# Minimum cost of reducing Array by merging any adjacent elements repetitively

• Difficulty Level : Hard
• Last Updated : 14 Jun, 2021

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:

1. 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.
2. Add the result of merging two numbers in above steps and store the minimum of them.
3. Since there are many subproblems that are repeating, hence we use memoization to store the values in a NxN matrix.
4. 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 â‰¤ 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 ``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 > 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

 ``

Output:

`40`

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

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