Generate Complete Binary Tree in such a way that sum of non-leaf nodes is minimum

Given an array arr[] of size N, the task is to generate a Complete Binary Tree in such a way that sum of the non-leaf nodes is minimum, whereas values of the leaf node corresponds to the array elements in an In-order Traversal of the tree and value of each non-leaf node corresponds to the product of the largest leaf value in the left sub-tree and right sub-tree

Examples:

Input: arr[] = {1, 2, 3, 4}
Output: 20
Explanation:
Please refer below for explanation

Input: arr[] = {5, 2, 3}
Output: 21



Approach:
To remove a number arr[i], it needs a cost a * b, where b >= a and also an element of the array. To minimize the cost of removal, the idea is to minimize b. To compute the non-leaf node there are two candidates, that is the first largest number on the left and the first largest number on the right. The cost to remove arr[i] is a * min(left, right). It can be further decomposed as to find the next greater element in the array, on the left and one right.

Refer: Next greater element

Below is the implementation of the above approach:

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// C++ implementation to find the
// minimum cost tree
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find minimum cost tree
int MinCostTree(int arr[], int n)
{
    int ans = 0;
  
    // Stack
    vector<int> st = { INT_MAX };
  
    // Loop to traverse the array elements
    for (int i = 0; i < n; i++) {
          
        // Keep array elements 
        // in decreasing order by poping out
        // the elements from stack till the top
        // element is less than current element
        while (st.back() <= arr[i]) {
              
            // Get top element
            int x = st.back();
  
            // Remove it
            st.pop_back();
  
            // Get the minimum cost to remove x
            ans += x * min(st.back(), arr[i]);
        }
  
        // Push current element
        st.push_back(arr[i]);
    }
  
    // Find cost for all remaining elements
    for (int i = 2; i < st.size(); i++)
        ans += st[i] * st[i - 1];
  
    return ans;
}
  
// Driver Code
int main()
{
    int arr[] = { 5, 2, 3 };
  
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Function call
    cout << MinCostTree(arr, n);
  
    return 0;
}
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// Java implementation to find the
// minimum cost tree
import java.util.*;
  
class GFG{
  
// Function to find minimum cost tree
static int MinCostTree(int arr[], int n)
{
    int ans = 0;
  
    // Stack
    Vector<Integer> st = new Vector<Integer>();
    st.add(Integer.MAX_VALUE);
  
    // Loop to traverse the array elements
    for (int i = 0; i < n; i++) {
          
        // Keep array elements 
        // in decreasing order by poping out
        // the elements from stack till the top
        // element is less than current element
        while (st.get(st.size()-1) <= arr[i]) {
              
            // Get top element
            int x = st.get(st.size()-1);
  
            // Remove it
            st.remove(st.size()-1);
  
            // Get the minimum cost to remove x
            ans += x * Math.min(st.get(st.size()-1), arr[i]);
        }
  
        // Push current element
        st.add(arr[i]);
    }
  
    // Find cost for all remaining elements
    for (int i = 2; i < st.size(); i++)
        ans += st.get(i) * st.get(i-1);
  
    return ans;
}
  
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 5, 2, 3 };
  
    int n = arr.length;
  
    // Function call
    System.out.print(MinCostTree(arr, n));
}
}
  
// This code is contributed by sapnasingh4991
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# Python3 implementation to find the
# minimum cost tree
  
# Function to find minimum cost tree
def MinCostTree(arr, n):
      
    ans = 0
    st = [2**32]
      
    # Loop to traverse the array elements
    for i in range(n):
          
        # Keep array elements 
        # in decreasing order by poping out
        # the elements from stack till the top
        # element is less than current element
        while (st[-1] <= arr[i]):
              
            # Get top element
            x = st[-1]
              
            # Remove it
            st.pop()
              
            # Get the minimum cost to remove x
            ans += x * min(st[-1], arr[i])
              
        # Push current element
        st.append(arr[i])
          
    # Find cost for all remaining elements
    for i in range(2,len(st)):
        ans += st[i] * st[i - 1]
          
    return ans
      
# Driver Code
arr = [5, 2, 3]
  
n = len(arr)
  
# Function call
print(MinCostTree(arr, n))
  
# This code is contributed by shubhamsingh10
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// C# implementation to find the
// minimum cost tree
using System;
using System.Collections.Generic;
  
class GFG
{
  
// Function to find minimum cost tree
static int MinCostTree(int []arr, int n)
{
    int ans = 0;
  
    // Stack
    List<int> st = new List<int>();
    st.Add(int.MaxValue);
  
    // Loop to traverse the array elements
    for (int i = 0; i < n; i++) {
          
        // Keep array elements 
        // in decreasing order by poping out
        // the elements from stack till the top
        // element is less than current element
        while (st[st.Count-1] <= arr[i]) {
              
            // Get top element
            int x = st[st.Count-1];
  
            // Remove it
            st.RemoveAt(st.Count-1);
  
            // Get the minimum cost to remove x
            ans += x * Math.Min(st[st.Count-1], arr[i]);
        }
  
        // Push current element
        st.Add(arr[i]);
    }
  
    // Find cost for all remaining elements
    for (int i = 2; i < st.Count; i++)
        ans += st[i] * st[i-1];
  
    return ans;
}
  
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 5, 2, 3 };
  
    int n = arr.Length;
  
    // Function call
    Console.Write(MinCostTree(arr, n));
}
}
  
// This code is contributed by 29AjayKumar
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Output:
21

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