Minimum cost of reducing Array by merging any adjacent elements repetitively

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Given an array arr[] of N numbers. We can merge two adjacent numbers into one and the cost of merging the two numbers is equal to the sum of the two values. The task is to find the total minimum cost of merging all the numbers.
Examples:

Input: arr[] = { 6, 4, 4, 6 }
Output: 40
Explanation:
Following is the optimal way of merging numbers to get the total minimum cost of merging of numbers:
1. Merge (6, 4), then array becomes, arr[] = {10, 4, 6} and cost = 10
2. Merge (4, 6), then array becomes, arr[] = {10, 10} and cost = 10
3. Merge (10, 10), then array becomes, arr[] = {20} and cost = 20
Hence the total cost is 10 + 10 + 20 = 40.
Input: arr[] = { 3, 2, 4, 1 }
Output: 20
Explanation:
Following is the optimal way of merging numbers to get the total minimum cost of merging of numbers:
1. Merge (3, 2), then array becomes, arr[] = {5, 4, 1} and cost = 5
2. Merge (4, 1), then array becomes, arr[] = {5, 5} and cost = 5
3. Merge (5, 5), then array becomes, arr[] = {10} and cost = 10
Hence the total cost is 5 + 5 + 10 = 20.

Approach: The problem can be solved using Dynamic Programming. Below are the steps:

The idea is to merge 2 consecutive numbers into every possible index i and then solve recursively for left and right parts of index i.
Add the result of merging two numbers in above steps and store the minimum of them.
Since there are many subproblems that are repeating, hence we use memoization to store the values in a NxN matrix.
The recurrence relation for the above problem statement is given as:

dp[i][j] = min(dp[i][j], (sum[i][j] + dp[i][k] + dp[k + 1][j])), for every (i ? k lt; j)

5. Now dp[1][N] will give the minimum total cost of merging all the numbers.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the total minimum
// cost of merging two consecutive numbers
int mergeTwoNumbers(vector<int>& numbers)
{
    int len, i, j, k;
 
    // Find the size of numbers[]
    int n = numbers.size();
 
    // If array is empty, return 0
    if (numbers.size() == 0) {
        return 0;
    }
 
    // To store the prefix Sum of
    // numbers array numbers[]
    vector<int> prefixSum(n + 1, 0);
 
    // Traverse numbers[] to find the
    // prefix sum
    for (int i = 1; i <= n; i++) {
        prefixSum[i] = prefixSum[i - 1]
                       + numbers[i - 1];
    }
 
    // dp table to memoised the value
    vector<vector<int> > dp(
        n + 1,
        vector<int>(n + 1));
 
    // For single numbers cost is zero
    for (int i = 1; i <= n; i++) {
        dp[i][i] = 0;
    }
 
    // Iterate for length >= 1
    for (len = 2; len <= n; len++) {
 
        for (i = 1; i <= n - len + 1; i++) {
            j = i + len - 1;
 
            // Find sum in range [i..j]
            int sum = prefixSum[j]
                      - prefixSum[i - 1];
 
            // Initialise dp[i][j] to INT_MAX
            dp[i][j] = INT_MAX;
 
            // Iterate for all possible
            // K to find the minimum cost
            for (k = i; k < j; k++) {
 
                // Update the minimum sum
                dp[i][j]
                    = min(dp[i][j],
                          dp[i][k]
                              + dp[k + 1][j]
                              + sum);
            }
        }
    }
 
    // Return the final minimum cost
    return dp[1][n];
}
 
// Driver Code
int main()
{
    // Given set of numbers
    vector<int> arr1 = { 6, 4, 4, 6 };
 
    // Function Call
    cout << mergeTwoNumbers(arr1)
         << endl;
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
 
// Function to find the total minimum
// cost of merging two consecutive numbers
static int mergeTwoNumbers(int []numbers)
{
    int len, i, j, k;
 
    // Find the size of numbers[]
    int n = numbers.length;
 
    // If array is empty, return 0
    if (numbers.length == 0)
    {
        return 0;
    }
 
    // To store the prefix Sum of
    // numbers array numbers[]
    int []prefixSum = new int[n + 1];
 
    // Traverse numbers[] to find the
    // prefix sum
    for (i = 1; i <= n; i++) 
    {
        prefixSum[i] = prefixSum[i - 1] +
                         numbers[i - 1];
    }
 
    // dp table to memoised the value
    int [][]dp = new int[n + 1][n + 1];
 
    // Iterate for length >= 1
    for (len = 2; len <= n; len++) 
    {
 
        for (i = 1; i <= n - len + 1; i++)
        {
            j = i + len - 1;
 
            // Find sum in range [i..j]
            int sum = prefixSum[j] - 
                      prefixSum[i - 1];
 
            // Initialise dp[i][j] to Integer.MAX_VALUE
            dp[i][j] = Integer.MAX_VALUE;
 
            // Iterate for all possible
            // K to find the minimum cost
            for (k = i; k < j; k++) 
            {
 
                // Update the minimum sum
                dp[i][j] = Math.min(dp[i][j], 
                                    dp[i][k] +
                                    dp[k + 1][j] + 
                                    sum);
            }
        }
    }
 
    // Return the final minimum cost
    return dp[1][n];
}
 
// Driver Code
public static void main(String[] args)
{
    // Given set of numbers
    int []arr1 = { 6, 4, 4, 6 };
 
    // Function Call
    System.out.print(mergeTwoNumbers(arr1) + "\n");
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 program for the above approach
import sys
 
