplied by 21 sum *= 21; // If '$' is present at the odd // position in the string else if (s[i] == '$' ) //'sum' is multiplied by 5 sum *= 5; } return sum; } // Driver code // Let the string 'str' be s$$e$ let str = "s$$e$" ; // Print result document.write(countStrings(str)); // This code is contributed by gauravrajput1 </script> |
输出：
2205
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