Count Pairs
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.
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.
Program:
#include <stdio.h>
int pairs(int elementlst[],int n,int z){
int count=0;
for(int i=0;i<n;i++){
int a=elementlst[i];
int id1=i;
int id2=i;
if(i==0){
while(elementlst[id2+1]==a)
id2+=1;
if(elementlst[id2+1]<=a+z && elementlst[id2+1]>=a-z)
count+=1;
}
else if(i<n-1){
while(elementlst[id2+1]==a)
id2+=1;
while(elementlst[id1-1]==a)
id1-=1;
if(((elementlst[id1-1]<=a+z) && (elementlst[id1-1]>=a-z)) || ((elementlst[id2+1]<=a+z) && (elementlst[id2+1]>=a-z)))
count+=1;
}
else{
while(elementlst[id1-1]==a)
id1-=1;
if(elementlst[id1-1]<=a+z && elementlst[id1-1]>=a-z)
count+=1;
}
}
return count;
}
int main() {
int n,z,swap=0;
scanf("%d",&n);
scanf("%d",&z);
int elementlst[n];
for(int i=0;i<n;i++){
scanf("%d",&elementlst[i]);
}
for (int c = 0 ; c < n ; c++)
{
for ( int d = 0 ; d < n - c - 1; d++)
{
if (elementlst[d] > elementlst[d+1]) /* For decreasing order use '<' instead of '>' */
{
swap = elementlst[d];
elementlst[d] = elementlst[d+1];
elementlst[d+1] = swap;
}
}
}
printf("%d",pairs(elementlst,n,z));
return 0;
}
