How do you solve Check If All 1's Are at Least Length K Places Away on LeetCode?

Check If All 1's Are at Least Length K Places Away (LeetCode #1437) 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: Using a single pass reduces time complexity significantly.

What is the time complexity of Check If All 1's Are at Least Length K Places Away?

The optimal solution for Check If All 1's Are at Least Length K Places Away runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only loop through the array once. The space complexity is O(1) since we only use a few variables for tracking.

What is the brute force solution for Check If All 1's Are at Least Length K Places Away?

The brute force approach for Check If All 1's Are at Least Length K Places Away is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute-force approach checks every pair of '1's in the array to see if they are at least 'k' places apart. 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 Check If All 1's Are at Least Length K Places Away?

Check If All 1's Are at Least Length K Places Away uses the following concepts: Array. The recommended patterns to study are: Array, Two Pointers.

What are the interview tips for Check If All 1's Are at Least Length K Places Away?

Always clarify edge cases with the interviewer. Explain your thought process as you code. Consider both time and space complexity in your solution.

What are common mistakes when solving Check If All 1's Are at Least Length K Places Away?

Not handling the case when 'k' is 0 correctly. Forgetting to check the first '1' against the last found index.

#1437

Check If All 1's Are at Least Length K Places Away

Easy

ArrayArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses a single pass through the array to track the last position of '1'. This reduces the time complexity significantly.

⚙️

Algorithm

4 steps

1Step 1: Initialize a variable to track the last index of '1' found.
2Step 2: Loop through the array. When a '1' is found, check the distance from the last '1'.
3Step 3: If the distance is less than 'k', return false. Update the last index of '1'.
4Step 4: If the loop completes without returning false, return true.

solution.py10 lines

1# Full working Python code
2
3def k_length_apart(nums, k):
4    last_index = -1
5    for i in range(len(nums)):
6        if nums[i] == 1:
7            if last_index != -1 and i - last_index < k:
8                return False
9            last_index = i
10    return True

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 only use a few variables for tracking.

1Using a single pass reduces time complexity significantly.
2Tracking the last index of '1' helps avoid unnecessary comparisons.

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