Sum of integers upto N with given unit digit (Set 2)

Given two integer N and D where 1 ? N ? 10¹⁸, the task is to find the sum of all the integers from 1 to N whose unit digit is D.
Examples:

Input: N = 30, D = 3 
Output: 39 
3 + 13 + 23 = 39
Input: N = 5, D = 7 
Output: 0 

Approach: In Set 1 we saw two basic approaches to find the required sum, but the complexity is O(N) which will take more time for larger N. Here’s an even efficient approach, suppose we are given N = 30 and D = 3: 

sum = 3 + 13 + 23 
sum = 3 + (10 + 3) + (20 + 3) 
sum = 3 * (3) + (10 + 20) 

From the above observation, we can find the sum following the steps below: 

• Decrement N until N % 10 != D.
• Find K = N / 10.
• Now, sum = (K + 1) * D + (((K * 10) + (10 * K * K)) / 2).

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int

// Function to return the required sum
ll getSum(ll n, int d)
{
    if (n < d)
        return 0;

    // Decrement N
    while (n % 10 != d)
        n--;

    ll k = n / 10;

    return (k + 1) * d + (k * 10 + 10 * k * k) / 2;
}

// Driver code
int main()
{
    ll n = 30;
    int d = 3;
    cout << getSum(n, d);
    return 0;
}

Java

// Java  implementation of the approach

import java.io.*;

class GFG {


// Function to return the required sum
static long getSum(long n, int d)
{
    if (n < d)
        return 0;

    // Decrement N
    while (n % 10 != d)
        n--;

    long k = n / 10;

    return (k + 1) * d + (k * 10 + 10 * k * k) / 2;
}

// Driver code

    public static void main (String[] args) {
     long n = 30;
    int d = 3;
    System.out.println(getSum(n, d));    }
}
//This code is contributed by inder_verma..

Python3

# Python3 implementation of the approach

# Function to return the required sum
def getSum(n, d) :

    if (n < d) :
        return 0

    # Decrement N
    while (n % 10 != d) :
        n -= 1

    k = n // 10

    return ((k + 1) * d +
            (k * 10 + 10 * k * k) // 2)

# Driver code
if __name__ == "__main__" :

    n = 30
    d = 3
    print(getSum(n, d))

# This code is contributed by Ryuga

C#

// C# implementation of the approach


class GFG {


// Function to return the required sum
static int getSum(int n, int d)
{
    if (n < d)
        return 0;

    // Decrement N
    while (n % 10 != d)
        n--;

    int k = n / 10;

    return (k + 1) * d + (k * 10 + 10 * k * k) / 2;
}

// Driver code

    public static void Main () {
    int n = 30;
    int d = 3;
    System.Console.WriteLine(getSum(n, d)); }
}
//This code is contributed by mits.

PHP

<?php
// PHP implementation of the approach

// Function to return the required sum
function getSum($n, $d)
{
    if ($n < $d)
        return 0;

    // Decrement N
    while ($n % 10 != $d)
        $n--;

    $k = (int)($n / 10);

    return ($k + 1) * $d +
           ($k * 10 + 10 * $k * $k) / 2;
}

// Driver code
$n = 30;
$d = 3;
echo getSum($n, $d);

// This code is contributed by mits
?>

Javascript

<script>

// java script  implementation of the approach

// Function to return the required sum
function getSum(n, d)
{
    if (n < d)
        return 0;

    // Decrement N
    while (n % 10 != d)
        n--;

    k = parseInt(n / 10);

    return (k + 1) * d +
        (k * 10 + 10 * k * k) / 2;
}

// Driver code
let n = 30;
let d = 3;
document.write( getSum(n, d));

// This code is contributed
// by bobby
</script>

Output:

39

Time Complexity: O(n) //since one traversal of the loop till the number is required to complete all operations hence the overall time required by the algorithm is linear

Auxiliary Space: O(1) // since no extra array is used so the space taken by the algorithm is constant