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# Number of ways to partition a string into two balanced subsequences

Given a string ‘S’ consisting of open and closed brackets, the task is find the number of ways in which each character of ‘S’ can be assigned to either a string ‘X’ or string ‘Y’ (both initially empty) such that the strings formed by X and Y are balanced. It can be assumed that ‘S’ is itself balanced.

Examples:

```Input: S = "(())"
Output: 6
Valid assignments are :
X = "(())" and Y = "" [All characters in X]
X = "" and Y = "(())" [Nothing in X]
X = "()" and Y = "()" [1st and 3rd characters in X]
X = "()" and Y = "()" [2nd and 3rd characters in X]
X = "()" and Y = "()" [2nd and 4th characters in X]
X = "()" and Y = "()" [1st and 4th characters in X]

Input: S = "()()"
Output: 4
X = "()()", Y = ""
X = "()", Y = "()"  [1st and 2nd in X]
X = "()", Y = ""  [1st and 4th in X]
X = "", Y = "()()"```

A simple approach:

We can generate every possible way of assigning the characters, and check if the strings formed are balanced or not. There are 2n assignments, valid or invalid, and it takes O(n) time to check if the strings formed are balanced or not. Therefore the time complexity of this approach is O(n * 2n).

An efficient approach (Dynamic programming):

We can solve this problem in a more efficient manner using Dynamic Programming. We can describe the current state of assignment using three variables: the index i of the character to be assigned, and the strings formed by X and Y up to that state. Passing the whole strings to function calls will result in high memory requirements, so we can replace them with count variables cx and cy. We will increment the count variable for every opening bracket and decrement it for every closing bracket. The time and space complexity of this approach is O(n3).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of``// the above approach``#include ``using` `namespace` `std;` `// For maximum length of input string``const` `int` `MAX = 10;` `// Declaring the DP table``int` `F[MAX][MAX][MAX];` `// Function to calculate the``// number of valid assignments``int` `noOfAssignments(string& S, ``int``& n, ``int` `i,``                    ``int` `c_x, ``int` `c_y)``{``    ``if` `(F[i][c_x][c_y] != -1)``        ``return` `F[i][c_x][c_y];` `    ``if` `(i == n) {` `        ``// Return 1 if both``        ``// subsequences are balanced``        ``F[i][c_x][c_y] = !c_x && !c_y;``        ``return` `F[i][c_x][c_y];``    ``}` `    ``// Increment the count``    ``// if it an opening bracket``    ``if` `(S[i] == ``'('``) {``        ``F[i][c_x][c_y]``            ``= noOfAssignments(S, n, i + 1,``                              ``c_x + 1, c_y)``              ``+ noOfAssignments(S, n, i + 1,``                                ``c_x, c_y + 1);``        ``return` `F[i][c_x][c_y];``    ``}` `    ``F[i][c_x][c_y] = 0;` `    ``// Decrement the count``    ``// if it a closing bracket``    ``if` `(c_x)``        ``F[i][c_x][c_y]``            ``+= noOfAssignments(S, n, i + 1,``                               ``c_x - 1, c_y);` `    ``if` `(c_y)``        ``F[i][c_x][c_y]``            ``+= noOfAssignments(S, n, i + 1,``                               ``c_x, c_y - 1);` `    ``return` `F[i][c_x][c_y];``}` `// Driver code``int` `main()``{``    ``string S = ``"(())"``;``    ``int` `n = S.length();` `    ``// Initializing the DP table``    ``memset``(F, -1, ``sizeof``(F));` `    ``// Initial value for c_x``    ``// and c_y is zero``    ``cout << noOfAssignments(S, n, 0, 0, 0);` `    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``class` `GFG``{` `    ``// For maximum length of input string``    ``static` `int` `MAX = ``10``;` `    ``// Declaring the DP table``    ``static` `int``[][][] F = ``new` `int``[MAX][MAX][MAX];` `    ``// Function to calculate the``    ``// number of valid assignments``    ``static` `int` `noOfAssignments(String s, ``int` `n,``                               ``int` `i, ``int` `c_x, ``int` `c_y)``    ``{``        ``if` `(F[i][c_x][c_y] != -``1``)``            ``return` `F[i][c_x][c_y];``        ``if` `(i == n)``        ``{` `            ``// Return 1 if both``            ``// subsequences are balanced``            ``F[i][c_x][c_y] = (c_x == ``0` `&&``                              ``c_y == ``0``) ? ``1` `: ``0``;``            ``return` `F[i][c_x][c_y];``        ``}` `        ``// Increment the count``        ``// if it an opening bracket``        ``if` `(s.charAt(i) == ``'('``)``        ``{``            ``F[i][c_x][c_y] = noOfAssignments(s, n, i + ``1``,``                                             ``c_x + ``1``, c_y) +``                             ``noOfAssignments(s, n, i + ``1``,``                                             ``c_x, c_y + ``1``);``            ``return` `F[i][c_x][c_y];``        ``}` `        ``F[i][c_x][c_y] = ``0``;` `        ``// Decrement the count``        ``// if it a closing bracket``        ``if` `(c_x != ``0``)``            ``F[i][c_x][c_y] += noOfAssignments(s, n, i + ``1``,``                                              ``c_x - ``1``, c_y);` `        ``if` `(c_y != ``0``)``            ``F[i][c_x][c_y] += noOfAssignments(s, n, i + ``1``,``                                              ``c_x, c_y - ``1``);` `        ``return` `F[i][c_x][c_y];``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String s = ``"(())"``;``        ``int` `n = s.length();` `        ``// Initializing the DP table``        ``for` `(``int` `i = ``0``; i < MAX; i++)``            ``for` `(``int` `j = ``0``; j < MAX; j++)``                ``for` `(``int` `k = ``0``; k < MAX; k++)``                    ``F[i][j][k] = -``1``;` `        ``// Initial value for c_x``        ``// and c_y is zero``        ``System.out.println(noOfAssignments(s, n, ``0``, ``0``, ``0``));``    ``}``}` `// This code is contributed by``// sanjeev2552`

## Python3

 `# Python3 implementation of above approach` `# For maximum length of input string``MAX` `=` `10` `# Declaring the DP table``F ``=` `[[[``-``1` `for` `i ``in` `range``(``MAX``)]``          ``for` `j ``in` `range``(``MAX``)]``          ``for` `k ``in` `range``(``MAX``)]` `# Function to calculate the number``# of valid assignments``def` `noOfAssignments(S, n, i, c_x, c_y):` `    ``if` `F[i][c_x][c_y] !