Sum of integers upto N with given unit digit

Given two integer N and D, 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 
 

Naive approach: 
 

• Traverse from 1 to N.
• If the unit digit of the number is D add the number to the sum.
• Finally print the value of sum.

Below is the implementation of the above approach:
 

// 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(int n, int d)
{
    ll sum = 0;
    for (int i = 1; i <= n; i++) {

        // If the unit digit is d
        if (i % 10 == d)
            sum += i;
    }
    return sum;
}

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

// Java implementation of the approach
import java.util.*;

class solution
{

// Function to return the required sum
static long getSum(int n, int d)
{
    long sum = 0;
    for (int i = 1; i <= n; i++) {

        // If the unit digit is d
        if (i % 10 == d)
            sum += i;
    }
    return sum;
}

// Driver code
public static void main(String args[])
{
    int n = 30, d = 3;
    System.out.println(getSum(n, d));
   
}
}

# Python3 implementation of the approach 

# Function to return the required sum 
def getSum(n, d) :
    sum = 0; 
    for i in range(n + 1) : 

        # If the unit digit is d 
        if (i % 10 == d) :
            sum += i 
    return sum

# Driver code 
if __name__ == "__main__" :

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

# This code is contributed by Ryuga

// C# implementation of the approach
using System;
class gfg
{
  // Function to return the required sum
 public static int getSum(int n, int d)
 {
    int sum = 0;
    for (int i = 1; i <= n; i++) {

        // If the unit digit is d
        if (i % 10 == d)
            sum += i;
    }
    return sum;
}

// Driver code
 public static int Main()
 { 
    int n = 30, d = 3;
    Console.WriteLine( getSum(n, d));
    return 0;
 }
}

<?php
// PHP implementation of the approach

// Function to return the required sum
function getSum($n, $d)
{
    $sum = 0;
    for ($i = 1; $i <= $n; $i++) 
    {

        // If the unit digit is d
        if ($i % 10 == $d)
            $sum += $i;
    }
    return $sum;
}

// Driver code
$n = 30;
$d = 3;
echo getSum($n, $d);
    
// This code is contributed by Sachin
?>

<script>

// java script implementation of the approach

// Function to return the required sum
function getSum(n, d)
{
    let sum = 0;
    for (let i = 1; i <= n; i++)
    {

        // If the unit digit is d
        if (i % 10 == d)
            sum += i;
    }
    return sum;
}

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

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

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, the space taken by the algorithm is constant
 

Efficient approach: While D < N update sum = sum + D and D = D + 10. Print the sum in the end.
Below is the implementation of the above approach: 
 

// 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(int n, int d)
{
    ll sum = 0;
    while (d <= n)
    {
        sum += d;
        d += 10;
    }
    return sum;
}

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

// Java implementation of the approach
class Solution
{

 
// Function to return the required sum
static long getSum(int n, int d)
{
    long sum = 0;
    while (d <= n) {
        sum += d;
        d += 10;
    }
    return sum;
}
 
// Driver code
public static void main(String args[])
{
    int n = 30, d = 3;
    System.out.print(getSum(n, d));
    
}
}
//contributed by Arnab Kundu

# Python3 implementation of the approach 
# Function to return the required sum 
def getSum(n, d): 
    sum = 0
    while (d <= n): 
        sum += d
        d += 10
    return sum

# Driver code 
n = 30
d = 3
print(getSum(n, d))

# This code is contributed 
# by sahishelangia

// C# implementation of the approach
using System;
class GFG
{

// Function to return the required sum
static long getSum(int n, int d)
{
    long sum = 0;
    while (d <= n) 
    {
        sum += d;
        d += 10;
    }
    return sum;
}

// Driver code
public static void Main()
{
    int n = 30, d = 3;
    Console.Write(getSum(n, d));
}
}

// This code is contributed 
// by Akanksha Rai

<?php
// PHP implementation of the approach
// Function to return the required sum
function getSum($n, $d)
{
    $sum = 0;
    while ($d <= $n) 
    {
        $sum += $d;
        $d += 10;
    }
    return $sum;
}

// Driver code
$n = 30; $d = 3;
echo(getSum($n, $d));
    
// This Code is contributed
// by Mukul Singh
?>

<script>
// java script  implementation of the approach
// Function to return the required sum
function getSum(n, d)
{
    let sum = 0;
    while (d <= n) 
    {
        sum += d;
        d += 10;
    }
    return sum;
}

// Driver code
let n = 30;
let d = 3;
document.write(getSum(n, d));
   
// This code is contributed
// by sravan kumar
</script>

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