How do you solve Three Consecutive Odds on LeetCode?

Three Consecutive Odds (LeetCode #1550) 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: Consecutive elements can be tracked using a counter.

What is the brute force solution for Three Consecutive Odds?

The brute force approach for Three Consecutive Odds is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible triplet of consecutive numbers in the array to see if they are all odd. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Three Consecutive Odds?

Three Consecutive Odds uses the following concepts: Array. The recommended patterns to study are: Array, Two Pointers.

What are the interview tips for Three Consecutive Odds?

Explain your thought process clearly while coding. Consider edge cases and test them in your explanation. Optimize your solution after presenting the brute force approach.

What are common mistakes when solving Three Consecutive Odds?

Not resetting the counter when an even number is found. Overlooking edge cases where the array is too short.

#1550

Three Consecutive Odds

Easy

ArrayArrayTwo Pointers

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)

This approach uses a single loop to track consecutive odd numbers, making it much more efficient. We only need to check each number once.

⚙️

Algorithm

4 steps

1Step 1: Initialize a counter to track consecutive odd numbers.
2Step 2: Loop through the array and check if each number is odd.
3Step 3: If it is odd, increment the counter. If it is even, reset the counter to zero.
4Step 4: If the counter reaches 3 at any point, return true. If the loop ends without reaching 3, return false.

solution.py10 lines

1def threeConsecutiveOdds(arr):
2    count = 0
3    for num in arr:
4        if num % 2 != 0:
5            count += 1
6            if count == 3:
7                return True
8        else:
9            count = 0
10    return False

ℹ

Complexity note: The time complexity is O(n) because we only loop through the array once. The space complexity is O(1) since we use a constant amount of extra space.

1Consecutive elements can be tracked using a counter.
2Checking parity (odd/even) is a simple modulus operation.

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