Minimum cost to merge all elements of List

Given a list of N integers, the task is to merge all the elements of the list into one with the minimum possible cost. The rule for merging is as follows:
Choose any two adjacent elements of the list with values say X and Y and merge them into a single element with value (X + Y) paying a total cost of (X + Y).

Examples:

Input: arr[] = {1, 3, 7}
Output: 15
All possible ways of merging:
a) {1, 3, 7} (cost = 0) -> {4, 7} (cost = 0+4 = 4) -> 11 (cost = 4+11 = 15)
b) {1, 3, 7} (cost = 0) -> {1, 10} (cost = 0+10= 10) -> 11 (cost = 10+11 = 21)
Thus, ans = 15

Input: arr[] = {1, 2, 3, 4}
Output: 19

Approach: This problem can be solved using dynamic programming. Let’s define the states of the dp first. DP[l][r] will be the minimum cost of merging the subarray arr[l…r] into one.
Now, lets look at the recurrence relation:



DP[l][r] = min(S(l, l) + S(l + 1, r) + DP[l][l] + DP[l + 1][r], S(l, l + 1) + S(l + 2, r) + DP[l][l + 1] + DP[l + 2][r], …, S(l, r – 1) + S(r, r) + DP[l][r – 1] + DP[r][r]) = S(l, r) + min(DP[l][l] + DP[l + 1][r], DP[l][l + 1] + DP[l + 2][r], …, DP[l][r – 1] + DP[r][r])
where S(x, y) is the sum of all the elements of the subarray arr[x…y]

The time complexity of the approach will be O(N3) as there are N * N states in total and each state takes O(N) time to solve in general.

Below is the implementation of the above approach:

CPP

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
#define N 401
  
// To store the states of DP
int dp[N][N];
bool v[N][N];
  
// Function to return the minimum merge cost
int minMergeCost(int i, int j, int* arr)
{
  
    // Base case
    if (i == j)
        return 0;
  
    // If the state has been solved before
    if (v[i][j])
        return dp[i][j];
  
    // Marking the state as solved
    v[i][j] = 1;
    int& x = dp[i][j];
  
    // Reference to dp[i][j]
    x = INT_MAX;
  
    // To store the sum of all the
    // elements in the subarray arr[i...j]
    int tot = 0;
    for (int k = i; k <= j; k++)
        tot += arr[k];
  
    // Loop to iterate the recurrence
    for (int k = i + 1; k <= j; k++) {
        x = min(x, tot + minMergeCost(i, k - 1, arr)
                       + minMergeCost(k, j, arr));
    }
  
    // Returning the solved value
    return x;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 3, 7 };
    int n = sizeof(arr) / sizeof(int);
  
    cout << minMergeCost(0, n - 1, arr);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG
{
  
static final int N = 401;
  
// To store the states of DP
static int [][]dp = new int[N][N];
static boolean [][]v = new boolean[N][N];
  
// Function to return the minimum merge cost
static int minMergeCost(int i, int j, int[] arr)
{
  
    // Base case
    if (i == j)
        return 0;
  
    // If the state has been solved before
    if (v[i][j])
        return dp[i][j];
  
    // Marking the state as solved
    v[i][j] = true;
    int x = dp[i][j];
  
    // Reference to dp[i][j]
    x = Integer.MAX_VALUE;
  
    // To store the sum of all the
    // elements in the subarray arr[i...j]
    int tot = 0;
    for (int k = i; k <= j; k++)
        tot += arr[k];
  
    // Loop to iterate the recurrence
    for (int k = i + 1; k <= j; k++) 
    {
        x = Math.min(x, tot + minMergeCost(i, k - 1, arr)
                    + minMergeCost(k, j, arr));
    }
  
    // Returning the solved value
    return x;
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 3, 7 };
    int n = arr.length;
  
    System.out.print(minMergeCost(0, n - 1, arr));
}
}
  
// This code is contributed by PrinciRaj1992

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Python3

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# Python3 implementation of the approach
import sys
  
N = 401;
  
# To store the states of DP
dp = [[0 for i in range(N)] for j in range(N)];
v = [[False for i in range(N)] for j in range(N)];
  
# Function to return the minimum merge cost
def minMergeCost(i, j, arr):
  
    # Base case
    if (i == j):
        return 0;
  
    # If the state has been solved before
    if (v[i][j]):
        return dp[i][j];
  
    # Marking the state as solved
    v[i][j] = True;
    x = dp[i][j];
  
    # Reference to dp[i][j]
    x = sys.maxsize;
  
    # To store the sum of all the
    # elements in the subarray arr[i...j]
    tot = 0;
    for k in range(i, j + 1):
        tot += arr[k];
  
    # Loop to iterate the recurrence
    for k in range(i + 1, j + 1):
        x = min(x, tot + minMergeCost(i, k - 1, arr) + \
                minMergeCost(k, j, arr));
      
    # Returning the solved value
    return x;
  
# Driver code
if __name__ == '__main__':
    arr = [ 1, 3, 7 ];
    n = len(arr);
  
    print(minMergeCost(0, n - 1, arr));
  
# This code is contributed by PrinciRaj1992

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
  
static readonly int N = 401;
  
// To store the states of DP
static int [,]dp = new int[N, N];
static bool [,]v = new bool[N, N];
  
// Function to return the minimum merge cost
static int minMergeCost(int i, int j, int[] arr)
{
  
    // Base case
    if (i == j)
        return 0;
  
    // If the state has been solved before
    if (v[i, j])
        return dp[i, j];
  
    // Marking the state as solved
    v[i, j] = true;
    int x = dp[i, j];
  
    // Reference to dp[i,j]
    x = int.MaxValue;
  
    // To store the sum of all the
    // elements in the subarray arr[i...j]
    int tot = 0;
    for (int k = i; k <= j; k++)
        tot += arr[k];
  
    // Loop to iterate the recurrence
    for (int k = i + 1; k <= j; k++) 
    {
        x = Math.Min(x, tot + minMergeCost(i, k - 1, arr)
                    + minMergeCost(k, j, arr));
    }
  
    // Returning the solved value
    return x;
}
  
// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 3, 7 };
    int n = arr.Length;
  
    Console.Write(minMergeCost(0, n - 1, arr));
}
}
  
// This code is contributed by 29AjayKumar

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Output:

15

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