# Count pairs of parentheses sequences such that parentheses are balanced

Given N bracket sequences, the task is to find the number of pairs of bracket sequences by joining which can be obtained a balanced bracket sequence as a whole. A bracket parentheses sequence can only be a part of a single pair.

**Examples:**

Input:{ ")())", ")", "((", "((", "(", ")", ")"}Output:2 Bracket sequence {1, 3} and {5, 6}Input:{"()", "(())", "(())", "()"}Output:2 Since all brackets are balanced, hence we can form 2 pairs with 4.

**Approach:** The following steps can be followed to solve the above problem:

- Count required opening and closing brackets, of individuals.
- If required closing brackets > 0 and opening brackets is 0, then hash the bracket’s required closing number.
- Similarly, if required opening brackets > 0 and closing brackets is 0, then hash the bracket’s required opening number.
- Count the balanced bracket sequences.
- Add (number of balanced bracket sequences/2) to the number of pairs.
- For every number of sequences which requires same number of opening brackets, min(hash[open], hash[close]) will be added to the number of pairs

Below is the implementation of the above approach:

## C++

`// C++ program to count the number of pairs ` `// of balanced parentheses ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to count the number of pairs ` `int` `countPairs(string bracks[], ` `int` `num) ` `{ ` ` ` ` ` `// Hashing function to count the ` ` ` `// opening and closing brackets ` ` ` `unordered_map<` `int` `, ` `int` `> open, close; ` ` ` ` ` `int` `cnt = 0; ` ` ` ` ` `// Traverse for all bracket sequences ` ` ` `for` `(` `int` `i = 0; i < num; i++) { ` ` ` ` ` `// Get the string ` ` ` `string s = bracks[i]; ` ` ` ` ` `int` `l = s.length(); ` ` ` ` ` `// Counts the opening and closing required ` ` ` `int` `op = 0, cl = 0; ` ` ` ` ` `// Traverse in the string ` ` ` `for` `(` `int` `j = 0; j < l; j++) { ` ` ` ` ` `// If it is a opening bracket ` ` ` `if` `(s[j] == ` `'('` `) ` ` ` `op++; ` ` ` `else` `// Closing bracket ` ` ` `{ ` ` ` ` ` `// If openings are there, then close it ` ` ` `if` `(op) ` ` ` `op--; ` ` ` `else` `// Else increase count of closing ` ` ` `cl++; ` ` ` `} ` ` ` `} ` ` ` ` ` `// If requirements of openings ` ` ` `// are there and no closing ` ` ` `if` `(op && !cl) ` ` ` `open[op]++; ` ` ` ` ` `// If requirements of closing ` ` ` `// are there and no opening ` ` ` `if` `(cl && !op) ` ` ` `close[cl]++; ` ` ` ` ` `// Perfect ` ` ` `if` `(!op && !cl) ` ` ` `cnt++; ` ` ` `} ` ` ` ` ` `// Divide by two since two ` ` ` `// perfect makes one pair ` ` ` `cnt = cnt / 2; ` ` ` ` ` `// Traverse in the open and find ` ` ` `// corresponding minimum ` ` ` `for` `(` `auto` `it : open) ` ` ` `cnt += min(it.second, close[it.first]); ` ` ` ` ` `return` `cnt; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `string bracks[] = { ` `")())"` `, ` `")"` `, ` `"(("` `, ` `"(("` `, ` `"("` `, ` `")"` `, ` `")"` `}; ` ` ` `int` `num = ` `sizeof` `(bracks) / ` `sizeof` `(bracks[0]); ` ` ` ` ` `cout << countPairs(bracks, num); ` `} ` |

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## Python3

`# Python3 program to count the number of pairs ` `# of balanced parentheses ` `import` `math as mt ` ` ` `# Function to count the number of pairs ` `def` `countPairs(bracks, num): ` ` ` ` ` `# Hashing function to count the ` ` ` `# opening and closing brackets ` ` ` `openn` `=` `dict` `() ` ` ` `close` `=` `dict` `() ` ` ` ` ` `cnt ` `=` `0` ` ` ` ` `# Traverse for all bracket sequences ` ` ` `for` `i ` `in` `range` `(num): ` ` ` ` ` `# Get the string ` ` ` `s ` `=` `bracks[i] ` ` ` ` ` `l ` `=` `len` `(s) ` ` ` ` ` `# Counts the opening and closing required ` ` ` `op,cl ` `=` `0` `,` `0` ` ` ` ` `# Traverse in the string ` ` ` `for` `j ` `in` `range` `(l): ` ` ` `# If it is a opening bracket ` ` ` `if` `(s[j] ` `=` `=` `'('` `): ` ` ` `op` `+` `=` `1` ` ` `else` `: ` `# Closing bracket ` ` ` ` ` ` ` `# If openings are there, then close it ` ` ` `if` `(op): ` ` ` `op` `-` `=` `1` ` ` `else` `: ` `# Else increase count of closing ` ` ` `cl` `+` `=` `1` ` ` ` ` ` ` ` ` `# If requirements of openings ` ` ` `# are there and no closing ` ` ` `if` `(op ` `and` `cl` `=` `=` `0` `): ` ` ` `if` `op ` `in` `openn.keys(): ` ` ` `openn[op]` `+` `=` `1` ` ` `else` `: ` ` ` `openn[op]` `=` `1` ` ` ` ` ` ` `# If requirements of closing ` ` ` `# are there and no opening ` ` ` `if` `(cl ` `and` `op` `=` `=` `0` `): ` ` ` `if` `cl ` `in` `openn.keys(): ` ` ` `close[cl]` `+` `=` `1` ` ` `else` `: ` ` ` `close[cl]` `=` `1` ` ` ` ` ` ` `# Perfect ` ` ` `if` `(op` `=` `=` `0` `and` `cl` `=` `=` `0` `): ` ` ` `cnt` `+` `=` `1` ` ` ` ` ` ` `# Divide by two since two ` ` ` `# perfect makes one pair ` ` ` `cnt ` `=` `cnt ` `/` `/` `2` ` ` ` ` `# Traverse in the open and find ` ` ` `# corresponding minimum ` ` ` `for` `it ` `in` `openn: ` ` ` `cnt ` `+` `=` `min` `(openn[it], close[it]) ` ` ` ` ` `return` `cnt ` ` ` ` ` `# Driver Code ` `bracks` `=` `[` `")())"` `, ` `")"` `, ` `"(("` `, ` `"(("` `, ` `"("` `, ` `")"` `, ` `")"` `] ` `num ` `=` `len` `(bracks) ` ` ` `print` `(countPairs(bracks, num)) ` ` ` `#This code is contributed by Mohit kumar 29 ` |

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**Output:**

2

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