Given an array of positive integers (may contain duplicates), the task is to find the number of triplets whose product is equal to a given number t.
Examples:
Input: arr = [1, 31, 3, 1, 93, 3, 31, 1, 93]
t = 93
Output: 18
Input: arr = [4, 2, 4, 2, 3, 1]
t = 8
Output: 4
[(4, 2, 1), (4, 2, 1), (2, 4, 1), (4, 2, 1)]
Naive Approach: The easiest way to solve this is to compare each possible triplet with t and increment count if their product is equal to t.
Below is the implementation of above approach:
C++
// C++ program for above implementation #include<iostream>using namespace std ;int main(){ // The target value for which // we have to find the solution int target = 93 ; int arr[] = {1, 31, 3, 1, 93, 3, 31, 1, 93}; int length = sizeof(arr) / sizeof(arr[0]) ; // This variable contains the total // count of triplets found int totalCount = 0 ; // Loop from the first to the third //last integer in the list for(int i = 0 ; i < length - 2; i++) { // Check if arr[i] is a factor // of target or not. If not, // skip to the next element if (target % arr[i] == 0) { for (int j = i + 1 ; j < length - 1; j++) { // Check if the pair (arr[i], arr[j]) // can be a part of triplet whose // product is equal to the target if (target % (arr[i] * arr[j]) == 0) { // Find the remaining // element of the triplet int toFind = target / (arr[i] * arr[j]) ; for(int k = j + 1 ; k < length ; k++ ) { // If element is found. increment // the total count of the triplets if (arr[k] == toFind) { totalCount ++ ; } } } } } }cout << "Total number of triplets found : " << totalCount ; return 0 ; }// This code is contributed by ANKITRAI1 |
Java
// Java program for above implementation class GFG{public static void main(String[] args){ // The target value for which // we have to find the solution int target = 93 ; int[] arr = {1, 31, 3, 1, 93, 3, 31, 1, 93}; int length = arr.length; // This variable contains the total // count of triplets found int totalCount = 0 ; // Loop from the first to the third //last integer in the list for(int i = 0 ; i < length - 2; i++) { // Check if arr[i] is a factor // of target or not. If not, // skip to the next element if (target % arr[i] == 0) { for (int j = i + 1 ; j < length - 1; j++) { // Check if the pair (arr[i], arr[j]) // can be a part of triplet whose // product is equal to the target if (target % (arr[i] * arr[j]) == 0) { // Find the remaining // element of the triplet int toFind = target / (arr[i] * arr[j]); for(int k = j + 1 ; k < length ; k++ ) { // If element is found. increment // the total count of the triplets if (arr[k] == toFind) { totalCount ++ ; } } } } } } System.out.println("Total number of triplets found : " + totalCount);}}// This code is contributed by mits |
Python3
# Python program for above implementation# The target value for which we have# to find the solutiontarget = 93arr = [1, 31, 3, 1, 93, 3, 31, 1, 93]length = len(arr)# This variable contains the total# count of triplets foundtotalCount = 0# Loop from the first to the third# last integer in the listfor i in range(length - 2): # Check if arr[i] is a factor of target # or not. If not, skip to the next element if target % arr[i] == 0: for j in range(i + 1, length - 1): # Check if the pair (arr[i], arr[j]) can be # a part of triplet whose product is equal # to the target if target % (arr[i] * arr[j]) == 0: # Find the remaining element of the triplet toFind = target // (arr[i] * arr[j]) for k in range(j + 1, length): # If element is found. increment the # total count of the triplets if arr[k] == toFind: totalCount += 1print ('Total number of triplets found: ', totalCount) |
C#
// C# program for above implementation using System;class GFG{public static void Main(){ // The target value for which // we have to find the solution int target = 93 ; int[] arr = {1, 31, 3, 1, 93, 3, 31, 1, 93}; int length = arr.Length; // This variable contains the total // count of triplets found int totalCount = 0 ; // Loop from the first to the third //last integer in the list for(int i = 0 ; i < length - 2; i++) { // Check if arr[i] is a factor // of target or not. If not, // skip to the next element if (target % arr[i] == 0) { for (int j = i + 1 ; j < length - 1; j++) { // Check if the pair (arr[i], arr[j]) // can be a part of triplet whose // product is equal to the target if (target % (arr[i] * arr[j]) == 0) { // Find the remaining // element of the triplet int toFind = target / (arr[i] * arr[j]); for(int k = j + 1 ; k < length ; k++ ) { // If element is found. increment // the total count of the triplets if (arr[k] == toFind) { totalCount ++ ; } } } } } } Console.Write("Total number of triplets found : " + totalCount);}} |
