# GEEK02 - Editorial

Practice

Contest

Author: Bhuvnesh Jain

Tester: Bhuvnesh Jain

Editorialist: Bhuvnesh Jain

EASY

# Prerequisites

Looping Techniques, Greedy Algorithms

# Problem

Find the smallest number such that the sum of the digits is N and it is divisible by {10}^{N}.

# Explanation

To make a number divisible by {10}^{N}, we need at least N zeros at the end of the number. To make the number smallest, we append exactly N zeros to the end of the number. Now, we need to ensure the sum of the digits is N. For this, we will try to make the length of the number as small as possible to get the answer. Thus we keep on inserting 9 into the number till the sum doesn’t exceed N. If we have any remainder left, then we keep it as the first digit (most significant one) so that the resulting number is minimised.

The approach works well for all subtasks but there are 2 corner cases:

1. The first is that the final number may not fit into the dtata types present in C++/Java. Since, we only need to output the number, we can use strings to store the answer.
2. The only corner case where the answer is 0 is N = 0.
3. There are no cases where the answer doesn't exist.

# Time Complexity

O(N), per test case.

# Space Complexity

O(1) or O(N), if storing the answer in the string.