Minimize Cost to reduce the Array to a single element by given operations

Given an array a[] consisting of N integers and an integer K, the task is to find the minimum cost to reduce the given array to a single element by choosing any pair of consecutive array elements and repace them by (a[i] + a[i+1]) for a cost K * (a[i] + a[i+1]).

Examples:

Input: a[] = {1, 2, 3}, K = 2 
Output: 18 
Explanation: 
Repacing {1, 2} by 3 modifies the array to {3, 3}. Cost 2 * 3 = 6 
Repacing {3, 3} by 6 modifies the array to {6}. Cost 2 * 6 = 12 
Therefore, the total cost is 18

Input: a[] = {4, 5, 6, 7}, K = 3 
Output: 132

Naive Approach: 
The simplest solution is to split the array into two halves, for every index and compute the cost of the two halves recursively and finally add their respective costs.



Below is the implementation of the above approach:

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// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
#define inf 10000009
  
// Function to combine the sum of the two halves
int Combine(int a[], int i, int j)
{
    int sum = 0;
  
    // Calculate the sum from i to j
    for (int l = i; l <= j; l++)
        sum += a[l];
  
    return sum;
}
  
// Function to minimize the cost to
// reduce the array to a single element
int minCost(int a[], int i, int j, int k)
{
    if (i >= j) 
    {
  
        // Base case
        // If n = 1 or n = 0
        return 0;
    }
  
    // Intialize cost to maximum value
    int best_cost = inf;
  
    // Iterate through all possible indices
    // and find the best index
    // to combine the subproblems
    for (int pos = i; pos < j; pos++)
    {
  
        // Compute left subproblem
        int left = minCost(a, i, pos, k);
  
        // Compute right subproblem
        int right = minCost(a, pos + 1, j, k);
  
        // Calculate the best cost
        best_cost = min(best_cost, left + right + 
                                   k * Combine(a, i, j));
    }
  
    // Return the answer
    return best_cost;
}
  
// Driver code
int main()
{
    int n = 4;
    int a[] = { 4, 5, 6, 7 };
    int k = 3;
  
    cout << minCost(a, 0, n - 1, k) << endl;
    return 0;
}
  
// This code is contributed by PrinciRaj1992
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// Java Program to implement
// the above approach
import java.io.*;
  
class GFG {
  
    static int inf = 10000009;
  
    // Function to minimize the cost to
    // reduce the array to a single element
    public static int minCost(int a[], int i,
                              int j, int k)
    {
        if (i >= j) {
  
            // Base case
            // If n = 1 or n = 0
            return 0;
        }
  
        // Intialize cost to maximum value
        int best_cost = inf;
  
        // Iterate through all possible indices
        // and find the best index
        // to combine the subproblems
        for (int pos = i; pos < j; pos++) {
  
            // Compute left subproblem
            int left = minCost(a, i, pos, k);
  
            // Compute right subproblem
            int right = minCost(a, pos + 1, j, k);
  
            // Calculate the best  cost
            best_cost = Math.min(
                best_cost,
                left + right + k * Combine(a, i, j));
        }
  
        // Return the answer
        return best_cost;
    }
  
    // Function to combine the sum of the two halves
    public static int Combine(int a[], int i, int j)
    {
        int sum = 0;
  
        // Calculate the sum from i to j
        for (int l = i; l <= j; l++)
            sum += a[l];
  
        return sum;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int n = 4;
        int a[] = { 4, 5, 6, 7 };
        int k = 3;
  
        System.out.println(minCost(a, 0, n - 1, k));
    }
}
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# Python3 Program to implement
# the above approach
inf = 10000009;
  
# Function to minimize the cost to
# reduce the array to a single element
def minCost(a, i, j, k):
    if (i >= j):
        
        # Base case
        # If n = 1 or n = 0
        return 0;
  
    # Intialize cost to maximum value
    best_cost = inf;
  
    # Iterate through all possible indices
    # and find the best index
    # to combine the subproblems
    for pos in range(i, j):
        
        # Compute left subproblem
        left = minCost(a, i, pos, k);
  
        # Compute right subproblem
        right = minCost(a, pos + 1, j, k);
  
        # Calculate the best cost
        best_cost = min(best_cost, 
                        left + right + 
                        k * Combine(a, i, j));
  
    # Return the answer
    return best_cost;
  
# Function to combine 
# the sum of the two halves
def Combine(a, i, j):
    sum = 0;
  
    # Calculate the sum from i to j
    for l in range(i, j + 1):
        sum += a[l];
  
    return sum;
  
# Driver code
if __name__ == '__main__':
    n = 4;
    a = [4, 5, 6, 7];
    k = 3;
  
    print(minCost(a, 0, n - 1, k));
  
# This code is contributed by Amit Katiyar
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// C# Program to implement
// the above approach
using System;
class GFG{
  
  static int inf = 10000009;
  