``=` `-``1``:``        ``return` `F[i][c_x][c_y]` `    ``if` `i ``=``=` `n:` `        ``# Return 1 if both subsequences are balanced``        ``F[i][c_x][c_y] ``=` `not` `c_x ``and` `not` `c_y``        ``return` `F[i][c_x][c_y]` `    ``# Increment the count if``    ``# it is an opening bracket``    ``if` `S[i] ``=``=` `'('``:``        ``F[i][c_x][c_y] ``=` `\``            ``noOfAssignments(S, n, i ``+` `1``, c_x ``+` `1``, c_y) ``+` `\``            ``noOfAssignments(S, n, i ``+` `1``, c_x, c_y ``+` `1``)``        ` `        ``return` `F[i][c_x][c_y]` `    ``F[i][c_x][c_y] ``=` `0` `    ``# Decrement the count``    ``# if it a closing bracket``    ``if` `c_x:``        ``F[i][c_x][c_y] ``+``=` `\``            ``noOfAssignments(S, n, i ``+` `1``, c_x ``-` `1``, c_y)` `    ``if` `c_y:``        ``F[i][c_x][c_y] ``+``=` `\``            ``noOfAssignments(S, n, i ``+` `1``, c_x, c_y ``-` `1``)` `    ``return` `F[i][c_x][c_y]` `# Driver code``if` `__name__ ``=``=` `"__main__"``:` `    ``S ``=` `"(())"``    ``n ``=` `len``(S)` `    ``# Initial value for c_x and c_y is zero``    ``print``(noOfAssignments(S, n, ``0``, ``0``, ``0``))` `# This code is contributed by Rituraj Jain`

## C#

 `// C# implementation of the above approach``using` `System;``    ` `class` `GFG``{` `    ``// For maximum length of input string``    ``static` `int` `MAX = 10;` `    ``// Declaring the DP table``    ``static` `int``[,,] F = ``new` `int``[MAX, MAX, MAX];` `    ``// Function to calculate the``    ``// number of valid assignments``    ``static` `int` `noOfAssignments(String s, ``int` `n,``                         ``int` `i, ``int` `c_x, ``int` `c_y)``    ``{``        ``if` `(F[i, c_x, c_y] != -1)``            ``return` `F[i, c_x, c_y];``            ` `        ``if` `(i == n)``        ``{` `            ``// Return 1 if both``            ``// subsequences are balanced``            ``F[i, c_x, c_y] = (c_x == 0 &&``                              ``c_y == 0) ? 1 : 0;``            ``return` `F[i, c_x, c_y];``        ``}` `        ``// Increment the count``        ``// if it an opening bracket``        ``if` `(s[i] == ``'('``)``        ``{``            ``F[i, c_x, c_y] = noOfAssignments(s, n, i + 1,``                                             ``c_x + 1, c_y) +``                             ``noOfAssignments(s, n, i + 1,``                                             ``c_x, c_y + 1);``            ``return` `F[i, c_x, c_y];``        ``}` `        ``F[i, c_x, c_y] = 0;` `        ``// Decrement the count``        ``// if it a closing bracket``        ``if` `(c_x != 0)``            ``F[i, c_x, c_y] += noOfAssignments(s, n, i + 1,``                                              ``c_x - 1, c_y);` `        ``if` `(c_y != 0)``            ``F[i, c_x, c_y] += noOfAssignments(s, n, i + 1,``                                              ``c_x, c_y - 1);` `        ``return` `F[i, c_x, c_y];``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``String s = ``"(())"``;``        ``int` `n = s.Length;` `        ``// Initializing the DP table``        ``for` `(``int` `i = 0; i < MAX; i++)``            ``for` `(``int` `j = 0; j < MAX; j++)``                ``for` `(``int` `k = 0; k < MAX; k++)``                    ``F[i, j, k] = -1;` `        ``// Initial value for c_x``        ``// and c_y is zero``        ``Console.WriteLine(noOfAssignments(s, n, 0, 0, 0));``    ``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output

`6`

Another approach : Using DP Tabulation method ( Iterative approach )

In this approach we use Dp to store computation of subproblems and get the desired output without the help of recursion.

Steps to solve this problem :

• Create a 3D table F to store the solution of the subproblems and initialize it with 0.
