Count Triplets HackerRank solution in Java with Explanation (Dictionaries and Hashmaps Problem)

Java solution for Count Triplets HackerRank Problem

Problem Description :

You are given an array and you need to find number of tripets of indices (i, j, k) such that the elements at those indices are in geometric progression for a given common ratio r and i < j < k.

Example 1 :

Input :

arr = [1, 4, 16, 64], r = 4

Output :

There are [1, 4, 16] and [4, 16, 64] at indices (0, 1, 2) and (1, 2, 3). So answer is 2.

Explanation :

r is 4 so we have to multiply by 4.

1 * 4 = 4
4 * 4 = 16
16 * 4 = 64

Example 2 :

arr = [1, 2, 2, 4], r = 2

Output :

There are 2 triplets in satisfying our criteria, whose indices are (0, 1, 3) and (0, 2, 3).

See full description on Hackerrank :

• HackerRank Problem Description : Count Triplets 
  

In this solution, we will use HashMap. So lets see solution.

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;

public class Solution {

    // Complete the countTriplets function below.
    private static long countTriplets(List<Long> arr, long ratio) {
        Map<Long, Long> leftMap = new HashMap<>();
        Map<Long, Long> rightMap = new HashMap<>();

        for (long number : arr) {
            rightMap.put(number, rightMap.getOrDefault(number, 0L) + 1);
        }

        long totalTriplets = 0;

        for (int i = 0; i < arr.size(); i++) {
            long current = arr.get(i);
            long left = 0, right = 0;

            rightMap.put(current, rightMap.getOrDefault(current, 0L) - 1);

            if (leftMap.containsKey(current / ratio) && current % ratio == 0) {
                left = leftMap.get(current / ratio);
            }

            if (rightMap.containsKey(current * ratio))
                right = rightMap.get(current * ratio);

            totalTriplets += left * right;

            leftMap.put(current, leftMap.getOrDefault(current, 0L) + 1);

        }
        return totalTriplets;
    }
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        String[] nr = bufferedReader.readLine().replaceAll("\\s+$", "").split(" ");

        int n = Integer.parseInt(nr[0]);

        long r = Long.parseLong(nr[1]);

        List<Long> arr = Stream.of(bufferedReader.readLine().replaceAll("\\s+$", "").split(" "))
            .map(Long::parseLong)
            .collect(toList());

        long ans = countTriplets(arr, r);

        bufferedWriter.write(String.valueOf(ans));
        bufferedWriter.newLine();

        bufferedReader.close();
        bufferedWriter.close();
    }
}

Solution Explanation :

• We have given Array as arr and Ratio as r. 
• So we will loop through entire array, First put all array integer on rightMap with value frequency. As we going forward, put left index integers of current index in leftMap and substarct current index value in rightMap.
• getOrDefault() method of map used to get the value mapped with specified key. If no value is mapped with the provided key then the default value is returned (in our case 0).
• We have to check only 3 indices, so in left variable assign (current / r) value and in right variable assign (current * r) value.
• arr = [1, 4, 16, 64], r = 4
• So if we are on index 1 value 4 then left become (current / r) = 1 and right become (current * 4) = 16, so we find 3 indices that matches our condition.
• Last we check, multiple of right and left integer.
• Return totalTriplets.

Output Explanation :

arr = [1, 5, 5, 25, 125], r = 5

• leftMap = [], rightMap = [1 = 1, 5 = 2, 25 = 1, 125 = 1], totalTriplets = 0
  
• i = 0
• left = 0, right = 0, current = 1
• Decrement current value in map | rightMap = [1 = 0, 5 = 2, 25 = 1, 125 = 1] 
• leftMap.containsKey(0) && 1 % 5 == 0 becomes false
• rightMap.containsKey(5) becomes true
• right = 2  
• totalTriplets += left*right | 0 = 0*2 
• leftMap = [1 = 1]
  
  
• i = 1, leftMap = [1 = 1], rightMap = [1 = 0, 5 = 2, 25 = 1, 125 = 1], totalTriplets = 0
• left = 0, right = 0, current = 5
• rightMap = [1 = 0, 5 = 1, 25 = 1, 125 = 1]
• leftMap contains key 1 && 5 % 5 == 0 becomes true
• left =1 
• rightMap containsKey(25) becomes true
• right = 1    
• totalTriplets = 1*1+0
  
• leftMap = [1 = 1, 5 = 1]
  
  
• i = 2, leftMap = [1 = 1, 5 = 1], rightMap = [1 = 0, 5 = 1, 25 = 1, 125 = 1], totalTriplets = 1
• left = 0, right = 0, current = 5
• rightMap = [1 = 0, 5 = 0, 25 = 1, 125 = 1]
• leftMap contains key 1 && 5 % 5 == 0 becomes true
• left =1 
• rightMap containsKey(25) becomes true
• right = 1    
• totalTriplets = 1*1+1 = 2
  
• leftMap = [1 = 1, 5 = 2]
  
  
• i = 3, leftMap = [1 = 1, 5 = 2], rightMap = [1 = 0, 5 = 0, 25 = 1, 125 = 1], totalTriplets = 2
• left = 0, right = 0, current = 25
• rightMap = [1 = 0, 5 = 0, 25 = 0, 125 = 1]
• leftMap contains key 5 && 25 % 5 == 0 becomes true
• left = 2
  
• rightMap containsKey(125) becomes true
• right = 1    
• totalTriplets = 2*1+2 = 4
  
• leftMap = [1 = 1, 5 = 2, 25 = 1]
  
  
•  i = 4, leftMap = [1 = 1, 5 = 2, 25 = 1], rightMap = [1 = 0, 5 = 0, 25 = 0, 125 = 1], totalTriplets =4
• left = 0, right = 0, current = 125
• rightMap = [1 = 0, 5 = 0, 25 = 0, 125 = 0]
• leftMap contains key 25 && 125 % 5 == 0 becomes true
• left = 2
  
• rightMap containsKey(625) becomes false. 
  
• totalTriplets = 2*0+2 = 4
  
• leftMap = [1 = 1, 5 = 2, 25 = 1, 125 = 1]
  
  
•  Return totalTriplets = 4

 

Happy Learning. Happy Coding.

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