  // Function to minimize the cost to
  // reduce the array to a single element
  public static int minCost(int []a, int i,
                            int j, int k)
  {
    if (i >= j) 
    {
  
      // Base case
      // If n = 1 or n = 0
      return 0;
    }
  
    // Intialize cost to maximum value
    int best_cost = inf;
  
    // Iterate through all possible indices
    // and find the best index
    // to combine the subproblems
    for (int pos = i; pos < j; pos++)
    {
  
      // Compute left subproblem
      int left = minCost(a, i, pos, k);
  
      // Compute right subproblem
      int right = minCost(a, pos + 1, j, k);
  
      // Calculate the best  cost
      best_cost = Math.Min(best_cost,
                           left + right + 
                           k * Combine(a, i, j));
    }
  
    // Return the answer
    return best_cost;
  }
  
  // Function to combine the sum of the two halves
  public static int Combine(int []a, int i, int j)
  {
    int sum = 0;
  
    // Calculate the sum from i to j
    for (int l = i; l <= j; l++)
      sum += a[l];
  
    return sum;
  }
  
  // Driver code
  public static void Main(String[] args)
  {
    int n = 4;
    int []a = { 4, 5, 6, 7 };
    int k = 3;
  
    Console.WriteLine(minCost(a, 0, n - 1, k));
  }
}
  
// This code is contributed by Rohit_ranjan
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Output: 
132

Time Complexity: O(2N
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use the concept of Dynamic Programming. Follow the steps below to solve the problem:

Below is the implementation of the above approach:

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// C++ program for the above approach 
#include <bits/stdc++.h>
using namespace std;
  
int inf = 10000000; 
  
// Function to generate the cost using 
// Prefix Sum Array technique 
vector<int> preprocess(vector<int> a, int n) 
    vector<int> p(n); 
    p[0] = a[0]; 
      
    for(int i = 1; i < n; i++) 
    {
        p[i] = p[i - 1] + a[i]; 
    }
    return p; 
}
  
// Function to combine the sum of the
// two subproblems 
int Combine(vector<int> p, int i, int j) 
    if (i == 0) 
        return p[j]; 
    else
        return p[j] - p[i - 1]; 
  
// Function to minimize the cost to 
// add the array elements to a single element 
int minCost(vector<int> a, int i, int j, int k, 
            vector<int> prefix, vector<vector<int>> dp) 
    if (i >= j) 
        return 0; 
  
    // Check if the value is 
    // already stored in the array 
    if (dp[i][j] != -1) 
        return dp[i][j]; 
  
    int best_cost = inf; 
    for(int pos = i; pos < j; pos++)
    
          
        // Compute left subproblem 
        int left = minCost(a, i, pos, 
                           k, prefix, dp); 
  
        // Compute left subproblem 
        int right = minCost(a, pos + 1, j, 
                            k, prefix, dp); 
  
        // Calculate minimum cost 
        best_cost = min(best_cost, left + right +
                       (k * Combine(prefix, i, j))); 
    
      
    // Store the answer to 
    // avoid recalculation 
    return dp[i][j] = best_cost; 
  
// Driver code    
int main()
{
    int n = 4; 
  
    vector<int> a = { 4, 5, 6, 7 }; 
  
    int k = 3; 
  
    // Initialise dp array 
    vector<vector<int>> dp;
    dp.resize(n + 1, vector<int>(n + 1));
    for(int i = 0; i < n + 1; i++)
    {
        for(int j = 0; j < n + 1; j++)
        {
            dp[i][j] = -1;
        }
    }
  
    // Preprocessing the array 
    vector<int> prefix = preprocess(a, n); 
      
    cout << minCost(a, 0, n - 1, k, prefix, dp)
         << endl;
  
    return 0;
}
  
// This code is contributed by divyeshrabadiya07
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// Java Program for the above approach
import java.util.*;
public class Main {
  
    static int inf = 10000000;
  
    // Function to minimize the cost to
    // add the array elements to a single element
    public static int minCost(int a[], int i, int j, int k,
                              int[] prefix, int[][] dp)
    {
        if (i >= j)
            return 0;
  
        // Check if the value is
        // already stored in the array
        if (dp[i][j] != -1)
            return dp[i][j];
  
        int best_cost = inf;
        for (int pos = i; pos < j; pos++) {
  
            // Compute left subproblem
            int left = minCost(a, i, pos, k, prefix, dp);
  
            // Compute left subproblem
            int right
                = minCost(a, pos + 1, j, k, prefix, dp);
  
            // Calculate minimum cost
            best_cost = Math.min(
                best_cost,
                left + right + (k * Combine(prefix, i, j)));
        }
  
        // Store the answer to
        // avoid recalculation
        return dp[i][j] = best_cost;
    }
  
