Check if every pair of 1 in the array is at least K length apart from each other

Given a binary array and an integer K, check if every pair of 1 in the array is at least K length apart from each other. Return true if the condition holds, otherwise return false.

Examples:

Input: arr = [1, 0, 0, 0, 1, 0, 0, 1, 0, 0], K = 2. 
Output: True 
Explanation:
Every 1 in the array is at least K distance apart from each other.

Input: arr= [1, 0, 1, 0, 1, 1], K = 1 
Output: False 
Explanation:
The fifth 1 and sixth 1 are not apart from each other. Hence, the output is false. 

Approach:
To solve the problem mentioned above we have to check the distance between each pair of adjacent 1. Find the first position of 1 then iterate through the rest of the array and increment distance if it’s 0 otherwise perform a check operation if the distance is less than k and reset the count to 0 again.
Below is the implementation of the above approach:

C++

// C++ implementation to Check if every pair of 1 in
// the array is at least K length from each other

#include <bits/stdc++.h>
using namespace std;

// Function to check distance
bool kLengthApart(vector<int>& nums, int k)
{
    // Find first position of 1
    int pos = 0, count = 0;

    while (pos < nums.size() && nums[pos] == 0)
        pos++;

    // Iterate through the rest of array
    for (int i = pos + 1; i < nums.size(); i++) {
        // Increment distance if its 0
        if (nums[i] == 0)
            count++;

        // Check if the distance is less than k
        else {
            if (count < k)
                return false;

            // Reset count to 0
            count = 0;
        }
    }

    // Return the result
    return true;
}

// Driver code
int main()
{
    vector<int> nums = { 1, 0, 0, 0, 1, 0, 0, 1, 0, 0 };
    int k = 2;

    bool ans = kLengthApart(nums, k);
    if (ans == 1)
        cout << "True" << endl;

    else
        cout << "False" << endl;

    return 0;
}

Java

// Java implementation to check if
// every pair of 1 in the array is
// at least K length from each other
class Main{

// Function to check distance
public static boolean kLengthApart(int[] nums,
                                   int k)
{

    // Find first position of 1
    int pos = 0, count = 0;

    while (pos < nums.length && nums[pos] == 0)
        pos++;

    // Iterate through the rest of array
    for(int i = pos + 1; i < nums.length; i++)
    {

       // Increment distance if its 0
       if (nums[i] == 0)
           count++;

       // Check if the distance is less than k
       else
       {
           if (count < k)
               return false;

           // Reset count to 0
           count = 0;
       }
    }

    // Return the result
    return true;
}

// Driver Code
public static void main(String[] args)
{
    int[] nums = { 1, 0, 0, 0, 1,
                   0, 0, 1, 0, 0 };
    int k = 2;
    boolean ans = kLengthApart(nums, k);

    if (ans)
        System.out.println("True");

    else
        System.out.println("False");
}
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 implementation to check if
# every pair of 1 in the array is
# at least K length from each other

# Function to check distance
def kLengthApart(nums, k):

    # Find first position of 1
    pos = 0
    count = 0

    while (pos < len(nums) and nums[pos] == 0):
        pos += 1

    # Iterate through the rest of list
    for i in range(pos + 1, len(nums)):

        # Increment distance if its 0
        if nums[i] == 0:
            count += 1

        # Check if the distance is less than k
        else :
            if count < k:
                return False

            # Reset count to 0
            count = 0

        # Return the result
    return True

# Driver Code
if __name__ == "__main__":

    nums = [ 1, 0, 0, 0, 1, 0, 0, 1, 0, 0 ]
    k = 2

    print(kLengthApart(nums, k))

# This code is contributed by rutvik_56

C#

// C# implementation to check if
// every pair of 1 in the array is
// at least K length from each other
using System;

class GFG{

// Function to check distance
public static bool kLengthApart(int[] nums,
                                int k)
{

    // Find first position of 1
    int pos = 0, count = 0;

    while (pos < nums.Length && nums[pos] == 0)
        pos++;

    // Iterate through the rest of array
    for(int i = pos + 1; i < nums.Length; i++)
    {

       // Increment distance if its 0
       if (nums[i] == 0)
           count++;

       // Check if the distance is
       // less than k
       else
       {
           if (count < k)
               return false;

           // Reset count to 0
           count = 0;
       }
    }

    // Return the result
    return true;
}

// Driver Code
public static void Main()
{
    int[] nums = { 1, 0, 0, 0, 1,
                   0, 0, 1, 0, 0 };
    int k = 2;
    bool ans = kLengthApart(nums, k);

    if (ans)
        Console.Write("True");
    else
        Console.Write("False");
}
}

// This code is contributed by chitranayal

Javascript

<script>

// JavaScript implementation to
// Check if every pair of 1 in
// the array is at least K length
// from each other

// Function to check distance
function kLengthApart(nums, k)
{
    // Find first position of 1
    var pos = 0, count = 0;
    var i;
    while (pos < nums.length && nums[pos] == 0)
        pos++;

    // Iterate through the rest of array
    for (i = pos + 1; i < nums.length; i++) {
        // Increment distance if its 0
        if (nums[i] == 0)
            count++;

        // Check if the distance is less than k
        else {
            if (count < k)
                return false;

            // Reset count to 0
            count = 0;
        }
    }

    // Return the result
    return true;
}

// Driver code
    var nums = [1, 0, 0, 0, 1, 0, 0, 1, 0, 0];
    var k = 2;

    var ans = kLengthApart(nums, k);
    if (ans == 1)
        document.write("True");

    else
        document.write("False");

</script>

Output:

True

Time Complexity: O(n) 
Auxiliary Space: O(1)