How do you solve Count of Smaller Numbers After Self on LeetCode?

Count of Smaller Numbers After Self (LeetCode #315) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using sorting techniques can significantly reduce time complexity.

What is the time complexity of Count of Smaller Numbers After Self?

The optimal solution for Count of Smaller Numbers After Self runs in O(n log n) time and uses O(n) space. This complexity is due to the merge sort process, which divides the array and merges it while counting smaller elements efficiently.

What is the brute force solution for Count of Smaller Numbers After Self?

The brute force approach for Count of Smaller Numbers After Self is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each element in the array and counting how many elements to its right are smaller. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Count of Smaller Numbers After Self?

Always clarify the problem requirements and constraints. Discuss your thought process and approach before coding. Test your solution with edge cases after implementation.

What are common mistakes when solving Count of Smaller Numbers After Self?

Not considering the original indices when counting smaller elements. Overlooking edge cases like duplicates or single-element arrays.

#315

Count of Smaller Numbers After Self

Hard

ArrayBinary SearchDivide and ConquerBinary Indexed TreeSegment TreeMerge SortOrdered SetMerge SortDivide and Conquer

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(n)

Using a modified merge sort allows us to count smaller elements efficiently while sorting the array. This approach leverages the divide-and-conquer strategy to maintain order and count inversions.

⚙️

Algorithm

3 steps

1Step 1: Create a helper function that performs merge sort and counts smaller elements.
2Step 2: During the merge step, count how many elements from the right half are smaller than elements from the left half.
3Step 3: Store the counts in an auxiliary array that corresponds to the original indices.

solution.py27 lines

1# Full working Python code
2
3def countSmaller(nums):
4    def merge_sort(enum):
5        if len(enum) <= 1:
6            return enum, [0] * len(enum)
7        mid = len(enum) // 2
8        left, left_counts = merge_sort(enum[:mid])
9        right, right_counts = merge_sort(enum[mid:])
10        merged = []
11        counts = [0] * len(enum)
12        j = 0
13        for i in range(len(left)):
14            while j < len(right) and right[j][0] < left[i][0]:
15                merged.append(right[j])
16                j += 1
17            counts[left[i][1]] += j
18            merged.append(left[i])
19        merged.extend(right[j:])
20        return merged, counts
21
22    indexed_nums = list(enumerate(nums))
23    _, counts = merge_sort(indexed_nums)
24    return counts
25
26# Example usage
27print(countSmaller([5, 2, 6, 1]))  # Output: [2, 1, 1, 0]

ℹ

Complexity note: This complexity is due to the merge sort process, which divides the array and merges it while counting smaller elements efficiently.

1Using sorting techniques can significantly reduce time complexity.
2Understanding how to count inversions during sorting is crucial.

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