**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[0], min = nums[0], 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
```

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