# Minimum operation required to make a balanced sequence

A balanced sequence is defined as a string in which for every opening bracket there are 2 continuous closing brackets. Thus {}}, {{}}}}, {{}}{}}}} are balanced whereas }}{, {} are not balanced.

Now given a sequence of brackets (‘{‘ and ‘}’) and you can perform only one operation on that sequence i.e. either insert an opening or closing bracket at any position. You have to tell the minimum number of operations required to make the given sequence balanced.

Input: str = “{}}”
Output: 0

Input: str = “{}{}}}”
Output: 3
The updated sequence will be “{}}{}}{}}”.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: For making a balanced sequence, two continuous closing brackets are required for every opening bracket. There can be 3 cases:

1. When current character is an opening bracket: If the previous character is not a closing bracket then simply insert the opening bracket to stack else there is a need of one closing bracket that will cost one operation.
2. If the stack is empty and current character is a closing bracket: In this case, one opening bracket is required that will cost one operation and insert that opening bracket to the stack.
3. If stack is not empty but current character is a closing bracket: Here, only the count of closing brackets is required. If it is 2 then remove one opening bracket from stack else increment the count of closing bracket.

At the end of the string, if the stack is not empty then the required count of closing brackets will be ((2 * size of the stack) – current count closing brackets).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the minimum operations required ` `int` `minOperations(string s, ``int` `len) ` `{ ` `    ``int` `operationCnt = 0; ` `    ``stack<``char``> st; ` `    ``int` `cntClosing = 0; ` `    ``for` `(``int` `i = 0; i < len; i++) { ` ` `  `        ``// Condition where we got only one closing ` `        ``// bracket instead of 2, here we have to ` `        ``// add one more closing bracket to ` `        ``// make the sequence balanced ` `        ``if` `(s[i] == ``'{'``) { ` `            ``if` `(cntClosing > 0) { ` ` `  `                ``// Add closing bracket that ` `                ``// costs us one operation ` `                ``operationCnt++; ` ` `  `                ``// Remove the top opening bracket because ` `                ``// we got the 1 opening and 2 ` `                ``// continuous closing brackets ` `                ``st.pop(); ` `            ``} ` ` `  `            ``// Inserting the opening bracket to stack ` `            ``st.push(s[i]); ` ` `  `            ``// After making the sequence balanced ` `            ``// closing is now set to 0 ` `            ``cntClosing = 0; ` `        ``} ` `        ``else` `if` `(st.empty()) { ` ` `  `            ``// Case when there is no opening bracket ` `            ``// the sequence starts with a closing bracket ` `            ``// and one opening bracket is required ` `            ``// Now we have one opening and one closing bracket ` `            ``st.push(``'{'``); ` ` `  `            ``// Add opening bracket that ` `            ``// costs us one operation ` `            ``operationCnt++; ` ` `  `            ``// Assigning 1 to cntClosing because ` `            ``// we have one closing bracket ` `            ``cntClosing = 1; ` `        ``} ` `        ``else` `{ ` `            ``cntClosing = (cntClosing + 1) % 2; ` ` `  `            ``// Case where we got two continuous closing brackets ` `            ``// Need to pop one opening bracket from stack top ` `            ``if` `(cntClosing == 0) { ` `                ``st.pop(); ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Condition where stack is not empty ` `    ``// This is the case where we have ` `    ``// only opening brackets ` `    ``// (st.size() * 2) will give us the total ` `    ``// closing bracket needed ` `    ``// cntClosing is the count of closing ` `    ``// bracket that we already have ` `    ``operationCnt += st.size() * 2 - cntClosing; ` `    ``return` `operationCnt; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string str = ``"}}{"``; ` `    ``int` `len = str.length(); ` ` `  `    ``cout << minOperations(str, len); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG{ ` ` `  `// Function to return the ` `// minimum operations required ` `static` `int` `minOperations(String s, ``int` `len) ` `{ ` `    ``int` `operationCnt = ``0``; ` `    ``Stack st = ``new` `Stack(); ` `    ``int` `cntClosing = ``0``; ` `     `  `    ``for``(``int` `i = ``0``; i < len; i++) ` `    ``{ ` `         `  `       ``// Condition where we got only one  ` `       ``// closing bracket instead of 2,  ` `       ``// here we have to add one more  ` `       ``// closing bracket to make the  ` `       ``// sequence balanced ` `       ``if` `(s.charAt(i) == ``'{'``)  ` `       ``{ ` `           ``if` `(cntClosing > ``0``) ` `           ``{ ` `                `  `               ``// Add closing bracket that ` `               ``// costs us one operation ` `               ``operationCnt++; ` `                `  `               ``// Remove the top opening bracket  ` `               ``// because we got the 1 opening  ` `               ``// and 2 continuous closing brackets ` `               ``st.pop(); ` `           ``} ` `            `  `           ``// Inserting the opening bracket to stack ` `           ``st.add(s.charAt(i)); ` `            `  `           ``// After making the sequence balanced ` `           ``// closing is now set to 0 ` `           ``cntClosing = ``0``; ` `       ``} ` `       ``else` `if` `(st.isEmpty()) ` `       ``{ ` `            `  `           ``// Case when there is no opening  ` `           ``// bracket the sequence starts  ` `           ``// with a closing bracket and  ` `           ``// one opening bracket is required ` `           ``// Now we have one opening and one ` `           ``// closing bracket ` `           ``st.add(``'{'``); ` `            `  `           ``// Add opening bracket that ` `           ``// costs us one operation ` `           ``operationCnt++; ` `            `  `           ``// Assigning 1 to cntClosing because ` `           ``// we have one closing bracket ` `            ``cntClosing = ``1``; ` `       ``} ` `       ``else`  `       ``{ ` `           ``cntClosing = (cntClosing + ``1``) % ``2``; ` `            `  `           ``// Case where we got two continuous  ` `           ``// closing brackets need to pop one ` `           ``// opening bracket from stack top ` `           ``if` `(cntClosing == ``0``) ` `           ``{ ` `               ``st.pop(); ` `           ``} ` `       ``} ` `    ``} ` ` `  `    ``// Condition where stack is not empty ` `    ``// This is the case where we have only ` `    ``// opening brackets (st.size() * 2)  ` `    ``// will give us the total closing  ` `    ``// bracket needed cntClosing is the ` `    ``// count of closing bracket that ` `    ``// we already have ` `    ``operationCnt += st.size() * ``2` `- cntClosing; ` ` `  `    ``return` `operationCnt; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``String str = ``"}}{"``; ` `    ``int` `len = str.length(); ` ` `  `    ``System.out.print(minOperations(str, len)); ` `} ` `} ` ` `  `// This code is contributed by amal kumar choubey `

Output:

```3
```

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