LeetCode #1437 — EASY

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

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

Given an binary array nums and an integer k, return true if all 1's are at least k places away from each other, otherwise return false.

Example 1:

Input: nums = [1,0,0,0,1,0,0,1], k = 2
Output: true
Explanation: Each of the 1s are at least 2 places away from each other.

Example 2:

Input: nums = [1,0,0,1,0,1], k = 2
Output: false
Explanation: The second 1 and third 1 are only one apart from each other.

Constraints:

1 <= nums.length <= 10⁵
0 <= k <= nums.length
nums[i] is 0 or 1

Step 01

Brute Force Baseline

Problem summary: Given an binary array nums and an integer k, return true if all 1's are at least k places away from each other, otherwise return false.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[1,0,0,0,1,0,0,1]
2

Example 2

[1,0,0,1,0,1]
2

Core Insight

What unlocks the optimal approach

Each time you find a number 1, check whether or not it is K or more places away from the next one. If it's not, return false.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1437: Check If All 1's Are at Least Length K Places Away
class Solution {
    public boolean kLengthApart(int[] nums, int k) {
        int j = -(k + 1);
        for (int i = 0; i < nums.length; ++i) {
            if (nums[i] == 1) {
                if (i - j - 1 < k) {
                    return false;
                }
                j = i;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #1437: Check If All 1's Are at Least Length K Places Away
func kLengthApart(nums []int, k int) bool {
	j := -(k + 1)
	for i, x := range nums {
		if x == 1 {
			if i-j-1 < k {
				return false
			}
			j = i
		}
	}
	return true
}

# Accepted solution for LeetCode #1437: Check If All 1's Are at Least Length K Places Away
class Solution:
    def kLengthApart(self, nums: List[int], k: int) -> bool:
        j = -inf
        for i, x in enumerate(nums):
            if x:
                if i - j - 1 < k:
                    return False
                j = i
        return True

// Accepted solution for LeetCode #1437: Check If All 1's Are at Least Length K Places Away
impl Solution {
    pub fn k_length_apart(nums: Vec<i32>, k: i32) -> bool {
        let mut j = -(k + 1);
        for (i, &x) in nums.iter().enumerate() {
            if x == 1 {
                if (i as i32) - j - 1 < k {
                    return false;
                }
                j = i as i32;
            }
        }
        true
    }
}

// Accepted solution for LeetCode #1437: Check If All 1's Are at Least Length K Places Away
function kLengthApart(nums: number[], k: number): boolean {
    let j = -(k + 1);
    for (let i = 0; i < nums.length; ++i) {
        if (nums[i] === 1) {
            if (i - j - 1 < k) {
                return false;
            }
            j = i;
        }
    }
    return true;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.