[LeetCode January Challange] Day 25 - Check If All 1's Are at Least Length K Places Away

Given an array nums of 0s and 1s 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.

Example 3:

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

Example 4:

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

Constraints:

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

---

Solution

Time complexity : O(n)
Space complexity : O(1)

class Solution {
public:
    bool kLengthApart(vector<int>& nums, int k) {
        int last_idx = -1;
        for (int i=0; i<nums.size(); ++i) {
            if (nums[i]) {
                if (last_idx != -1 && i-last_idx <= k)
                    return false;
                last_idx = i;
            }
        }
        return true;
    }
};