PHP
<?php// PHP program for above implementation // The target value for which // we have to find the solution $target = 93 ;$arr = array(1, 31, 3, 1, 93, 3, 31, 1, 93);$length = sizeof($arr);// This variable contains the// total count of triplets found $totalCount = 0 ;// Loop from the first to the // third last integer in the list for($i = 0 ; $i < $length - 2; $i++){ // Check if arr[i] is a factor // of target or not. If not, // skip to the next element if ($target % $arr[$i] == 0) { for ($j = $i + 1 ; $j < $length - 1; $j++) { // Check if the pair (arr[i], arr[j]) // can be a part of triplet whose // product is equal to the target if ($target % ($arr[$i] * $arr[$j]) == 0) { // Find the remaining // element of the triplet $toFind = $target / ($arr[$i] * $arr[$j]) ; for($k = $j + 1 ; $k < $length ; $k++ ) { // If element is found. increment // the total count of the triplets if ($arr[$k] == $toFind) { $totalCount ++ ; } } } } }}echo ("Total number of triplets found : ");echo ($totalCount);// This code is contributed // by Shivi_Aggarwal?> |
Javascript
<script>// Javascript program for above implementation // The target value for which // we have to find the solution var target = 93;var arr = [ 1, 31, 3, 1, 93, 3, 31, 1, 93 ];var length = arr.length;// This variable contains the total // count of triplets found var totalCount = 0;// Loop from the first to the third // last integer in the list for(var i = 0; i < length - 2; i++){ // Check if arr[i] is a factor // of target or not. If not, // skip to the next element if (target % arr[i] == 0) { for(var j = i + 1; j < length - 1; j++) { // Check if the pair (arr[i], arr[j]) // can be a part of triplet whose // product is equal to the target if (target % (arr[i] * arr[j]) == 0) { // Find the remaining // element of the triplet var toFind = target / (arr[i] * arr[j]); for(var k = j + 1; k < length; k++) { // If element is found. increment // the total count of the triplets if (arr[k] == toFind) { totalCount ++; } } } } }}document.write("Total number of triplets found : " + totalCount);// This code is contributed by rutvik_56</script> |
Output:
Total number of triplets found: 18
Time Complexity: O(n3)
Auxiliary Space: O(1)
Efficient Approach:
- Remove the numbers which are not the factors of t from the array.
- Then sort the array so that we don’t have to verify the index of each number to avoid additional counting of pairs.
- Then store the number of times each number appears in a dictionary count.
- Use two loops to find the first two numbers of a valid triplet by checking if their product divides t
- Find the third number of the triplet and check if we have already seen the triplet to avoid duplicate calculations
- Count the total possible combinations of that triplet such that they occur in the same order (all pairs should follow the order (x, y, z) to avoid repetitions)
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// This function returns the total number of// combinations of the triplet (x, y, z) possible in the// given listint Combinations(int x, int y, int z, map<int, int> count){ int valx = count[x]; int valy = count[y]; int valz = count[z]; if (x == y) { if (y == z) { return (valx * (valy - 1) * (valz - 2)) / 6; } else { return ((((valy - 1) * valx) / 2) * valz); } } else if (y == z) { return valx * (((valz - 1) * valy) / 2); } else { return (valx * valy * valz); }}// Driver codeint main(){ // Value for which solution has to be found int target = 93; int ar[] = { 1, 31, 3, 1, 93, 3, 31, 1, 93 }; // length of the array int N = sizeof(ar) / sizeof(ar[0]); // Create a list of integers from arr which // contains only factors of the target // Using list comprehension vector<int> list; for (int i = 0; i < N; i++) if (ar[i] != 0 && target % ar[i] == 0) list.push_back(ar[i]); // Sort the list sort(list.begin(), list.end()); int length = list.size(); int arr[length]; // List to Array Conversion std::copy(list.begin(), list.end(), arr); // Initialize the Map with the default value map<int, int> count; set<string> tripletSeen; // Count the number of times a value is present in // the list and store it in a Map for further // use for (int val : list) count[val]++; // Used to store the total number of triplets int totalCount = 0; for (int i = 0; i < length - 2; i++) { for (int j = i + 1; j < length - 1; j++) { // Check if the pair (arr[i], arr[j]) can be // a part of triplet whose product is equal // to the target if (target % (arr[i] * arr[j]) == 0) { int toFind = target / (arr[i] * arr[j]); // This condition makes sure that a // solution is not repeated int a[] = { arr[i], arr[j], toFind }; sort(a, a + 3); string str = to_string(a[0]) + "#" + to_string(a[1]) + "#" + to_string(a[2]); if (toFind >= arr[i] && toFind >= arr[j] && (tripletSeen.find(str) == tripletSeen.end())) { tripletSeen.insert(str); totalCount += Combinations( arr[i], arr[j], toFind, count); } } } } cout << "Total number of triplets found: " << totalCount << endl;}// This code is contributed by phasing17 |
