Is Sum of Sortable Integers easy or hard?

The problem is typically categorized as Hard because recognizing the correct digit manipulation strategy is not obvious at first glance. Brute force permutation approaches are too slow, so the challenge is identifying the frequency-counting optimization that reduces the complexity to linear time per number.

Sum of Sortable Integers Python/Java solution

Most implementations convert each number into digits, compute digit frequencies, and verify the sortable condition. The logic is identical across languages: iterate digits, update a 10-element count array, and rebuild digits in ascending order before adding valid numbers to the sum.

How to solve Sum of Sortable Integers in O(n)?

Process each integer independently and count digit frequencies using a fixed array of size 10. Rebuild the digits in ascending order using the counts and validate the sortable property. Since digit counting is linear in the number of digits, the total complexity becomes O(n · d).

What is the best approach for Sum of Sortable Integers?

Digit frequency counting is typically the best approach. Count how many times each digit (0–9) appears, then reconstruct the digits in non‑decreasing order to verify the sortable condition. This runs in O(n · d) time with constant extra space, making it faster than explicitly sorting digits for every number.

Is Sum of Sortable Integers asked at Google/Amazon/Meta?

Problems involving digit manipulation, sorting checks, and frequency counting appear frequently in interviews at companies like Google, Amazon, and Meta. Variations often test whether candidates recognize when sorting can be replaced by counting or greedy reconstruction.

What data structure is used in Sum of Sortable Integers?

The core data structure is a fixed-size frequency array for digits 0 through 9. This structure allows constant-time updates while scanning digits and enables quick reconstruction of sorted digit sequences.

What is the time complexity of Sum of Sortable Integers?

The optimal solution runs in O(n · d) time where n is the number of integers and d is the number of digits per integer. Using digit frequency counting avoids sorting, which would otherwise cost O(d log d) per number. Space complexity stays O(1) because the digit range is fixed.

Description Hints 4 Solution Video 15:57

3886. Sum of Sortable Integers

Hard55.3% acceptance

You are given an integer array nums of length n.

An integer k is called sortable if k divides n and you can sort nums in non-decreasing order by sequentially performing the following operations:

Partition nums into consecutive subarrays of length k.
Cyclically rotate each subarray independently any number of times to the left or to the right.

Return an integer denoting the sum of all possible sortable integers k.

Example 1:

Input: nums = [3,1,2]

Output: 3

Explanation:

For n = 3, possible divisors are 1 and 3.
For k = 1: each subarray has one element. No rotation can sort the array.
For k = 3: the single subarray [3, 1, 2] can be rotated once to produce [1, 2, 3], which is sorted.
Only k = 3 is sortable. Hence, the answer is 3.

Example 2:

Input: nums = [7,6,5]

Output: 0

Explanation:

For n = 3, possible divisors are 1 and 3.
For k = 1: each subarray has one element. No rotation can sort the array.
For k = 3: the single subarray [7, 6, 5] cannot be rotated into non-decreasing order.
No k is sortable. Hence, the answer is 0.

Example 3:

Input: nums = [5,8]

Output: 3

Explanation:

For n = 2, possible divisors are 1 and 2.
Since [5, 8] is already sorted, every divisor is sortable. Hence, the answer is 1 + 2 = 3.

Constraints:

1 <= n == nums.length <= 10⁵
1 <= nums[i] <= 10⁵

Topics

Hints open one at a time

Hint 1

First, observe that k must divide n, since the array is partitioned into subarrays of length k. Iterate over all divisors of n.

Hint 2

Let s be the sorted version of the array. For each divisor k, check every block [i, i+k).

Hint 3

A block can be rotated to match its sorted counterpart if the sorted segment s[i..i+k-1] appears as a cyclic rotation of nums[i..i+k-1].

Hint 4

You can test this by checking whether the sorted segment exists inside two concatenated copies of the block. If all blocks satisfy this, then k is sortable.

15:57

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1class Solution:
2    def sortableIntegers(self, nums: list[int]) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

[3,1,2]

Input

[7,6,5]

Input

[5,8]

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