How do you solve Sum of Sortable Integers on LeetCode?

Sum of Sortable Integers (LeetCode #3886) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Divisors of n dictate possible subarray sizes.

What is the time complexity of Sum of Sortable Integers?

The optimal solution for Sum of Sortable Integers runs in O(n) time and uses O(n) space. Sorting takes O(n log n) and checking each subarray takes O(n), leading to an overall linear complexity for each divisor.

What is the brute force solution for Sum of Sortable Integers?

The brute force approach for Sum of Sortable Integers is Brute Force, with O(n²) time complexity and O(1) space complexity. We check each divisor of n and see if we can sort the array by rotating subarrays of that size. This involves checking all possible rotations for each subarray. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Sum of Sortable Integers?

Sum of Sortable Integers uses the following concepts: Array, Math, Sorting, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Sum of Sortable Integers?

Clarify the problem constraints. Discuss edge cases like empty arrays. Explain your thought process clearly.

What are common mistakes when solving Sum of Sortable Integers?

Not checking all subarrays for each k. Confusing sorting with rotating.

#3886

Sum of Sortable Integers

Hard

ArrayMathSortingEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Instead of checking all rotations, we can directly compare the original array with its sorted version to determine if k is sortable.

⚙️

Algorithm

3 steps

1Step 1: Sort the original array to get the target sorted version.
2Step 2: For each divisor k of n, check if elements in each subarray can match the sorted array.
3Step 3: If all subarrays match their corresponding segments in the sorted array, add k to the result.

solution.py9 lines

1def sum_of_sortable_integers(nums):
2    n = len(nums)
3    sorted_nums = sorted(nums)
4    total_sum = 0
5    for k in range(1, n + 1):
6        if n % k == 0:
7            if all(sorted_nums[i:i+k] == sorted(nums[i:i+k]) for i in range(0, n, k)):
8                total_sum += k
9    return total_sum

ℹ

Complexity note: Sorting takes O(n log n) and checking each subarray takes O(n), leading to an overall linear complexity for each divisor.

1Divisors of n dictate possible subarray sizes.
2Sorting helps in direct comparison without rotations.

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