    // Function to generate the cost using
    // Prefix Sum Array technique
    public static int[] preprocess(int[] a, int n)
    {
        int p[] = new int[n];
        p[0] = a[0];
        for (int i = 1; i < n; i++)
            p[i] = p[i - 1] + a[i];
        return p;
    }
  
    // Function to combine the sum of the two subproblems
    public static int Combine(int[] p, int i, int j)
    {
        if (i == 0)
            return p[j];
        else
            return p[j] - p[i - 1];
    }
  
    // Driver Code
    public static void main(String args[])
    {
        int n = 4;
  
        int a[] = { 4, 5, 6, 7 };
  
        int k = 3;
  
        // Initialise dp array
        int dp[][] = new int[n + 1][n + 1];
        for (int i[] : dp)
            Arrays.fill(i, -1);
  
        // Preprocessing the array
        int prefix[] = preprocess(a, n);
  
        System.out.println(
            minCost(a, 0, n - 1, k, prefix, dp));
    }
}
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# Python3 program for the above approach
inf = 10000000
  
# Function to minimize the cost to
# add the array elements to a single element
def minCost(a, i, j, k, prefix, dp):
      
    if (i >= j):
        return 0
  
    # Check if the value is
    # already stored in the array
    if (dp[i][j] != -1):
        return dp[i][j]
  
    best_cost = inf
    for pos in range(i, j):
  
        # Compute left subproblem
        left = minCost(a, i, pos,
                       k, prefix, dp)
  
        # Compute left subproblem
        right = minCost(a, pos + 1, j, 
                        k, prefix, dp)
  
        # Calculate minimum cost
        best_cost = min(best_cost,
                        left + right + 
                          (k * Combine(prefix, i, j)))
  
    # Store the answer to
    # avoid recalculation
    dp[i][j] = best_cost
      
    return dp[i][j]
  
# Function to generate the cost using
# Prefix Sum Array technique
def preprocess(a, n):
      
    p = [0] * n
    p[0] = a[0]
      
    for i in range(1, n):
        p[i] = p[i - 1] + a[i]
          
    return p
  
# Function to combine the sum
# of the two subproblems
def Combine(p, i, j):
      
    if (i == 0):
        return p[j]
    else:
        return p[j] - p[i - 1]
  
# Driver Code
if __name__ == "__main__":
      
    n = 4
    a = [ 4, 5, 6, 7 ]
    k = 3
  
    # Initialise dp array
    dp = [[-1 for x in range (n + 1)]
              for y in range (n + 1)]
  
    # Preprocessing the array
    prefix = preprocess(a, n)
  
    print(minCost(a, 0, n - 1, k, prefix, dp))
  
# This code is contributed by chitranayal
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// C# Program for the above approach
using System;
class GFG{
  
  static int inf = 10000000;
  
  // Function to minimize the cost to
  // add the array elements to a single element
  public static int minCost(int []a, int i, int j, int k,
                            int[] prefix, int[,] dp)
  {
    if (i >= j)
      return 0;
  
    // Check if the value is
    // already stored in the array
    if (dp[i, j] != -1)
      return dp[i, j];
  
    int best_cost = inf;
    for (int pos = i; pos < j; pos++) 
    {
  
      // Compute left subproblem
      int left = minCost(a, i, pos, k, prefix, dp);
  
      // Compute left subproblem
      int right = minCost(a, pos + 1, j, 
                          k, prefix, dp);
  
      // Calculate minimum cost
      best_cost = Math.Min(best_cost, left + right + 
                          (k * Combine(prefix, i, j)));
    }
  
    // Store the answer to
    // avoid recalculation
    return dp[i, j] = best_cost;
  }
  
  // Function to generate the cost using
  // Prefix Sum Array technique
  public static int[] preprocess(int[] a, int n)
  {
    int []p = new int[n];
    p[0] = a[0];
    for (int i = 1; i < n; i++)
      p[i] = p[i - 1] + a[i];
    return p;
  }
  
  // Function to combine the sum of the two subproblems
  public static int Combine(int[] p, int i, int j)
  {
    if (i == 0)
      return p[j];
    else
      return p[j] - p[i - 1];
  }
  
  // Driver Code
  public static void Main(String []args)
  {
    int n = 4;
  
    int []a = { 4, 5, 6, 7 };
  
    int k = 3;
  
    // Initialise dp array
    int [,]dp = new int[n + 1, n + 1];
    for(int i = 0; i < n + 1; i++)
    {
      for (int j = 0; j < n + 1; j++)
      {
        dp[i, j] = -1;
      }
    }
    // Preprocessing the array
    int []prefix = preprocess(a, n);
  
    Console.WriteLine(minCost(a, 0, n - 1, k, 
                              prefix, dp));
  }
}
  
// This code is contributed by sapnasingh4991
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Output: 
132

Time Complexity: O(N2
Auxiliary Space: O(N2)
 

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