How do you solve Minimum Right Shifts to Sort the Array on LeetCode?

Minimum Right Shifts to Sort the Array (LeetCode #2855) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The array must have a single pivot point for it to be sortable by right shifts.

What is the time complexity of Minimum Right Shifts to Sort the Array?

The optimal solution for Minimum Right Shifts to Sort the Array runs in O(n) time and uses O(1) space. This complexity is achieved by scanning the array once to find the minimum element and checking sorted conditions, which is linear.

What is the brute force solution for Minimum Right Shifts to Sort the Array?

The brute force approach for Minimum Right Shifts to Sort the Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves simulating each right shift and checking if the array becomes sorted after each shift. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Right Shifts to Sort the Array?

Minimum Right Shifts to Sort the Array uses the following concepts: Array. The recommended patterns to study are: Array.

What are the interview tips for Minimum Right Shifts to Sort the Array?

Always consider edge cases, such as already sorted arrays or arrays with only one element. Explain your thought process clearly; it shows your understanding of the problem. Practice identifying patterns in array manipulation problems, like finding pivots or using two pointers.

What are common mistakes when solving Minimum Right Shifts to Sort the Array?

Not checking if the array is already sorted before performing shifts. Failing to recognize that multiple pivot points indicate impossibility.

#2855

Minimum Right Shifts to Sort the Array

Easy

ArrayArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach identifies the pivot point where the array is rotated. If there is only one such pivot, we can calculate the number of shifts needed to sort the array.

⚙️

Algorithm

3 steps

1Step 1: Find the index of the smallest element in the array (this is the pivot).
2Step 2: Check if the array is sorted from the pivot to the end and from the start to the pivot.
3Step 3: If the array is sorted, return the index of the pivot as the number of shifts; otherwise, return -1.

solution.py6 lines

1def min_right_shifts(nums):
2    n = len(nums)
3    min_index = nums.index(min(nums))
4    if (min_index == 0 and all(nums[i] <= nums[i + 1] for i in range(n - 1))) or (min_index > 0 and all(nums[i] <= nums[i + 1] for i in range(min_index - 1)) and all(nums[i] <= nums[i + 1] for i in range(min_index, n - 1))):
5        return min_index
6    return -1

ℹ

Complexity note: This complexity is achieved by scanning the array once to find the minimum element and checking sorted conditions, which is linear.

1The array must have a single pivot point for it to be sortable by right shifts.
2If there are multiple points where nums[i] < nums[i-1], sorting is impossible.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.