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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 \(nums\).

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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