Java
// java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;class GFG { // This function returns the total number of // combinations of the triplet (x, y, z) possible in the // given list static int Combinations(int x, int y, int z, HashMap<Integer, Integer> count) { int valx = count.getOrDefault(x, 0); int valy = count.getOrDefault(y, 0); int valz = count.getOrDefault(z, 0); if (x == y) { if (y == z) { return (valx * (valy - 1) * (valz - 2)) / 6; } else { return ((((valy - 1) * valx) / 2) * valz); } } else if (y == z) { return valx * (((valz - 1) * valy) / 2); } else { return (valx * valy * valz); } } // Driver code public static void main(String[] args) { // Value for which solution has to be found int target = 93; int ar[] = { 1, 31, 3, 1, 93, 3, 31, 1, 93 }; // length of the array int N = ar.length; // Create a list of integers from arr which // contains only factors of the target // Using list comprehension ArrayList<Integer> list = new ArrayList<>(); for (int i = 0; i < N; i++) if (ar[i] != 0 && target % ar[i] == 0) list.add(ar[i]); // Sort the list Collections.sort(list); int length = list.size(); // ArrayList to Array Conversion int[] arr = list.stream().mapToInt(i -> i).toArray(); // Initialize the Map with the default value HashMap<Integer, Integer> count = new HashMap<>(); HashSet<String> tripletSeen = new HashSet<>(); // Count the number of times a value is present in // the list and store it in a Map for further // use for (int val : list) count.put(val, count.getOrDefault(val, 0) + 1); // Used to store the total number of triplets int totalCount = 0; for (int i = 0; i < length - 2; i++) { for (int j = i + 1; j < length - 1; j++) { // Check if the pair (arr[i], arr[j]) can be // a part of triplet whose product is equal // to the target if (target % (arr[i] * arr[j]) == 0) { int toFind = target / (arr[i] * arr[j]); // This condition makes sure that a // solution is not repeated int a[] = { arr[i], arr[j], toFind }; Arrays.sort(a); String str = (a[0] + "#" + a[1] + "#" + a[2]); if (toFind >= arr[i] && toFind >= arr[j] && tripletSeen.contains(str) == false) { tripletSeen.add(str); totalCount += Combinations( arr[i], arr[j], toFind, count); } } } } System.out.println( "Total number of triplets found: " + totalCount); }}// This code is contributed by Kingash. |
Python3
# Python3 code to find the number of triplets# whose product is equal to a given number# in quadratic time# This function is used to initialize# a dictionary with a default valuefrom collections import defaultdict# Value for which solution has to be foundtarget = 93arr = [1, 31, 3, 1, 93, 3, 31, 1, 93]# Create a list of integers from arr which# contains only factors of the target# Using list comprehensionarr = [x for x in arr if x != 0 and target % x == 0]# Sort the listarr.sort()length = len(arr)# Initialize the dictionary with the default valuetripletSeen = defaultdict(lambda : False)count = defaultdict(lambda : 0)# Count the number of times a value is present in# the list and store it in a dictionary for further usefor key in arr: count[key] += 1# Used to store the total number of tripletstotalCount = 0# This function returns the total number of combinations# of the triplet (x, y, z) possible in the given listdef Combinations(x, y, z): if x == y: if y == z: return (count[x]*(count[y]-1)*(count[z]-2)) // 6 else: return ((((count[y]-1)*count[x]) // 2)*count[z]) elif y == z: return count[x]*(((count[z]-1)*count[y]) // 2) else: return (count[x] * count[y] * count[z])for i in range(length - 2): for j in range(i + 1, length - 1): # Check if the pair (arr[i], arr[j]) can be a # part of triplet whose product is equal to the target if target % (arr[i] * arr[j]) == 0: toFind = target // (arr[i] * arr[j]) # This condition makes sure that a solution is not repeated if (toFind >= arr[i] and toFind >= arr[j] and tripletSeen[(arr[i], arr[j], toFind)] == False): tripletSeen[(arr[i], arr[j], toFind)] = True totalCount += Combinations(arr[i], arr[j], toFind)print ('Total number of triplets found: ', totalCount) |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG { static int getValue(Dictionary<int, int> dict, int val, int defaultVal) { if (dict.ContainsKey(val)) return dict[val]; return defaultVal; } // This function returns the total number of // combinations of the triplet (x, y, z) possible in the // given list static int Combinations(int x, int y, int z, Dictionary<int, int> count) { int valx = getValue(count, x, 0); int valy = getValue(count, y, 0); int valz = getValue(count, z, 0); if (x == y) { if (y == z) { return (valx * (valy - 1) * (valz - 2)) / 6; } else { return ((((valy - 1) * valx) / 2) * valz); } } else if (y == z) { return valx * (((valz - 1) * valy) / 2); } else { return (valx * valy * valz); } } // Driver code public static void Main(string[] args) { // Value for which solution has to be found int target = 93; int[] ar = { 1, 31, 3, 1, 93, 3, 31, 1, 93 }; // length of the array int N = ar.Length; // Create a list of integers from arr which // contains only factors of the target // Using list comprehension List<int> list = new List<int>(); for (int i = 0; i < N; i++) if (ar[i] != 0 && target % ar[i] == 0) list.Add(ar[i]); // Sort the list list.Sort(); int length = list.Count; // List to Array Conversion int[] arr = list.ToArray(); // Initialize the Map with the default value Dictionary<int, int> count = new Dictionary<int, int>(); HashSet<string> tripletSeen = new HashSet<string>(); // Count the number of times a value is present in // the list and store it in a Map for further // use foreach(int val in list) count[val] = getValue(count, val, 0) + 1; // Used to store the total number of triplets int totalCount = 0; for (int i = 0; i < length - 2; i++) { for (int j = i + 1; j < length - 1; j++) { // Check if the pair (arr[i], arr[j]) can be // a part of triplet whose product is equal // to the target if (target % (arr[i] * arr[j]) == 0) { int toFind = target / (arr[i] * arr[j]); // This condition makes sure that a // solution is not repeated int[] a = { arr[i], arr[j], toFind }; Array.Sort(a); string str = (a[0] + "#" + a[1] + "#" + a[2]); if (toFind >= arr[i] && toFind >= arr[j] && tripletSeen.Contains(str) == false) { tripletSeen.Add(str); totalCount += Combinations( arr[i], arr[j], toFind, count); } } } } Console.WriteLine("Total number of triplets found: " + totalCount); }}// This code is contributed by phasing17 |
Javascript
// JavaScript code to find the number of triplets// whose product is equal to a given number// in quadratic timefunction getValue(dict, val, defaultVal) { if (dict.hasOwnProperty(val)) return dict[val]; return defaultVal; }// Value for which solution has to be foundlet target = 93let arr = [1, 31, 3, 1, 93, 3, 31, 1, 93]// Create a list of integers from arr which// contains only factors of the target// Using list comprehensionlet arr1 = []for (var i = 0; i < arr.length; i++){ if (arr[i] != 0) if (target % arr[i] == 0) arr1.push(arr[i]) }arr = arr1// Sort the listarr.sort()let length = arr.length// Initialize the dictionary with the default valuelet tripletSeen = new Set();let count = {}// Count the number of times a value is present in// the list and store it in a dictionary for further usefor (var key of arr) count[key] = getValue(count, key, 0) + 1;// Used to store the total number of tripletslet totalCount = 0// This function returns the total number of combinations// of the triplet (x, y, z) possible in the given listfunction Combinations(x, y, z){ let valx = getValue(count, x, 0); let valy = getValue(count, y, 0); let valz = getValue(count, z, 0); if (x == y) { if (y == z) { return Math.floor((valx * (valy - 1) * (valz - 2)) / 6); } else { return (Math.floor(((valy - 1) * valx) / 2) * valz); } } else if (y == z) { return valx * Math.floor(((valz - 1) * valy) / 2); } else { return (valx * valy * valz); } }for (var i = 0; i < (length - 2); i++){ for (var j = i + 1; j < length - 1; j++) { // Check if the pair (arr[i], arr[j]) can be a // part of triplet whose product is equal to the target if (target % (arr[i] * arr[j]) == 0) { let toFind = Math.floor(target / (arr[i] * arr[j])) let str = (arr[i] + "#" + arr[j] + "#" + toFind); // This condition makes sure that a solution is not repeated if (toFind >= arr[i] && toFind >= arr[j] && !tripletSeen.has(str)) { tripletSeen.add(str) totalCount += Combinations(arr[i], arr[j], toFind) } } }}console.log('Total number of triplets found: ', totalCount)// This code is contributed by phasing17 |
Output:
Total number of triplets found: 18
Time Complexity: O(n2)
Auxiliary Space: O(n) as using hashmap
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