Given an integer N, the task is to find the minimum cost to merge all the numbers from 1 to N where the cost of merging two set of numbers A and B is equal to the product of the product of the numbers in the respective sets.
Examples:
Input: N = 4
Output: 32
Merging {1} and {2} costs 1 * 2 = 2
Merging {1, 2} and {3} costs 2 * 3 = 6
Merge{1, 2, 3} and {4} costs 6 * 4 = 24
Hence, the minimal cost is 2 + 6 + 24 = 32Input: N = 2
Output: 2
Approach:
- The first approach that comes in our mind is sorting. We take first two smallest elements and add them, then continue adding to the rest of the elements in the sorted array. But it fails when the current running sum exceeds the next smallest value in the array coming next.
Take N = 5,
If we take the sorting approach, then-
Merge {1} and {2} - 1 * 2 = 2
Merge {1, 2} and {3} - 2 * 3 = 6
Merge{1, 2, 3} and {4} - 6 * 4 = 24
Merge{1, 2, 3, 4} and {5} - 24 * 5 = 120
Total sum = 152
But optimal way is,
Merge {1} and {2} - 1 * 2 = 2
Merge {1, 2} and {3} - 2 * 3 = 6
Merge {4} and {5} - 4 * 5 = 20
Merge {1, 2, 3} and {4, 5} - 6 * 20 = 120
Total sum = 148
This is the minimal answer.
- So, the correct approach to solve this problem is the Min-heap based approach. Initially, we push all the numbers from 1 to N into the Min-Heap.
- At every iteration, we extract the minimum and the second minimum element from the Min-Heap and insert their product back into it. This ensures that the addition cost generated will be minimum.
- We keep on repeating the above step until there is only one element remaining in the Min-Heap. The calculated sum till that instant gives us the required answer.
Below is the implementation of the above approach:
C++
// C++ program to find the Minimum // cost to merge numbers // from 1 to N.#include <bits/stdc++.h>using namespace std;// Function returns the // minimum costint GetMinCost(int N){ // Min Heap representation priority_queue<int, vector<int>, greater<int> > pq; // Add all elements to heap for (int i = 1; i <= N; i++) { pq.push(i); } int cost = 0; while (pq.size() > 1) { // First minimum int mini = pq.top(); pq.pop(); // Second minimum int secondmini = pq.top(); pq.pop(); // Multiply them int current = mini * secondmini; // Add to the cost cost += current; // Push the product into the // heap again pq.push(current); } // Return the optimal cost return cost;}// Driver codeint main(){ int N = 5; cout << GetMinCost(N);} |
Java
// Java program for the above approach import java.util.*;class GFG {// Function returns the // minimum cost static int GetMinCost(int N){ // Min Heap representation PriorityQueue<Integer> pq; pq = new PriorityQueue<>(); // Add all elements to heap for(int i = 1; i <= N; i++) { pq.add(i); } int cost = 0; while (pq.size() > 1) { // First minimum int mini = pq.remove(); // Second minimum int secondmini = pq.remove(); // Multiply them int current = mini * secondmini; // Add to the cost cost += current; // Push the product into the // heap again pq.add(current); } // Return the optimal cost return cost;}// Driver Codepublic static void main(String args[]){ int N = 5; // Function call System.out.println(GetMinCost(N));}}// This code is contributed by rutvik_56 |
Python3
# python3 program to find the Minimum # cost to merge numbers # from 1 to N.# Function returns the # minimum costdef GetMinCost(N): # Min Heap representation pq = [] # Add all elements to heap for i in range(1, N+1, 1): pq.append(i) pq.sort(reverse = False) cost = 0 while (len(pq) > 1): # First minimum mini = pq[0] pq.remove(pq[0]) # Second minimum secondmini = pq[0] pq.remove(pq[0]) # Multiply them current = mini * secondmini # Add to the cost cost += current # Push the product into the # heap again pq.append(current) pq.sort(reverse = False) # Return the optimal cost return cost# Driver codeif __name__ == '__main__': N = 5 print(GetMinCost(N))# This code is contributed by Bhupendra_Singh |
C#
// C# program to find the Minimum // cost to merge numbers // from 1 to N. using System;using System.Collections.Generic;class GFG{// Function returns the // minimum cost static int GetMinCost(int N) { // Min Heap representation List<int> pq = new List<int>(); // Add all elements to heap for(int i = 1; i <= N; i++) { pq.Add(i); } int cost = 0; pq.Sort(); while (pq.Count > 1) { // First minimum int mini = pq[0]; pq.RemoveAt(0); // Second minimum int secondmini = pq[0]; pq.RemoveAt(0); // Multiply them int current = mini * secondmini; // Add to the cost cost += current; // Push the product into the // heap again pq.Add(current); pq.Sort(); } // Return the optimal cost return cost; } // Driver codestatic void Main(){ int N = 5; Console.WriteLine(GetMinCost(N));}}// This code is contributed by divyeshrabadiya07 |
Javascript
<script> // Javascript program for the above approach // Function returns the // minimum cost function GetMinCost(N) { // Min Heap representation let pq = []; // Add all elements to heap for(let i = 1; i <= N; i++) { pq.push(i); } pq.sort(function(a, b){return a - b}); let cost = 0; while (pq.length > 1) { // First minimum let mini = pq[0]; pq.shift(); // Second minimum let secondmini = pq[0]; pq.shift(); // Multiply them let current = mini * secondmini; // Add to the cost cost += current; // Push the product into the // heap again pq.push(current); pq.sort(function(a, b){return a - b}); } // Return the optimal cost return cost; } let N = 5; // Function call document.write(GetMinCost(N)); // This code is contributed by decode2207.</script> |
Output:
148
Time Complexity: O(NlogN)
Auxiliary Space: O(N)
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