Given a string str. The task is to find the product of ASCII values of characters in the string.
Examples:
Input: str = “IS”
Output: 6059
73 * 83 = 6059Input: str = “GfG”
Output: 514182
The idea is to start with iterating through characters of the string and multiply their ASCII values to a variable namely, prod. Hence, return prod after the complete iteration of the string.
Note: If the string is large, the program may cause segmentation fault because of the limited size of an int.
Implementation:
// C++ program to find product // of ASCII value of characters // in string #include <bits/stdc++.h> using namespace std;
// Function to find product // of ASCII value of characters // in string long long productAscii(string str)
{ long long prod = 1;
// Traverse string to find the product
for ( int i = 0; i < str.length(); i++) {
prod *= ( int )str[i];
}
// Return the product
return prod;
} // Driver code int main()
{ string str = "GfG" ;
cout << productAscii(str);
return 0;
} |
// Java program to find product // of ASCII value of characters // in string class GFG
{ // Function to find product // of ASCII value of characters // in string static long productAscii(String str)
{ long prod = 1 ;
// Traverse string to find the product
for ( int i = 0 ; i < str.length(); i++)
{
prod *= str.charAt(i);
}
// Return the product
return prod;
} // Driver Code public static void main(String[] args)
{ String str = "GfG" ;
System.out.println(productAscii(str));
} } // This code is contributed by Bilal |
# Python3 program to find product # of ASCII value of characters # in string # Function to find product # of ASCII value of characters # in string def productAscii( str ):
prod = 1
# Traverse string to find the product
for i in range ( 0 , len ( str )):
prod = prod * ord ( str [i])
# Return the product
return prod
# Driver code if __name__ = = '__main__' :
str = "GfG"
print (productAscii( str ))
# This code is contributed by # Sanjit_Prasad |
// C# program to find product // of ASCII value of characters // in string using System;
class GFG
{ // Function to find product // of ASCII value of characters // in string static long productAscii(String str)
{ long prod = 1;
// Traverse string to find the product
for ( int i = 0; i < str.Length; i++)
{
prod *= str[i];
}
// Return the product
return prod;
} // Driver Code static public void Main ()
{ String str = "GfG" ;
Console.Write(productAscii(str));
} } // This code is contributed by Raj |
<script> // javascript program to find product // of ASCII value of characters // in string // Function to find product // of ASCII value of characters // in string function productAscii(str)
{ var prod = 1;
// Traverse string to find the product
for (i = 0; i < str.length; i++)
{
prod *= str.charAt(i).charCodeAt(0);
}
// Return the product
return prod;
} // Driver Code str = "GfG" ;
document.write(productAscii(str)); // This code contributed by shikhasingrajput </script> |
<?php // PHP program to find product // of ASCII value of characters // in string // Function to find product // of ASCII value of characters // in string function productAscii( $str )
{ $prod = 1;
// Traverse string to find the product
for ( $i = 0; $i < strlen ( $str ); $i ++)
{
$prod *= ord( $str [ $i ]);
}
// Return the product
return $prod ;
} // Driver code $str = "GfG" ;
echo productAscii( $str );
// This code is contributed // by ChitraNayal ?> |
514182
Time Complexity: O(N), where N is the length of string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.