# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

 `// C++ implementation of the approach ` `#include ` `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; ` `} `

## Java

 `// 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 `

## Python3

 `# 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 `

## C#

 `// 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 `

Output:

```15
```

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