Good Pairs Codechef Java Solution (EQPAIR)

Problem

• You are given an array A of length N.

A pair $(i, j)$ $(1\le i\lt j \le N)$ is said to be good if $gcd(A_i, A_j) = lcm(A_i, A_j)$. Here $gcd$GCD denotes the greatest common divisor and $lcm$LCM denotes  Least Common Multiple

• Find the total number of good pairs present in the given array.

Input Format

• The first line of input will contain a single integer T, denoting the number of test cases.

• The first line of each test case contains an integer N- the length of the array A.

• The second line of each test case contains N space-separated integers A1,A2,....AN

Output Format

• For each test case, output on a new line the total number of such good pairs possible in the given array.

Constraints

• $1\le T\le 100$
• $1 \leq N \leq 10^5$
• $1 \leq A_i \leq 10^9$
• The sum of $N$ over all test cases won't exceed $3 \cdot 10^5$.

Input 

• 5 2 5 9 5 1 5 5 5 9 8 2 5 5 2 9 9 9 12 4 12 12 18 18 5 12 15 10 5 9

Output

• 0 3 5 2 0

Explanation:

Test case $1$: No good pair is present.

Test case $2$: The good pairs $(i, j)$ are: $(2, 3), (3, 4),$ and $(2, 4)$. To elaborate, $gcd(A_2, A_3) = lcm(A_2, A_3) = 5$.

Test case $3$: The good pairs $(i, j)$ are: $(1, 4), (2, 3), (5, 6), (6, 7),$ and $(5, 7)$. To elaborate, $gcd(A_1, A_4) = lcm(A_1, A_4) = 2$.

Test case $4$: The good pair $(i, j)$ are: $(1, 2),$ and $(3, 4)$. To elaborate, $gcd(A_3, A_4) = lcm(A_3, A_4) = 18$.

Test case $5$: No good pair is present.

/* package codechef; // don't place package name! */ import java.util.*; import java.lang.*; import java.io.*; /* Name of the class has to be "Main" only if the class is public. */ class Codechef { public static void main (String[] args) throws java.lang.Exception { try { Scanner sc = new Scanner(System.in); long t = sc.nextInt(); HashMap<Long,Long> map = new HashMap<>(); while(t>=1 && t<=1000) { int n = sc.nextInt(); long a[] = new long[n]; for (int i=0;i<n ;i++ ) { a[(int) i] = sc.nextInt(); } for (int i=0;i<n ;i++ ) { if(map.containsKey(a[i])) { map.put(a[i],map.get(a[i])+1); } else{ map.put(a[i], (long) 1); } } long sum=0; for(Map.Entry e : map.entrySet()) { Long k = (Long)e.getValue(); if(k>1) { sum = sum +(k*(k-1))/2; } } System.out.println(sum); map.clear(); } } catch(Exception e) { return; } } }