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