LeetCode Challenge Day 63 — 1437. Check If All 1's Are at Least Length K Places Away

Hey folks 👋

This is Day 63 of my LeetCode streak 🚀
Today’s problem is 1437 — Check If All 1's Are at Least Length K Places Away.

A clean and efficient check:
Make sure every pair of 1s in the binary array has at least k zeros between them.

💡 Intuition

The idea is simple:

Whenever we encounter a 1, we only need to check how far it is from the previous 1.

• If the gap is less than k, return false
• Otherwise, update the last-seen position

This avoids scanning extra ranges or constructing substrings — just a distance check.

📌 Approach

1. Start with prev = -k - 1 so the first 1 always passes the check.
2. Traverse the array:
   • When you see a 1 at index i:
     • Check if i - prev - 1 < k
     • If true → not enough zeros → return false
     • Else, update prev = i
3. If no violations occur, return true.

A single left-to-right pass solves it.

📈 Complexity

• Time Complexity: O(n) — one scan
• Space Complexity: O(1) — constant extra space

🧑‍💻 Code (JavaScript)

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {boolean}
 */
var kLengthApart = function(nums, k) {
    // prev initialized so the first 1 always passes the check
    let prev = -k - 1;
    
    for (let i = 0; i < nums.length; i++) {
        if (nums[i] === 1) {
            if (i - prev - 1 < k) return false;
            prev = i;
        }
    }
    return true;
};

🎯 Example

Input:
nums = [1,0,0,0,1,0,0], k = 2
Output:
true

Because the distance between the two 1s is exactly 3 → satisfies k = 2.

🧠 Reflection

This problem highlights how a single pointer and a simple invariant can solve what looks like a spacing constraint problem.

Key takeaways:

• Track what matters (the last seen 1)

• Validate only the required condition

• Avoid extra data structures

See you tomorrow for Day 64! 🚀
Happy Coding 👨‍💻✨