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1437. Check If All 1's Are at Least Length K Places Away

Description

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

Solutions

Solution 1: Simulation

We can iterate through the array \(\textit{nums}\) and use a variable \(j\) to record the index of the previous \(1\). When the element at the current position \(i\) is \(1\), we just need to check if \(i - j - 1\) is less than \(k\). If it is less than \(k\), it means there exists a pair of \(1\)s with fewer than \(k\) zeros between them, so we return \(\text{false}\). Otherwise, we update \(j\) to \(i\) and continue iterating through the array.

After the iteration is complete, we return \(\text{true}\).

The time complexity is \(O(n)\), where \(n\) is the length of the array \(\textit{nums}\). The space complexity is \(O(1)\).

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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

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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;
    }
}

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

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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
}

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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;
}

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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
    }
}

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