# Divisible Numbers - Amazon Top Interview Questions

### Problem Statement :

```You are given integers n, a, b, and c. Return the nth (0-indexed) term of the sorted sequence of integers divisible by a, b or c.

Note that by convention the first term of any sequence always starts with 1.

Example 1

Input

n = 8
a = 2
b = 5
c = 7

Output

12

Explanation

The first 9 terms of the sequence are [1, 2, 4, 5, 6, 7, 8, 10, 12]```

### Solution :

```                        ```Solution in C++ :

class Solution {
public:
int lcm(int x, int y) {
return x * y / __gcd(x, y);
}

int a, b, c;

int lesser(int n) {
// how many elements less than it?
return n / a + n / b + n / c - n / lcm(a, b) - n / lcm(b, c) - n / lcm(a, c) +
n / lcm(a, lcm(b, c));
}

int solve(int n, int a, int b, int c) {
this->a = a, this->b = b, this->c = c;

int l = 1, r = INT_MAX;

while (l < r) {
int guess = l + (r - l) / 2;

// i want the first value with >=n elements lesser than it!
if (lesser(guess) >= n) {
r = guess;
} else {
l = guess + 1;
// guess is not the answer! +1 it
}
}

return l;
}
};

int solve(int n, int a, int b, int c) {
return (new Solution())->solve(n, a, b, c);
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(int n, int a, int b, int c) {
int x = 1, y = 1, z = 1;
int ans = 1;

while (n-- > 0) {
int xx = a * x;
int yy = b * y;
int zz = c * z;

ans = Math.min(xx, Math.min(yy, zz));

if (ans == xx) {
x++;
}

if (ans == yy) {
y++;
}

if (ans == zz) {
z++;
}
}

return ans;
}
}```
```

```                        ```Solution in Python :

class Solution:
def find_points(self, goal, a, b, c, ab, ac, bc, abc):
total = 0

for item in (a, b, c, abc):
total += goal // item

for item in (ab, ac, bc):
total -= goal // item

def solve(self, n, a, b, c):

if n == 0:
return 1

lcm_ab = a * b // gcd(a, b)
lcm_ac = a * c // gcd(a, c)
lcm_bc = b * c // gcd(b, c)
lcm_abc = lcm_ab * c // gcd(lcm_ab, c)

points = self.find_points(lcm_abc, a, b, c, lcm_ab, lcm_ac, lcm_bc, lcm_abc)
total = n // points * lcm_abc

if n % points == 0:

goal = n % points

approx = int(lcm_abc / points * goal)
approx -= min(approx % a, approx % b, approx % c)
calc = 0

while calc != goal:
calc = self.find_points(approx, a, b, c, lcm_ab, lcm_ac, lcm_bc, lcm_abc)

if calc > goal:
approx -= 1
approx -= min(approx % a, approx % b, approx % c)
if calc < goal:
approx += min(a - approx % a, b - approx % b, c - approx % c)

```

## Tree: Postorder Traversal

Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the

## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your \$inOrder* func

## Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

## Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

## Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

## Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <