Maximum Balanced String Partitions
Given a balanced string str of size N with an equal number of L and R, the task is to find a maximum number X, such that a given string can be partitioned into X balanced substring. A string called to be balanced if the number of ‘L’s in the string equals the number of ‘R’s.
Examples:
Input : str = “LRLLRRLRRL”
Output : 4
Explanation: { “LR”, “LLRR”, “LR”, “RL”} are the possible partitions.Input : “LRRRRLLRLLRL”
Output : 3
Explanation: {“LR”, “RRRLLRLL”, “RL”} are the possible partitions.
Approach: The approach to solving this problem is to loop through the string and keep incrementing the count of L and R whenever encountered. Any instant when the respective counts of L and R become equal, a balanced parenthesis is formed. Thus the count of such instances gives the desired maximum possible partitions.
Below is the implementation of the above approach:
C++
// C++ program to find a maximum number X, such // that a given string can be partitioned // into X substrings that are each balanced #include <bits/stdc++.h> using namespace std; // Function to find a maximum number X, such // that a given string can be partitioned // into X substrings that are each balanced int BalancedPartition(string str, int n) { // If the size of the string is 0, // then anwer is zero if (n == 0) return 0; // variable that represents the // number of 'R's and 'L's int r = 0, l = 0; // To store maximum number of // possible partitions int ans = 0; for (int i = 0; i < n; i++) { // increment the variable r if the // character in the string is 'R' if (str[i] == 'R') { r++; } // increment the variable l if the // character in the string is 'L' else if (str[i] = 'L') { l++; } // if r and l are equal, // then increment ans if (r == l) { ans++; } } // Return the required answer return ans; } // Driver code int main() { string str = "LLRRRLLRRL"; int n = str.size(); // Function call cout << BalancedPartition(str, n) << endl; return 0; } |
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Java
// Java program to find a maximum number X, such // that a given String can be partitioned // into X subStrings that are each balanced import java.util.*; class GFG{ // Function to find a maximum number X, such // that a given String can be partitioned // into X subStrings that are each balanced static int BalancedPartition(String str, int n) { // If the size of the String is 0, // then anwer is zero if (n == 0) return 0; // Variable that represents the // number of 'R's and 'L's int r = 0, l = 0; // To store maximum number of // possible partitions int ans = 0; for(int i = 0; i < n; i++) { // Increment the variable r if the // character in the String is 'R' if (str.charAt(i) == 'R') { r++; } // Increment the variable l if the // character in the String is 'L' else if (str.charAt(i) == 'L') { l++; } // If r and l are equal, // then increment ans if (r == l) { ans++; } } // Return the required answer return ans; } // Driver code public static void main(String[] args) { String str = "LLRRRLLRRL"; int n = str.length(); // Function call System.out.print(BalancedPartition(str, n) + "\n"); } } // This code is contributed by Rajput-Ji |
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Python3
# Python3 program to find a maximum number X, # such that a given string can be partitioned # into X substrings that are each balanced # Function to find a maximum number X, such # that a given string can be partitioned # into X substrings that are each balanced def BalancedPartition(str1, n): # If the size of the string is 0, # then anwer is zero if (n == 0): return 0 # Variable that represents the # number of 'R's and 'L's r = 0 l = 0 # To store maximum number of # possible partitions ans = 0 for i in range(n): # Increment the variable r if the # character in the string is 'R' if (str1[i] == 'R'): r += 1 # Increment the variable l if the # character in the string is 'L' elif (str1[i] == 'L'): l += 1 # If r and l are equal, # then increment ans if (r == l): ans += 1 # Return the required answer return ans # Driver code if __name__ == '__main__': str1 = "LLRRRLLRRL" n = len(str1) # Function call print(BalancedPartition(str1, n)) # This code is contributed by Bhupendra_Singh |
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C#
// C# program to find a maximum number X, such // that a given String can be partitioned // into X subStrings that are each balanced using System; class GFG{ // Function to find a maximum number X, such // that a given String can be partitioned // into X subStrings that are each balanced static int BalancedPartition(string str, int n) { // If the size of the String is 0, // then anwer is zero if (n == 0) return 0; // Variable that represents the // number of 'R's and 'L's int r = 0, l = 0; // To store maximum number of // possible partitions int ans = 0; for(int i = 0; i < n; i++) { // Increment the variable r if the // character in the String is 'R' if (str[i] == 'R') { r++; } // Increment the variable l if the // character in the String is 'L' else if (str[i] == 'L') { l++; } // If r and l are equal, // then increment ans if (r == l) { ans++; } } // Return the required answer return ans; } // Driver code public static void Main() { string str = "LLRRRLLRRL"; int n = str.Length; // Function call Console.Write(BalancedPartition(str, n) + "\n"); } } // This code is contributed by Nidhi_Biet |
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Output:
4
Time Complexity: O(N)
Space Complexity: O(1)
