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 
        adj[N][2] ={{5000, 3000},
                    {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 (()). 
where '(' denotes as adj[i][1] and ')' denotes as adj[i][0]. 
Hence, for ()() sum is 3000+6000+1000+9000=19000.
and (()), sum is 3000+2000+8000+9000=220000. 
Thus answer is 19000

Input : N = 4 
        adj[N][2] = {{435, 111},
                     {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.
3. Denote ‘(‘ as adj[i][1] and ‘)’ as a adj[i][0].
4. Find the minimum sum of all possible correct bracket expressions.
5. Return the answer + adj[0][1].

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 <bits/stdc++.h>
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

<?php
// PHP 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_fill(0, 100,
      array_fill(0, 100, -1));

// Recursive function to check for
// correct bracket expression
function find($index, $openbrk, $n, $adj)
{
    global $MAX_VAL;
    global $dp;

    /// 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
$n = 4;
$adj = array(array(5000, 3000),
             array(6000, 2000),
             array(8000, 1000),
             array(9000, 6000));

echo find(1, 1, $n, $adj) + $adj[0][1];

// This code is contributed by ihritik
?>

Javascript

<script>
    // Javascript program to find the Minimum sum possible
    // of any bracket sequence of length N using
    // the given values for brackets

    let MAX_VAL = 10000000 ;

    // DP array
    let dp = new Array(100);
    for(let i = 0; i < 100; i++)
    {
        dp[i] = new Array(100);
        for(let j = 0; j < 100; j++)
        {
            dp[i][j] = 0;
        }
    }

    // Recursive function to check for
    // correct bracket expression
    function 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;
            }
            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];
    }

    let n = 4;
    let adj = [ [ 5000, 3000 ],
                [ 6000, 2000 ],
                [ 8000, 1000 ],
                [ 9000, 6000 ] ];

    for (let i = 0; i < dp.length; i ++)
      for (let j = 0; j < dp.length; j++)
        dp[i][j] = -1 ;


    document.write(find(1, 1, n, adj) + adj[0][1]);

    // This code is contributed by divyesh072019.
</script>

Output:

19000

Time Complexity: O(N²)