Largest triplet product in a stream
Given a stream of integers represented as arr[]. For each index i from 0 to n-1, print the multiplication of largest, second largest, third largest element of the subarray arr[0…i]. If i < 2 print -1.
Examples:
Input : arr[] = {1, 2, 3, 4, 5} Output :-1 -1 6 24 60 Explanation : for i = 2 only three elements are there {1, 2, 3} so answer is 6. For i = 3 largest three elements are {2, 3, 4} their product is 2*3*4 = 24 ....so on
We will use priority queue here.
- Insert arr[i] in the priority queue
- As the top element in priority queue is largest so pop it and store it as x. Now the top element in the priority queue will be the second largest element in subarray arr[0…i] pop it and store as y. Now the top element is third largest element in subarray arr[0…i] so pop it and store it as z.
- Print x*y*z
- Reinsert x, y, z.
Below is the implementation of the above approach:
C++
// C++ implementation of largest triplet // multiplication #include <bits/stdc++.h> using namespace std; // Prints the product of three largest numbers // in subarray arr[0..i] void LargestTripletMultiplication( int arr[], int n) { // call a priority queue priority_queue< int > q; // traversing the array for ( int i = 0; i < n; i++) { // pushing arr[i] in the array q.push(arr[i]); // if less than three elements are present // in array print -1 if (q.size() < 3) cout << "-1" << endl; else { // pop three largest elements int x = q.top(); q.pop(); int y = q.top(); q.pop(); int z = q.top(); q.pop(); // Reinsert x, y, z in priority_queue int ans = x * y * z; cout << ans << endl; q.push(x); q.push(y); q.push(z); } } return ; } // Driver Function int main() { int arr[] = { 1, 2, 3, 4, 5 }; int n = sizeof (arr) / sizeof (arr[0]); LargestTripletMultiplication(arr, n); return 0; } |
Java
// Java implementation of largest triplet // multiplication import java.util.Collections; import java.util.PriorityQueue; class GFG { // Prints the product of three largest numbers // in subarray arr[0..i] static void LargestTripletMultiplication( int arr[], int n) { // call a priority queue PriorityQueue<Integer> q = new PriorityQueue(Collections.reverseOrder()); // traversing the array for ( int i = 0 ; i < n; i++) { // pushing arr[i] in array q.add(arr[i]); // if less than three elements are present // in array print -1 if (q.size() < 3 ) System.out.println( "-1" ); else { // pop three largest elements int x = q.poll(); int y = q.poll(); int z = q.poll(); // Reinsert x, y, z in priority_queue int ans = x * y * z; System.out.println(ans); q.add(x); q.add(y); q.add(z); } } } // Driver code public static void main(String[] args) { int arr[] = { 1 , 2 , 3 , 4 , 5 }; int n = arr.length; LargestTripletMultiplication(arr, n); } } // This code is contributed by shubham96301 |
Python3
# Python3 implementation of largest triplet # multiplication from queue import PriorityQueue # Prints the product of three largest # numbers in subarray arr[0..i] def LargestTripletMultiplication(arr, n): # Call a priority queue q = PriorityQueue() # Traversing the array for i in range (n): # Pushing -arr[i] in array # to get max PriorityQueue q.put( - arr[i]) # If less than three elements # are present in array print -1 if (q.qsize() < 3 ): print ( - 1 ) else : # pop three largest elements x = q.get() y = q.get() z = q.get() # Reinsert x, y, z in # priority_queue ans = x * y * z print ( - ans) q.put(x); q.put(y); q.put(z); # Driver Code if __name__ = = '__main__' : arr = [ 1 , 2 , 3 , 4 , 5 ] n = len (arr) LargestTripletMultiplication(arr, n) # This code is contributed by math_lover |
C#
// C# implementation of largest triplet // multiplication using System; using System.Collections.Generic; public class GFG { // Prints the product of three largest numbers // in subarray arr[0..i] static void LargestTripletMultiplication( int []arr, int n) { // call a priority queue List< int > q = new List< int >(); // traversing the array for ( int i = 0; i < n; i++) { // pushing arr[i] in array q.Add(arr[i]); q.Sort(); q.Reverse(); // if less than three elements are present // in array print -1 if (q.Count < 3) Console.WriteLine( "-1" ); else { // pop three largest elements int x = q[0]; int y = q[1]; int z = q[2]; q.RemoveRange(0, 3); // Reinsert x, y, z in priority_queue int ans = x * y * z; Console.WriteLine(ans); q.Add(x); q.Add(y); q.Add(z); } } } // Driver code public static void Main(String[] args) { int []arr = { 1, 2, 3, 4, 5 }; int n = arr.Length; LargestTripletMultiplication(arr, n); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // javascript implementation of largest triplet // multiplication // Prints the product of three largest numbers // in subarray arr[0..i] function LargestTripletMultiplication(arr , n) { // call a priority queue var q = []; // traversing the array for (i = 0; i < n; i++) { // pushing arr[i] in array q.push(arr[i]); q.sort(); // if less than three elements are present // in array print -1 if (q.length < 3) document.write( "-1<br/>" ); else { // pop three largest elements q.sort(); var x = q.pop(); var y = q.pop(); var z = q.pop(); // Reinsert x, y, z in priority_queue var ans = x * y * z; document.write(ans+ "<br/>" ); q.push(x); q.push(y); q.push(z); q.sort(); q.reverse(); } } } // Driver code var arr = [ 1, 2, 3, 4, 5 ]; var n = arr.length; LargestTripletMultiplication(arr, n); // This code contributed by umadevi9616 </script> |
Output
-1 -1 6 24 60
Time Complexity: O(N * log N ).
Auxiliary Space: O(N), N is the length of the array.
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