# Arithmetic Sequence Permutation - Amazon Top Interview Questions

### Problem Statement :

```Given a list of integers nums, return whether you can rearrange the order of nums such that the difference between every consecutive two numbers is the same.

Constraints

n ≤ 100,000 where n is the length of nums

Example 1

Input
nums = [7, 1, 5, 3]

Output
True

Explanation
If we rearrange nums to [1, 3, 5, 7], then the difference between every two consecutive numbers is 2.

Example 2

Input
nums = [1, 5, 1, 5, 1, 5]

Output
False

Explanation
The difference between every consecutive two numbers alternates between 4 and -4.```

### Solution :

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

bool solve(vector<int>& nums) {
// Finding Max and Min
int max = nums, min = nums, max_i = -1, min_i = -1;
for (int i = 1; i < nums.size(); i++) {
if (nums[i] > max) {
max = nums[i];
max_i = i;
}
if (nums[i] < min) {
min = nums[i];
min_i = i;
}
}
if (max - min == 0)  // All numbers same, diff=0
return true;
if ((max - min) % (nums.size() - 1) != 0)  // Case not possible
return false;
// Check for duplicates now
vector<int> freqArr(max + 1, 0);
for (int i = 0; i < nums.size(); i++) {
freqArr[nums[i]] += 1;
if (freqArr[nums[i]] > 1) return false;
}
int min_diff = (max - min) / (nums.size() - 1);
for (int i = 0; i < nums.size(); i++) {
if (i == min_i || i == max_i) continue;
int d2 = nums[i] - min;
if (d2 % min_diff != 0) {
return false;
}
}
return true;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public boolean solve(int[] nums) {
Arrays.sort(nums);
for (int i = 2; i < nums.length; i++) {
if (nums[i] - nums[i - 1] != nums[i - 1] - nums[i - 2])
return false;
}
return true;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, nums):
minima = min(nums)
maxima = max(nums)

d = (maxima - minima) / (len(nums) - 1)
num = minima + d
nums_set = set(nums)
if d == 0:
return True
while num in nums_set:
num += d

if num > maxima:
return True
return False```
```

## Square-Ten Tree

The square-ten tree decomposition of an array is defined as follows: The lowest () level of the square-ten tree consists of single array elements in their natural order. The level (starting from ) of the square-ten tree consists of subsequent array subsegments of length in their natural order. Thus, the level contains subsegments of length , the level contains subsegments of length , the

## Balanced Forest

Greg has a tree of nodes containing integer data. He wants to insert a node with some non-zero integer value somewhere into the tree. His goal is to be able to cut two edges and have the values of each of the three new trees sum to the same amount. This is called a balanced forest. Being frugal, the data value he inserts should be minimal. Determine the minimal amount that a new node can have to a

## Jenny's Subtrees

Jenny loves experimenting with trees. Her favorite tree has n nodes connected by n - 1 edges, and each edge is ` unit in length. She wants to cut a subtree (i.e., a connected part of the original tree) of radius r from this tree by performing the following two steps: 1. Choose a node, x , from the tree. 2. Cut a subtree consisting of all nodes which are not further than r units from node x .

## Tree Coordinates

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## Array Pairs

Consider an array of n integers, A = [ a1, a2, . . . . an] . Find and print the total number of (i , j) pairs such that ai * aj <= max(ai, ai+1, . . . aj) where i < j. Input Format The first line contains an integer, n , denoting the number of elements in the array. The second line consists of n space-separated integers describing the respective values of a1, a2 , . . . an .

## Self Balancing Tree

An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