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:
- The first element of the bracket sequence can only be ‘(‘, hence value of adj[0][1] is only of use at index 0.
- Call a function to find a proper bracket expression using dp as discussed in this article.
- Denote ‘(‘ as adj[i][1] and ‘)’ as a adj[i][0].
- Find the minimum sum of all possible correct bracket expressions.
- 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]; // Recusive 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 alreaddy 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; } |
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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 alreaddy 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 } |
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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)]
# Recusive 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 alreaddy 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
[tabby title="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]; // Recusive 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 alreaddy 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_ } |
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Output:
19000
Time Complexity: O(N2)
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