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 sumll 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 codeint main(){ ll n = 30; int d = 3; cout << getSum(n, d); return 0;} |
Java
// Java implementation of the approachimport java.io.*;class GFG {// Function to return the required sumstatic 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 approachclass GFG {// Function to return the required sumstatic 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 sumfunction 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 sumfunction 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 codelet 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 so the space taken by the algorithm is constant
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