## Count Pairs Codevita 9 Solution 2020

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.

**SOLUTION**

**Program: **Count Pairs Codevita 9 Solution in C++

#include<bits/stdc++.h> using namespace std; typedef long long ll; void test() { ll n,k,l; scanf("%lld",&n); set<ll> s; // to remove duplicate elements from array map<ll,int> m; // to keep track of frequency of each element in given array for(int i=0;i<n;i++) {scanf("%lld",&l); s.insert(l); m[l]++; } //copying the set to array a n=s.size(); ll a[n],i=0,count=0; for(ll x : s) {a[i]=x; i++; } // loop to check (adjacent diff <=k ) if(a[1]-a[0]<=k) count+=m[a[0]]; for(int i=1;i<n-1;i++) { if( a[i+1]-a[i]<=k || abs(a[i]-a[i-1])<=k) count+=m[a[i]]; // when this condition satisfies we increment the count by that element frequency. } if(a[n-1]-a[n-2]<=k) count+=m[a[n-1]]; cout<<count; // printing output } int main() { test(); }

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