Count Pairs

Solution

Java Solution

import java.util.*;
public class Main {
    public static int pairs(int a[], int n , int z)
    {
        int c=0,i;
        for(i=0;i<n;i++)
        {
            int temp=a[i];
            int id1=i; int id2=i;
            if(i==0)
            {
                while(a[id2+1]==temp)
                id2++;
                if(a[id2+1]<=temp+z && a[id2+1]>=temp-z)
                c++;
            }
            else if(i<n-1)
            {
                while(a[id2+1]==temp)
                id2+=1;
                while(a[id1-1]==temp)
                id1-=1;
                if(((a[id1-1]<=temp+z) && (a[id1-1]>=temp-z)) || (
(a[id2+1]<=temp+z) && (a[id2+1]>=temp-z)))
                c+=1;
            }
            else
            {
                while(a[id1-1]==temp)
                id1-=1;
                if(a[id1-1]<=aa+z && a[id1-1]>=aa-z)
                c+=1;
            }
        }
        return c;
    }
    public static void main(String[] args) throws Exception {
        // Your code  here!
        Scanner sc= new Scanner(System.in);
        // System.out.println("XXXXXXXX")
        int n,k,i,s=0; 
        n=sc.nextInt();
        k=sc.nextInt();
        int a[]= new int [n];
        
        for(i=0;i<n;i++) a[i]=sc.nextInt();
        Arrays.sort(a);
        System.out.print(pairs(a,n,k));
    }
}

Output

3 2
1 3 5
3

C++ Solution

#include <bits/stdc++.h>
using namespace std;
int pairs(int elementlst[],int n,int z){
 int ans_count=0;
  for(int i=0;i<n;i++){
    int temp=elementlst[i];
    int id1=i;
    int id2=i;
    if(i==0){
      while(elementlst[id2+1]==temp)
         id2+=1;
      if(elementlst[id2+1]<=temp+z && elementlst[id2+1]>=temp-z)
     ans_count+=1;
    }
    else if(i<n-1){
      while(elementlst[id2+1]==temp)
        id2+=1;
      while(elementlst[id1-1]==temp)
        id1-=1;
      if(((elementlst[id1-1]<=temp+z) && (elementlst[id1-1]>=temp-z)) || ((elementlst[id2+1]<=temp+z) && (elementlst[id2+1]>=temp-z)))
     ans_count+=1;
    }
    else{
      while(elementlst[id1-1]==temp)
        id1-=1;
      if(elementlst[id1-1]<=temp+z && elementlst[id1-1]>=temp-z)
     ans_count+=1;
    }
  }
  return ans_count;
}
int main() {
int n,z;
cin>>n>>z;
int elementlst[n];
for(int i=0;i<n;i++){
    cin>>elementlst[i];
}
sort(elementlst,elementlst+n);
cout<<pairs(elementlst,n,z);
return 0;
}

You can also try this code with Online C++ Compiler

Run Code

Output

3 2
1 3 5
3

You can also try this code with Online C++ Compiler

Run Code

Python Solution

n, k = map(int, input().split())
a = set()
a = set(map(int, input().split()))
a = list(a)
if n==1:
    print("0")
else:
    a = sorted(a)
c = 0
for i in range(1, len(a)-1):
    if a[i]-a[i-1]>k and a[i+1]-a[i]>k :
        c += 1
if a[1]-a[0]>k:
    c+=1
if a[len(a)-1]-a[len(a)-2]>k:
    c +=1
print(n-c)

Output

3 2
1 3 5
3

Complexity Analysis

• Time Complexity: We have solved the given coding problem in O(N²) time complexity.
• Space Complexity: O(1) will be the space complexity because we haven't used any extra space other than variables.

FAQs

What is a copy constructor?

A copy function Object() is a member function that uses another object of the same class to initialize an object.

What’s the key difference between exception and error in java?

The error indicates a problem, which is usually caused by a lack of system resources, and our application isn't set up to identify such problems. Exceptions are mistakes that can occur both during execution and compilation. It's typically encountered in code produced by programmers. The Java Throwable class, which is part of the java package, has two subclasses: exception and error.

Conclusion

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