2855. Minimum Right Shifts to Sort the Array LeetCode Solution

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Table of Contents

  1. Problem Statement
  2. Complexity Analysis
  3. Minimum Right Shifts to Sort the Array solution in C++
  4. Minimum Right Shifts to Sort the Array solution in Java
  5. Minimum Right Shifts to Sort the Array solution in Python
  6. Additional Resources
2855. Minimum Right Shifts to Sort the Array LeetCode Solution image

Problem Statement of Minimum Right Shifts to Sort the Array

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.

Complexity Analysis

  • Time Complexity: O(n)
  • Space Complexity: O(1)
See also  324. Wiggle Sort II LeetCode Solution

2855. Minimum Right Shifts to Sort the Array LeetCode Solution in C++

class Solution {
 public:
  int minimumRightShifts(vector<int>& nums) {
    int count = 0;
    int pivot = -1;

    for (int i = 0; i + 1 < nums.size(); i++)
      if (nums[i] > nums[i + 1]) {
        ++count;
        pivot = i;
      }

    if (count == 0)
      return 0;
    if (count > 1 || nums.back() > nums.front())
      return -1;
    return nums.size() - 1 - pivot;
  }
};
/* code provided by PROGIEZ */

2855. Minimum Right Shifts to Sort the Array LeetCode Solution in Java

class Solution {
  public int minimumRightShifts(List<Integer> nums) {
    int pivot = -1;
    int count = 0;

    for (int i = 0; i + 1 < nums.size(); i++)
      if (nums.get(i) > nums.get(i + 1)) {
        ++count;
        pivot = i;
      }

    if (count == 0)
      return 0;
    if (count > 1 || nums.get(nums.size() - 1) > nums.get(0))
      return -1;
    return nums.size() - 1 - pivot;
  }
}
// code provided by PROGIEZ

2855. Minimum Right Shifts to Sort the Array LeetCode Solution in Python

class Solution:
  def minimumRightShifts(self, nums: list[int]) -> int:
    count = 0

    for i, (a, b) in enumerate(itertools.pairwise(nums)):
      if a > b:
        count += 1
        pivot = i

    if count == 0:
      return 0
    if count > 1 or nums[-1] > nums[0]:
      return -1
    return len(nums) - pivot - 1
# code by PROGIEZ

Additional Resources

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