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Balanced expressions such that given positions have opening brackets in C++?
In case of a given integer m and an array of positions ‘position[]’ (1 <= length(position[]) <= 2m), find the number of ways of proper bracket expressions that can be constructed of length 2m such that given positions have the opening bracket.
Note: position[] array is provided in the form of (1-based indexing) [0, 1, 1, 0]. Here 1 indicates the positions at which open bracket should be set. At positions in case of value 0, either opening or closing bracket can be set.
Examples
Input: n = 2, position[] = [1, 0, 1, 0] Output: 1 The only possibility is given below: [ ] [ ] In this case, recursive and recursion implementing memorization approach will be explained.
Algorithm
We have to mark all the positions with open brackets in the given array adj1(say) as 1.
We run a recursive loop, in this way −
If count of total brackets(opening brackets subracted from closing brackets) is smaller than zero, return 0.
If the index attains till m and if the total brackets is equal to 0, then a solution is available and return 1, otherwise return 0.
If the index value has 1 pre-assigned to it, we have to return the function recursively with index+1 and increment the total brackets.
Otherwise we have to return the function recursively by inserting open brackets at that position or index and incrementing total brackets by 1 + inserting closed brackets at that index and decrementing total brackets by 1 and move on to the next index till m.
Following is the Recursive solution in case of above algorithm −
Example
// C++ application of above method implementing Recursion
#include <bits/stdc++.h>
using namespace std;
// Function to locate or find Number of proper bracket expressions
int find(int index1, int openbrk1, int m, int adj1[]){
// If open-closed brackets less than 0
if (openbrk1 < 0)
return 0;
// If index reaches the end of expression
if (index1 == m) {
// If brackets are balanced
if (openbrk1 == 0)
return 1;
else
return 0;
}
// If the current index has assigned open bracket
if (adj1[index1] == 1) {
// We have to move forward increasing the
// length of open brackets
return find(index1 + 1, openbrk1 + 1, m, adj1);
}
else {
// We have to move forward by inserting open as well
// as closed brackets on that index
return find(index1 + 1, openbrk1 + 1, m, adj1)
+ find(index1 + 1, openbrk1 - 1, m, adj1);
}
}
// Driver Code
int main(){
int m = 2;
// Open brackets at position 1
int adj1[4] = { 1, 0, 0, 0 };
// Calling the find function to calculate the answer
cout << find(0, 0, 2 * m, adj1) << endl;
return 0;
}Output
2
Memorized Approach −Time complexity of the above algorithm can be improved or optimized by implementing Memorization.
The only thing to be performed is to implement an array to store the results of previous iterations so that there is no require to recursively call the same function more than once if the value is already computed.
Following is the required implementation
// C++ application of above method implementing memorization
#include <bits/stdc++.h>
using namespace std;
#define M 1000
// Function to locate or find Number of proper bracket expressions
int find(int index1, int openbrk1, int m,
int dp1[M][M], int adj1[]){
// If open-closed brackets is less than 0
if (openbrk1 < 0)
return 0;
// If index attains or reaches the end of expression
if (index1 == m) {
// If brackets are balanced
if (openbrk1 == 0)
return 1;
else
return 0;
}
// If already stored in dp1
if (dp1[index1][openbrk1] != -1)
return dp1[index1][openbrk1];
// If the current index has assigned open bracket
if (adj1[index1] == 1) {
// We have to move forward increasing the length of open brackets
dp1[index1][openbrk1] = find(index1 + 1,
openbrk1 + 1, m, dp1, adj1);
}
else {
// We have to move forward by inserting open as
// well as closed brackets on that index
dp1[index1][openbrk1] = find(index1 + 1, openbrk1 + 1, m, dp1, adj1) + find(index1 + 1, openbrk1 - 1, m, dp1, adj1);
}
// We have to return the answer
return dp1[index1][openbrk1];
}
// Driver Code
int main(){
// dp1 array to precalculate the answer
int dp1[M][M];
int m = 2;
memset(dp1, -1, sizeof(dp1));
// Open brackets at position 1
int adj1[4] = { 1, 0, 0, 0 };
// We have to call the find function to compute the answer
cout<< find(0, 0, 2 * m, dp1, adj1) << endl;
return 0;
}
Output
2
Time complexity: O(N2)
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