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# Minimum sum possible of any bracket sequence of length N

Given a number N representing the length of a bracket sequence consisting of brackets ‘(‘, ‘)’. The actual sequence is not known beforehand. Given the values of both brackets ‘(‘ and ‘)’ if placed at index in the expression.
The task is to find the minimum sum possible of any bracket sequence of length N using the above information.
Here adj[i][0] represents the value assigned to ‘)’ bracket at ith index and adj[i][1] represents the value assigned to ‘(‘ bracket at ith index.
Constraints

• There should be N/2 pairs made of brackets. That is, N/2 pairs of ‘(‘, ‘)’.
• Find minimum sum of proper bracket expression.
• Index starts from 0.

Examples

```Input : N = 4
{6000, 2000},
{8000, 1000},
{9000, 6000}}
Output : 19000
Assigning first index as '(' for proper
bracket expression is (_ _ _  .
Now all the possible bracket expressions are ()() and (()).
Hence, for ()() sum is 3000+6000+1000+9000=19000.
and (()), sum is 3000+2000+8000+9000=220000.

Input : N = 4
{43, 33},
{1241, 1111},
{234, 22}}
Output : 1499```

Algorithm

1. The first element of the bracket sequence can only be ‘(‘, hence value of adj[0][1] is only of use at index 0.
2. Call a function to find a proper bracket expression using dp as discussed in this article.
4. Find the minimum sum of all possible correct bracket expressions.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the Minimum sum possible``// of any bracket sequence of length N using``// the given values for brackets` `#include ``using` `namespace` `std;` `#define MAX_VAL 10000000` `// DP array``int` `dp[100][100];` `// Recursive function to check for``// correct bracket expression``int` `find(``int` `index, ``int` `openbrk, ``int` `n, ``int` `adj[][2])``{``    ``/// Not a proper bracket expression``    ``if` `(openbrk < 0)``        ``return` `MAX_VAL;` `    ``// If reaches at end``    ``if` `(index == n) {` `        ``/// If proper bracket expression``        ``if` `(openbrk == 0) {``            ``return` `0;``        ``}``        ``else` `// if not, return max``            ``return` `MAX_VAL;``    ``}` `    ``// If already visited``    ``if` `(dp[index][openbrk] != -1)``        ``return` `dp[index][openbrk];` `    ``// To find out minimum sum``    ``dp[index][openbrk] = min(adj[index][1] + find(index + 1,``                                                  ``openbrk + 1, n, adj),``                             ``adj[index][0] + find(index + 1,``                                                  ``openbrk - 1, n, adj));` `    ``return` `dp[index][openbrk];``}` `// Driver Code``int` `main()``{``    ``int` `n = 4;``    ``int` `adj[n][2] = { { 5000, 3000 },``                      ``{ 6000, 2000 },``                      ``{ 8000, 1000 },``                      ``{ 9000, 6000 } };` `    ``memset``(dp, -1, ``sizeof``(dp));` `    ``cout << find(1, 1, n, adj) + adj[0][1] << endl;` `    ``return` `0;``}`

## Java

 `// Java program to find the Minimum sum possible``// of any bracket sequence of length N using``// the given values for brackets` `public` `class` `GFG {` `    ``final` `static` `int` `MAX_VAL = ``10000000` `;` `    ``// DP array``    ``static` `int` `dp[][] = ``new` `int``[``100``][``100``];` `    ``// Recursive function to check for``    ``// correct bracket expression``    ``static` `int` `find(``int` `index, ``int` `openbrk, ``int` `n, ``int` `adj[][])``    ``{``        ``/// Not a proper bracket expression``        ``if` `(openbrk < ``0``)``            ``return` `MAX_VAL;` `        ``// If reaches at end``        ``if` `(index == n) {` `            ``/// If proper bracket expression``            ``if` `(openbrk == ``0``) {``                ``return` `0``;``            ``}``            ``else` `// if not, return max``                ``return` `MAX_VAL;``        ``}` `        ``// If already visited``        ``if` `(dp[index][openbrk] != -``1``)``            ``return` `dp[index][openbrk];` `        ``// To find out minimum sum``        ``dp[index][openbrk] = Math.min(adj[index][``1``] + find(index + ``1``,``                                                      ``openbrk + ``1``, n, adj),``                                 ``adj[index][``0``] + find(index + ``1``,``                                                      ``openbrk - ``1``, n, adj));` `        ``return` `dp[index][openbrk];``    ``}`  `// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``            ``int` `n = ``4``;``            ``int` `adj[][] = { { ``5000``, ``3000` `},``                              ``{ ``6000``, ``2000` `},``                              ``{ ``8000``, ``1000` `},``                              ``{ ``9000``, ``6000` `} };` `            ``for` `(``int` `i = ``0``; i < dp.length; i ++)``                ``for` `(``int` `j = ``0``; j < dp.length; j++)``                    ``dp[i][j] = -``1` `;``                `  `            ``System.out.println(find(``1``, ``1``, n, adj) + adj[``0``][``1``]);`  `    ``}``    ``// This code is contributed by ANKITRAI1``}`

