# Minimum cost to merge numbers from 1 to N

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 = 32

Input: N = 2
Output: 2

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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
• 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 ` `using` `namespace` `std; ` ` `  `// Function returns the  ` `// minimum cost ` `int` `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 code ` `int` `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 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 Code ` `public` `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 cost ` `def` `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 code ` `if` `__name__ ``=``=` `'__main__'``: ` `     `  `    ``N ``=` `5` `    ``print``(GetMinCost(N)) ` ` `  `# This code is contributed by Bhupendra_Singh `

Output:

```148
```

Time Complexity: O(NlogN)
Auxiliary Space: O(N)

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Improved By : bgangwar59, rutvik_56