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

• Last Updated : 27 Sep, 2022

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 ``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

 ``

Output:

`True`

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

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