Open In App

Program to find the product of ASCII values of characters in a string

Last Updated : 08 Feb, 2024
Improve
Improve
Like Article
Like
Save
Share
Report

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 = 6059

Input: 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++




// 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




// 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




# 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#




// 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


Javascript




<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
// 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
?>


Output

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.



Like Article
Suggest improvement
Previous
Next
Share your thoughts in the comments

Similar Reads