• Initialize the table F with base cases
• Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP
• Return the final solution stored in F[0][0][0]

Implementation :

## C++

 `#include ``using` `namespace` `std;` `int` `noOfAssignments(string& S, ``int``& n) {``    ``// For maximum length of input string``    ``const` `int` `MAX = n + 1;` `    ``// Declaring the DP table``    ``int` `F[MAX][MAX][MAX];``    ``memset``(F, 0, ``sizeof``(F));` `    ``// Base case: both subsequences are balanced``    ``for``(``int` `i=0; i<=n; i++) {``        ``F[i][0][0] = 1;``    ``}` `    ``// Filling up the table in a bottom-up manner``    ``for``(``int` `i=n-1; i>=0; i--) {``        ``for``(``int` `c_x=0; c_x<=n; c_x++) {``            ``for``(``int` `c_y=0; c_y<=n; c_y++) {``                ``// If current character is an opening bracket``                ``if``(S[i] == ``'('``) {``                    ``F[i][c_x][c_y] = F[i+1][c_x+1][c_y] + F[i+1][c_x][c_y+1];``                ``}``                ``// If current character is a closing bracket``                ``else` `{``                    ``F[i][c_x][c_y] = 0;``                    ``if``(c_x > 0) {``                        ``F[i][c_x][c_y] += F[i+1][c_x-1][c_y];``                    ``}``                    ``if``(c_y > 0) {``                        ``F[i][c_x][c_y] += F[i+1][c_x][c_y-1];``                    ``}``                ``}``            ``}``        ``}``    ``}` `    ``// Final result``    ``return` `F[0][0][0];``}` `int` `main()``{``    ``string S = ``"(())"``;``    ``int` `n = S.length();` `    ``cout << noOfAssignments(S, n);` `    ``return` `0;``}`

Output:

`6`

Time complexity : O(N^3)

Auxiliary Space : O(N^3)

Optimized Dynamic Programming approach:

We can create a prefix array to store the count variable ci for the substring S[0 : i + 1]. We can observe that the sum of c_x and c_y will always be equal to the count variable for the whole string. By exploiting this property, we can reduce our dynamic programming approach to two states. A prefix array can be created in linear complexity, so the time and space complexity of this approach is O(n2).

Implementation:

## C++

 `// C++ implementation of``// the above approach``#include ``using` `namespace` `std;` `// For maximum length of input string``const` `int` `MAX = 10;` `// Declaring the DP table``int` `F[MAX][MAX];` `// Declaring the prefix array``int` `C[MAX];` `// Function to calculate the``// number of valid assignments``int` `noOfAssignments(string& S, ``int``& n,``                    ``int` `i, ``int` `c_x)``{``    ``if` `(F[i][c_x] != -1)``        ``return` `F[i][c_x];` `    ``if` `(i == n) {` `        ``// Return 1 if X is``        ``// balanced.``        ``F[i][c_x] = !c_x;``        ``return` `F[i][c_x];``    ``}` `    ``int` `c_y = C[i] - c_x;` `    ``// Increment the count``    ``// if it an opening bracket``    ``if` `(S[i] == ``'('``) {``        ``F[i][c_x]``            ``= noOfAssignments(S, n, i + 1,``                              ``c_x + 1)``              ``+ noOfAssignments(S, n,``                                ``i + 1, c_x);``        ``return` `F[i][c_x];``    ``}` `    ``F[i][c_x] = 0;` `    ``// Decrement the count``    ``// if it a closing bracket``    ``if` `(c_x)``        ``F[i][c_x]``            ``+= noOfAssignments(S, n,``                               ``i + 1, c_x - 1);` `    ``if` `(c_y)``        ``F[i][c_x]``            ``+= noOfAssignments(S, n,``                               ``i + 1, c_x);` `    ``return` `F[i][c_x];``}` `// Driver code``int` `main()``{``    ``string S = ``"()"``;``    ``int` `n = S.length();` `    ``// Initializing the DP table``    ``memset``(F, -1, ``sizeof``(F));` `    ``C[0] = 0;` `    ``// Creating the prefix array``    ``for` `(``int` `i = 0; i < n; ++i)``        ``if` `(S[i] == ``'('``)``            ``C[i + 1] = C[i] + 1;``        ``else``            ``C[i + 1] = C[i] - 1;` `    ``// Initial value for c_x``    ``// and c_y is zero``    ``cout << noOfAssignments(S, n, 0, 0);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach` `public` `class` `GFG {` `// For maximum length of input string``    ``static` `int` `MAX = ``10``;` `// Declaring the DP table``    ``static` `int` `F[][] = ``new` `int``[MAX][MAX];` `// Declaring the prefix array``    ``static` `int` `C[] = ``new` `int``[MAX];` `// Function to calculate the``// number of valid assignments``    ``static` `int` `noOfAssignments(String S, ``int` `n, ``int` `i, ``int` `c_x) {``        ``if` `(F[i][c_x] != -``1``) {``            ``return` `F[i][c_x];``        ``}` `        ``if` `(i == n) {` `            ``// Return 1 if X is``            ``// balanced.``            ``if` `(c_x == ``1``) {``                ``F[i][c_x] = ``0``;``            ``} ``else` `{``                ``F[i][c_x] = ``1``;``            ``}` `            ``return` `F[i][c_x];``        ``}` `        ``int` `c_y = C[i] - c_x;` `        ``// Increment the count``        ``// if it an opening bracket``        ``if` `(S.charAt(i) == ``'('``) {``            ``F[i][c_x]``                    ``= noOfAssignments(S, n, i + ``1``,``                            ``c_x + ``1``)``                    ``+ noOfAssignments(S, n,``                            ``i + ``1``, c_x);``            ``return` `F[i][c_x];``        ``}` `        ``F[i][c_x] = ``0``;` `        ``// Decrement the count``        ``// if it a closing bracket``        ``if` `(c_x == ``1``) {``            ``F[i][c_x]``                    ``+= noOfAssignments(S, n,``                            ``i + ``1``, c_x - ``1``);``        ``}` `        ``if` `(c_y == ``1``) {``            ``F[i][c_x]``                    ``+= noOfAssignments(S, n,``                            ``i + ``1``, c_x);``        ``}` `        ``return` `F[i][c_x];``    ``}` `// Driver code``    ``public` `static` `void` `main(String[] args) {``        ``String S = ``"()"``;``        ``int` `n = S.length();` `        ``// Initializing the DP table``        ``for` `(``int` `i = ``0``; i < MAX; i++) {``            ``for` `(``int` `j = ``0``; j < MAX; j++) {``                ``F[i][j] = -``1``;``            ``}``        ``}` `        ``C[``0``] = ``0``;` `        ``// Creating the prefix array``        ``for` `(``int` `i = ``0``; i < n; ++i) {``            ``if` `(S.charAt(i) == ``'('``) {``                ``C[i + ``1``] = C[i] + ``1``;``            ``} ``else` `{``                ``C[i + ``1``] = C[i] - ``1``;``            ``}``        ``}` `        ``// Initial value for c_x``        ``// and c_y is zero``        ``System.out.println(noOfAssignments(S, n, ``0``, ``0``));` `    ``}``}``// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of above approach` `# For maximum length of input string``MAX` `=` `10` `# Declaring the DP table``F ``=` `[[``-``1` `for` `i ``in` `range``(``MAX``)]``         ``for` `j ``in` `range``(``MAX``)]` `# Declaring the prefix array``C ``=` `[``None``] ``*` `MAX` `# Function to calculate the``# number of valid assignments``def` `noOfAssignments(S, n, i, c_x):` `    ``if` `F[i][c_x] !``=` `-``1``:``        ``return` `F[i][c_x]` `    ``if` `i ``=``=` `n:` `        ``# Return 1 if X is balanced.