## Python3

 `# Python 3 program to find the Minimum sum``# possible of any bracket sequence of length``# N using the given values for brackets` `MAX_VAL ``=` `10000000` `# DP array``dp ``=` `[[``-``1` `for` `i ``in` `range``(``100``)]``          ``for` `i ``in` `range``(``100``)]` `# Recursive function to check for``# correct bracket expression``def` `find(index, openbrk, n, adj):``    ` `    ``# Not a proper bracket expression``    ``if` `(openbrk < ``0``):``        ``return` `MAX_VAL` `    ``# If reaches at end``    ``if` `(index ``=``=` `n):``        ` `        ``# If proper bracket expression``        ``if` `(openbrk ``=``=` `0``):``            ``return` `0``            ` `    ``# if not, return max``        ``else``:``            ``return` `MAX_VAL` `    ``# If already visited``    ``if` `(dp[index][openbrk] !``=` `-``1``):``        ``return` `dp[index][openbrk]` `    ``# To find out minimum sum``    ``dp[index][openbrk] ``=` `min``(adj[index][``1``] ``+` `find(index ``+` `1``,``                                             ``openbrk ``+` `1``, n, adj),``                             ``adj[index][``0``] ``+` `find(index ``+` `1``,``                                             ``openbrk ``-` `1``, n, adj))` `    ``return` `dp[index][openbrk]` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``n ``=` `4``;``    ``adj ``=` `[[``5000``, ``3000``],[``6000``, ``2000``],``           ``[``8000``, ``1000``],[``9000``, ``6000``]]` `    ``print``(find(``1``, ``1``, n, adj) ``+` `adj[``0``][``1``])``    ` `# This code is contributed by``# Sanjit_Prasad`

## C#

 `// C# program to find the Minimum sum possible``// of any bracket sequence of length N using``// the given values for brackets``using` `System;``  ` `class` `GFG``{``    ``public` `static` `int` `MAX_VAL = 10000000;``    ` `    ``// DP array``    ``public` `static` `int``[,] dp = ``new` `int``[100,100];``      ` `    ``// Recursive function to check for``    ``// correct bracket expression``    ``public` `static` `int` `find(``int` `index, ``int` `openbrk, ``int` `n, ``int``[,] adj)``    ``{``        ``/// Not a proper bracket expression``        ``if` `(openbrk < 0)``            ``return` `MAX_VAL;``      ` `        ``// If reaches at end``        ``if` `(index == n) {``      ` `            ``/// If proper bracket expression``            ``if` `(openbrk == 0) {``                ``return` `0;``            ``}``            ``else` `// if not, return max``                ``return` `MAX_VAL;``        ``}``      ` `        ``// If already visited``        ``if` `(dp[index,openbrk] != -1)``            ``return` `dp[index,openbrk];``      ` `        ``// To find out minimum sum``        ``dp[index,openbrk] = Math.Min(adj[index,1] + find(index + 1,``                                                      ``openbrk + 1, n, adj),``                                 ``adj[index,0] + find(index + 1,``                                                      ``openbrk - 1, n, adj));``      ` `        ``return` `dp[index,openbrk];``    ``}``      ` `    ``// Driver Code     ``    ` `    ``static` `void` `Main()``    ``{``        ``int` `n = 4;``         ` `        ``int``[,] adj = ``new` `int``[,]{``                            ``{ 5000, 3000 },``                            ``{ 6000, 2000 },``                            ``{ 8000, 1000 },``                            ``{ 9000, 6000 }``        ``};``      ` `        ``for``(``int` `i = 0; i < 100; i++)``            ``for``(``int` `j = 0; j < 100; j++)``                ``dp[i,j] = -1;``      ` `        ``Console.Write(find(1, 1, n, adj) + adj[0,1] + ``"\n"``);``    ``}``    ``//This code is contributed by DrRoot_``}`

## PHP

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## Javascript

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Output:

`19000`

Time Complexity: O(N2)