**Count Pairs**

**Problem Description**

**Given an array of integers A, and an integer K find number of happy elements.**

Element X is happy if there exists at least 1 element whose difference is less than K i.e. an element X is happy,

if there is another element in the range [X-K, X+K] other than X itself.

**Constraints**

1 <= N <= 10^5

0 <= K <= 10^5

0 <= A[i] <= 10^9

**Input**

First line contains two integers N and K where N is size of the array and K is a number as described above

Second line contains N integers separated by space.

**Output**

**Print a single integer denoting the total number of happy elements.**

**Time Limit**

1

**Examples**

**Example 1**

**Input**

6 3

5 5 7 9 15 2

**Output**

5

**Explanation**

Other than number 15, everyone has at least 1 element in the range [X-3, X+3].

Hence they are all happy elements. Since these five are in number, the output is 5.

**Example 2**

**Input**

3 2

1 3 5

**Output**

3

**Explanation**

All numbers have at least 1 element in the range [X-2, X+2]. Hence they are all happy elements.

Since these three are in number, the output is 3.

**Recommended: Please try your approach on your integrated development environment (IDE) first, before moving on to the solution.**

**Few words from CodingHumans : Don't Just copy paste the solution, try to analyze the problem and solve it without looking by taking the the solution as a hint or a reference . Your understanding of the solution matters.**

**HAPPY CODING**😁

**Solution**

**( C ++ )**

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If you have any doubts regarding this problem or need the solution in other programming languages then leave a comment down below .

can you send the same code in python

ReplyDeleteand does the ques is like from ex 1 for 5 the range [5-3=2,5+3=8] and 7 and 2 is in this range is this a proper approach

This solution works fine😊

ReplyDeleteimport java.util.*;

ReplyDeleteclass B{

public static void main(String ar[])

{

Scanner sc = new Scanner(System.in);

int n = sc.nextInt();

int k = sc.nextInt();

int count=0,i;

int arr[] = new int[n];

B b = new B();

for(i=0;i=l) && (ar[j]<= h))

{

return 1;

}

}

}

return 0;

}

}