What is the time complexity of How Many Numbers Are Smaller Than the Current Number?

The optimal solution for How Many Numbers Are Smaller Than the Current Number runs in O(n²) time and uses O(1) space. This complexity arises because we use two nested loops to compare each element with every other element in the array.

What algorithm or data structure is used to solve How Many Numbers Are Smaller Than the Current Number?

How Many Numbers Are Smaller Than the Current Number uses the following concepts: Array, Hash Table, Sorting, Counting Sort. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for How Many Numbers Are Smaller Than the Current Number?

Explain your thought process clearly while coding Consider edge cases like all elements being the same Practice similar problems to recognize patterns.

What are common mistakes when solving How Many Numbers Are Smaller Than the Current Number?

Not considering duplicate numbers correctly Confusing the index of sorted array with the original array

How Many Numbers Are Smaller Than the Current Number — LeetCode #1365 (Easy)

Tags: Array, Hash Table, Sorting, Counting Sort

Related patterns: Hash Map, Array

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach involves checking each number in the array against every other number to count how many are smaller. This is straightforward but inefficient for larger arrays.

This complexity arises because we use two nested loops to compare each element with every other element in the array.

Step 1: Initialize an empty result array of the same length as nums.
Step 2: For each element nums[i], initialize a count to 0.
Step 3: Loop through the array again and for each nums[j], if nums[j] < nums[i], increment the count.
Step 4: Store the count in the result array at the index corresponding to nums[i].
Step 5: Return the result array.

For nums = [8,1,2,2,3]: 1. Initialize result = [] 2. For i=0 (nums[0]=8): count=0, j=0 (skip), j=1 (1<8, count=1), j=2 (2<8, count=2), j=3 (2<8, count=3), j=4 (3<8, count=4) → result=[4] 3. For i=1 (nums[1]=1): count=0, j=0 (8<1, skip), j=1 (skip), j=2 (2<1, skip), j=3 (2<1, skip), j=4 (3<1, skip) → result=[4,0] 4. For i=2 (nums[2]=2): count=0, j=0 (8<2, skip), j=1 (1<2, count=1), j=2 (skip), j=3 (2<2, skip), j=4 (3<2, skip) → result=[4,0,1] 5. For i=3 (nums[3]=2): count=0, j=0 (8<2, skip), j=1 (1<2, count=1), j=2 (2<2, skip), j=3 (skip), j=4 (3<2, skip) → result=[4,0,1,1] 6. For i=4 (nums[4]=3): count=0, j=0 (8<3, skip), j=1 (1<3, count=1), j=2 (2<3, count=2), j=3 (2<3, count=3), j=4 (skip) → result=[4,0,1,1,3].

Common Mistakes

Not considering duplicate numbers correctly
Confusing the index of sorted array with the original array

Interview Tips

Explain your thought process clearly while coding
Consider edge cases like all elements being the same
Practice similar problems to recognize patterns.

#1365

How Many Numbers Are Smaller Than the Current Number

Easy

Array↗Hash Table↗Sorting↗Counting Sort↗Hash Map↗Array↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n²)Space O(1)

The brute force approach involves checking each number in the array against every other number to count how many are smaller. This is straightforward but inefficient for larger arrays.

⚙️

Algorithm

5 steps

1Step 1: Initialize an empty result array of the same length as nums.
2Step 2: For each element nums[i], initialize a count to 0.
3Step 3: Loop through the array again and for each nums[j], if nums[j] < nums[i], increment the count.
4Step 4: Store the count in the result array at the index corresponding to nums[i].
5Step 5: Return the result array.

solution.py14 lines

1# Full working Python code
2
3def smallerNumbersThanCurrent(nums):
4    result = []
5    for i in range(len(nums)):
6        count = 0
7        for j in range(len(nums)):
8            if nums[j] < nums[i]:
9                count += 1
10        result.append(count)
11    return result
12
13# Example usage
14print(smallerNumbersThanCurrent([8,1,2,2,3]))

ℹ

Complexity note: This complexity arises because we use two nested loops to compare each element with every other element in the array.

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