Number of ways to partition a string into two balanced subsequences - GeeksforGeeks

21 Oct, 2021
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++
Java
Python3
C#
Javascript
// C++ implementation of// the above approach#include <bits/stdc++.h>using namespace std; // For maximum length of input stringconst int MAX = 10; // Declaring the DP tableint F[MAX][MAX][MAX]; // Function to calculate the// number of valid assignmentsint 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 codeint 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 implementation of the above approachclass 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 implementation of above approach # For maximum length of input stringMAX = 10 # Declaring the DP tableF = [[[-1 for i in range(MAX)]          for j in range(MAX)]          for k in range(MAX)] # Function to calculate the number# of valid assignmentsdef 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 codeif __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# implementation of the above approachusing 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
<script>// Javascript implementation of the above approach         // For maximum length of input string    let MAX = 10;         // Declaring the DP table    let F = new Array(MAX);    for(let i = 0; i < MAX; i++)    {        F[i] = new Array(MAX);        for(let j = 0; j < MAX; j++)        {            F[i][j] = new Array(MAX);            for(let k = 0; k < MAX; k++)            {                F[i][j][k] = -1;            }        }    }         // Function to calculate the    // number of valid assignments    function 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] = (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    let  s = "(())";    let n = s.length;         // Initial value for c_x    // and c_y is zero    document.write(noOfAssignments(s, n, 0, 0, 0)); // This code is contributed by rag2127</script>
6
 
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).  
C++
Java
Python3
C#
Javascript
// C++ implementation of// the above approach#include <bits/stdc++.h>using namespace std; // For maximum length of input stringconst int MAX = 10; // Declaring the DP tableint F[MAX][MAX]; // Declaring the prefix arrayint C[MAX]; // Function to calculate the// number of valid assignmentsint 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 codeint 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 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 implementation of above approach # For maximum length of input stringMAX = 10 # Declaring the DP tableF = [[-1 for i in range(MAX)]         for j in range(MAX)] # Declaring the prefix arrayC = [None] * MAX # Function to calculate the# number of valid assignmentsdef 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 codeif __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# 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.
<script>// Javascript implementation of the approach  // For maximum length of input stringvar MAX = 10;  // Declaring the DP tablevar F = Array.from(Array(MAX), ()=>Array(MAX).fill(0));  // Declaring the prefix arrayvar C = Array(MAX).fill(0);  // Function to calculate the// number of valid assignmentsfunction 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.            if (c_x == 1) {                F[i][c_x] = 0;            } else {                F[i][c_x] = 1;            }              return F[i][c_x];        }          var 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 codevar S = "()";var n = S.length; // Initializing the DP tablefor(var i = 0; i < MAX; i++) {    for(var j = 0; j < MAX; j++) {        F[i][j] = -1;    }} C[0] = 0; // Creating the prefix arrayfor (var 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 zerodocument.write(noOfAssignments(S, n, 0, 0) + "<br>"); // This code is contributed by rrrtnx.</script>
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rituraj_jain
sanjeev2552
princiraj1992
rag2127
rrrtnx
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