``        ``F[i][c_x] ``=` `not` `c_x``        ``return` `F[i][c_x]` `    ``c_y ``=` `C[i] ``-` `c_x` `    ``# Increment the count``    ``# if it is an opening bracket``    ``if` `S[i] ``=``=` `'('``:``        ``F[i][c_x] ``=` `\``            ``noOfAssignments(S, n, i ``+` `1``, c_x ``+` `1``) ``+` `\``            ``noOfAssignments(S, n, i ``+` `1``, c_x)``        ` `        ``return` `F[i][c_x]` `    ``F[i][c_x] ``=` `0` `    ``# Decrement the count if it is a closing bracket``    ``if` `c_x:``        ``F[i][c_x] ``+``=` `\``            ``noOfAssignments(S, n, i ``+` `1``, c_x ``-` `1``)` `    ``if` `c_y:``        ``F[i][c_x] ``+``=` `\``            ``noOfAssignments(S, n, i ``+` `1``, c_x)` `    ``return` `F[i][c_x]` `# Driver code``if` `__name__ ``=``=` `"__main__"``:` `    ``S ``=` `"()"``    ``n ``=` `len``(S)` `    ``C[``0``] ``=` `0` `    ``# Creating the prefix array``    ``for` `i ``in` `range``(``0``, n):``        ``if` `S[i] ``=``=` `'('``:``            ``C[i ``+` `1``] ``=` `C[i] ``+` `1``        ``else``:``            ``C[i ``+` `1``] ``=` `C[i] ``-` `1` `    ``# Initial value for c_x and c_y is zero``    ``print``(noOfAssignments(S, n, ``0``, ``0``))` `# This code is contributed by Rituraj Jain`

## C#

 `// C# implementation of the approach`` ` `using` `System;``public` `class` `GFG {`` ` `// For maximum length of input string``    ``static` `int` `MAX = 10;`` ` `// Declaring the DP table``    ``static` `int``[,] F = ``new` `int``[MAX,MAX];`` ` `// Declaring the prefix array``    ``static` `int``[] C = ``new` `int``[MAX];`` ` `// Function to calculate the``// number of valid assignments``    ``static` `int` `noOfAssignments(``string` `S, ``int` `n, ``int` `i, ``int` `c_x) {``        ``if` `(F[i,c_x] != -1) {``            ``return` `F[i,c_x];``        ``}`` ` `        ``if` `(i == n) {`` ` `            ``// Return 1 if X is``            ``// balanced.``            ``if` `(c_x == 1) {``                ``F[i,c_x] = 0;``            ``} ``else` `{``                ``F[i,c_x] = 1;``            ``}`` ` `            ``return` `F[i,c_x];``        ``}`` ` `        ``int` `c_y = C[i] - c_x;`` ` `        ``// Increment the count``        ``// if it an opening bracket``        ``if` `(S[i] == ``'('``) {``            ``F[i,c_x]``                    ``= noOfAssignments(S, n, i + 1,``                            ``c_x + 1)``                    ``+ noOfAssignments(S, n,``                            ``i + 1, c_x);``            ``return` `F[i,c_x];``        ``}`` ` `        ``F[i,c_x] = 0;`` ` `        ``// Decrement the count``        ``// if it a closing bracket``        ``if` `(c_x == 1) {``            ``F[i,c_x]``                    ``+= noOfAssignments(S, n,``                            ``i + 1, c_x - 1);``        ``}`` ` `        ``if` `(c_y == 1) {``            ``F[i,c_x]``                    ``+= noOfAssignments(S, n,``                            ``i + 1, c_x);``        ``}`` ` `        ``return` `F[i,c_x];``    ``}`` ` `// Driver code``    ``public` `static` `void` `Main() {``        ``string` `S = ``"()"``;``        ``int` `n = S.Length;`` ` `        ``// Initializing the DP table``        ``for` `(``int` `i = 0; i < MAX; i++) {``            ``for` `(``int` `j = 0; j < MAX; j++) {``                ``F[i,j] = -1;``            ``}``        ``}`` ` `        ``C[0] = 0;`` ` `        ``// Creating the prefix array``        ``for` `(``int` `i = 0; i < n; ++i) {``            ``if` `(S[i] == ``'('``) {``                ``C[i + 1] = C[i] + 1;``            ``} ``else` `{``                ``C[i + 1] = C[i] - 1;``            ``}``        ``}`` ` `        ``// Initial value for c_x``        ``// and c_y is zero``        ``Console.WriteLine(noOfAssignments(S, n, 0, 0));`` ` `    ``}``}``// This code is contributed by Ita_c.`

## Javascript

 ``

Output

`2`