# 2855. Minimum Right Shifts to Sort the Array

## Description

You are given a 0-indexed array nums of length n containing distinct positive integers. Return the minimum number of right shifts required to sort nums and -1 if this is not possible.

A right shift is defined as shifting the element at index i to index (i + 1) % n, for all indices.

Example 1:

Input: nums = [3,4,5,1,2]
Output: 2
Explanation:
After the first right shift, nums = [2,3,4,5,1].
After the second right shift, nums = [1,2,3,4,5].
Now nums is sorted; therefore the answer is 2.


Example 2:

Input: nums = [1,3,5]
Output: 0
Explanation: nums is already sorted therefore, the answer is 0.

Example 3:

Input: nums = [2,1,4]
Output: -1
Explanation: It's impossible to sort the array using right shifts.


Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i] <= 100
• nums contains distinct integers.

## Solutions

Solution 1: Direct Traversal

First, we use a pointer $i$ to traverse the array $nums$ from left to right, finding a continuous increasing sequence until $i$ reaches the end of the array or $nums[i - 1] > nums[i]$. Next, we use another pointer $k$ to traverse the array $nums$ from $i + 1$, finding a continuous increasing sequence until $k$ reaches the end of the array or $nums[k - 1] > nums[k]$ and $nums[k] > nums[0]$. If $k$ reaches the end of the array, it means the array is already increasing, so we return $n - i$; otherwise, we return $-1$.

The time complexity is $O(n)$, and the space complexity is $O(1)$. Here, $n$ is the length of the array $nums$.

• class Solution {
public int minimumRightShifts(List<Integer> nums) {
int n = nums.size();
int i = 1;
while (i < n && nums.get(i - 1) < nums.get(i)) {
++i;
}
int k = i + 1;
while (k < n && nums.get(k - 1) < nums.get(k) && nums.get(k) < nums.get(0)) {
++k;
}
return k < n ? -1 : n - i;
}
}

• class Solution {
public:
int minimumRightShifts(vector<int>& nums) {
int n = nums.size();
int i = 1;
while (i < n && nums[i - 1] < nums[i]) {
++i;
}
int k = i + 1;
while (k < n && nums[k - 1] < nums[k] && nums[k] < nums[0]) {
++k;
}
return k < n ? -1 : n - i;
}
};

• class Solution:
def minimumRightShifts(self, nums: List[int]) -> int:
n = len(nums)
i = 1
while i < n and nums[i - 1] < nums[i]:
i += 1
k = i + 1
while k < n and nums[k - 1] < nums[k] < nums[0]:
k += 1
return -1 if k < n else n - i


• func minimumRightShifts(nums []int) int {
n := len(nums)
i := 1
for i < n && nums[i-1] < nums[i] {
i++
}
k := i + 1
for k < n && nums[k-1] < nums[k] && nums[k] < nums[0] {
k++
}
if k < n {
return -1
}
return n - i
}

• function minimumRightShifts(nums: number[]): number {
const n = nums.length;
let i = 1;
while (i < n && nums[i - 1] < nums[i]) {
++i;
}
let k = i + 1;
while (k < n && nums[k - 1] < nums[k] && nums[k] < nums[0]) {
++k;
}
return k < n ? -1 : n - i;
}