# Function to find the total minimum
# cost of merging two consecutive numbers
def mergeTwoNumbers(numbers):
 
    # Find the size of numbers[]
    n = len(numbers)
 
    # If array is empty, return 0
    if (len(numbers) == 0):
        return 0
     
    # To store the prefix Sum of
    # numbers array numbers[]
    prefixSum = [0] * (n + 1)
 
    # Traverse numbers[] to find the
    # prefix sum
    for i in range(1, n + 1):
        prefixSum[i] = (prefixSum[i - 1] +
                          numbers[i - 1])
     
    # dp table to memoised the value
    dp = [[0 for i in range(n + 1)] 
             for j in range(n + 1)] 
 
    # For single numbers cost is zero
    for i in range(1, n + 1):
        dp[i][i] = 0
     
    # Iterate for length >= 1
    for p in range(2, n + 1):
        for i in range(1, n - p + 2):
            j = i + p - 1
 
            # Find sum in range [i..j]
            sum = prefixSum[j] - prefixSum[i - 1]
 
            # Initialise dp[i][j] to _MAX
            dp[i][j] = sys.maxsize
 
            # Iterate for all possible
            # K to find the minimum cost
            for k in range(i, j):
 
                # Update the minimum sum
                dp[i][j] = min(dp[i][j], 
                              (dp[i][k] +
                               dp[k + 1][j] + sum))
             
    # Return the final minimum cost
    return dp[1][n]
 
# Driver Code
 
# Given set of numbers
arr1 = [ 6, 4, 4, 6 ]
 
# Function call
print(mergeTwoNumbers(arr1))
 
# This code is contributed by sanjoy_62

C#

// C# program for the above approach 
using System;
class GFG{ 
 
// Function to find the total minimum 
// cost of merging two consecutive numbers 
static int mergeTwoNumbers(int []numbers) 
{ 
    int len, i, j, k; 
 
    // Find the size of numbers[] 
    int n = numbers.Length; 
 
    // If array is empty, return 0 
    if (numbers.Length == 0) 
    { 
        return 0; 
    } 
 
    // To store the prefix Sum of 
    // numbers array numbers[] 
    int []prefixSum = new int[n + 1]; 
 
    // Traverse numbers[] to find the 
    // prefix sum 
    for (i = 1; i <= n; i++) 
    { 
        prefixSum[i] = prefixSum[i - 1] + 
                         numbers[i - 1]; 
    } 
 
    // dp table to memoised the value 
    int [,]dp = new int[n + 1, n + 1]; 
 
    // Iterate for length >= 1 
    for (len = 2; len <= n; len++) 
    { 
 
        for (i = 1; i <= n - len + 1; i++) 
        { 
            j = i + len - 1; 
 
            // Find sum in range [i..j] 
            int sum = prefixSum[j] - 
                      prefixSum[i - 1]; 
 
            // Initialise dp[i,j] to int.MaxValue 
            dp[i,j] = int.MaxValue; 
 
            // Iterate for all possible 
            // K to find the minimum cost 
            for (k = i; k < j; k++) 
            { 
 
                // Update the minimum sum 
                dp[i, j] = Math.Min(dp[i, j], 
                                    dp[i, k] + 
                                    dp[k + 1, j] + 
                                    sum); 
            } 
        } 
    } 
 
    // Return the readonly minimum cost 
    return dp[1, n]; 
} 
 
// Driver Code 
public static void Main(String[] args) 
{ 
    // Given set of numbers 
    int []arr1 = { 6, 4, 4, 6 }; 
 
    // Function Call 
    Console.Write(mergeTwoNumbers(arr1) + "\n"); 
} 
} 
 
// This code is contributed by sapnasingh4991

Javascript

<script>
 
// JavaScript program for the above approach
 
// Function to find the total minimum
// cost of merging two consecutive numbers
function mergeTwoNumbers(numbers)
{
    var len, i, j, k;
 
    // Find the size of numbers[]
    var n = numbers.length;
 
    // If array is empty, return 0
    if (numbers.length == 0) {
        return 0;
    }
 
    // To store the prefix Sum of
    // numbers array numbers[]
    var prefixSum = Array(n+1).fill(0);
 
    // Traverse numbers[] to find the
    // prefix sum
    for (var i = 1; i <= n; i++) {
        prefixSum[i] = prefixSum[i - 1]
                       + numbers[i - 1];
    }
 
    // dp table to memoised the value
    var dp = Array.from(Array(n+1), ()=>Array(n+1));
 
    // For single numbers cost is zero
    for (var i = 1; i <= n; i++) {
        dp[i][i] = 0;
    }
 
    // Iterate for length >= 1
    for (len = 2; len <= n; len++) {
 
        for (i = 1; i <= n - len + 1; i++) {
            j = i + len - 1;
 
            // Find sum in range [i..j]
            var sum = prefixSum[j]
                      - prefixSum[i - 1];
 
            // Initialise dp[i][j] to INT_MAX
            dp[i][j] = 1000000000;
 
            // Iterate for all possible
            // K to find the minimum cost
            for (k = i; k < j; k++) {
 
                // Update the minimum sum
                dp[i][j]
                    = Math.min(dp[i][j],
                          dp[i][k]
                              + dp[k + 1][j]
                              + sum);
            }
        }
    }
 
    // Return the final minimum cost
    return dp[1][n];
}
 
// Driver Code
 
// Given set of numbers
var arr1 = [6, 4, 4, 6];
 
// Function Call
document.write( mergeTwoNumbers(arr1))
 
 
</script>

Output:

Time Complexity: O(N³)
Auxiliary Space Complexity: O(N²)

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