Given a list of strings arr[] of zeros and ones only and two integer N and M, where N is the number of 1’s and M is the number of 0’s. The task is to find the maximum number of strings from the given list of strings that can be constructured with given number of 0’s and 1’s.
Examples:
Input: arr[] = {“10”, “0001”, “11100”, “1”, “0”}, M = 5, N = 3
Output: 4
Explanation:
The 4 strings which can be formed using five 0’s and three 1’s are: “10”, “0001”, “1”, “0”Input: arr[] = {“10”, “00”, “000” “0001”, “111001”, “1”, “0”}, M = 3, N = 1
Output: 3
Explanation:
The 3 strings which can be formed using three 0’s and one 1’s are: “00”, “1”, “0”
Naive Approach: The idea is to generate all the combination of the given list of strings and check the count of zeros and ones satisfying the given condition. But the time complexity of this solution is exponential.
Time Complexity: O(2N), where N is the number of strings in the list.
Efficient Approach:
An efficient solution is given by using Dynamic Programming. The idea is to use recursion for generating all possible combinations and store the results for Overlapping Subproblems during recursion.
Below are the steps:
- The idea is to use 3D dp array(dp[M][N][i]) where N and M are the number of 1’s and 0’s respectively and i is the index of the string in the list.
- Find the number of 1’s and 0’s in the current string and check if the count of the zeros and ones is less than or equals to the given count N and M respectively.
- If above condition is true, then check whether current state value is stored in the dp table or not. If yes then return this value.
- Else recursively move for the next iteration by including and excluding the current string as:
// By including the current string x = 1 + recursive_function(M - zero, N - ones, arr, i + 1) // By excluding the current string y = recursive_function(M, N, arr, i + 1) // and update the dp table as: dp[M][N][i] = max(x, y)
- The maximum value of the above two recursive calls will give the maximum number with N 1’s and M 0’s for the current state.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // 3D dp table to store the state value int dp[100][100][100]; // Function that count the combination // of 0's and 1's from the given list // of string int countString(int m, int n, vector<string>& arr, int i) { // Base Case if count of 0's or 1's // becomes negative if (m < 0 || n < 0) { return INT_MIN; } // If index reaches out of bound if (i >= arr.size()) { return 0; } // Return the prestored result if (dp[m][n][i] != -1) { return dp[m][n][i]; } // Intialise count of 0's and 1's // to 0 for the current state int zero = 0, one = 0; // Calculate the number of 1's and // 0's in current string for (char c : arr[i]) { if (c == '0') { zero++; } else { one++; } } // Include the current string and // recurr for the next iteration int x = 1 + countString(m - zero, n - one, arr, i + 1); // Exclude the current string and // recurr for the next iteration int y = countString(m, n, arr, i + 1); // Update the maximum of the above // two states to the current dp state return dp[m][n][i] = max(x, y); } // Driver Code int main() { vector<string> arr = { "10", "0001", "1", "111001", "0" }; // N 0's and M 1's int N = 3, M = 5; // Intialise dp array to -1 memset(dp, -1, sizeof(dp)); // Function call cout << countString(M, N, arr, 0); } |
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Java
// Java program for the above approach class GFG{ // 3D dp table to store the state value static int [][][]dp = new int[100][100][100]; // Function that count the combination // of 0's and 1's from the given list // of String static int countString(int m, int n, String []arr, int i) { // Base Case if count of 0's or 1's // becomes negative if (m < 0 || n < 0) { return Integer.MIN_VALUE; } // If index reaches out of bound if (i >= arr.length) { return 0; } // Return the prestored result if (dp[m][n][i] != -1) { return dp[m][n][i]; } // Intialise count of 0's and 1's // to 0 for the current state int zero = 0, one = 0; // Calculate the number of 1's and // 0's in current String for (char c : arr[i].toCharArray()) { if (c == '0') { zero++; } else { one++; } } // Include the current String and // recurr for the next iteration int x = 1 + countString(m - zero, n - one, arr, i + 1); // Exclude the current String and // recurr for the next iteration int y = countString(m, n, arr, i + 1); // Update the maximum of the above // two states to the current dp state return dp[m][n][i] = Math.max(x, y); } // Driver Code public static void main(String[] args) { String []arr = { "10", "0001", "1", "111001", "0" }; // N 0's and M 1's int N = 3, M = 5; // Intialise dp array to -1 for(int i = 0;i<100;i++){ for(int j = 0;j<100;j++){ for(int l=0;l<100;l++) dp[i][j][l]=-1; } } // Function call System.out.print(countString(M, N, arr, 0)); } } // This code is contributed by 29AjayKumar |
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Python 3
# Python 3 program for the above approach import sys # 3D dp table to store the state value dp = [[[-1 for i in range(100)]for j in range(100)] for k in range(100)] # Function that count the combination # of 0's and 1's from the given list # of string def countString(m, n, arr, i): # Base Case if count of 0's or 1's # becomes negative if (m < 0 or n < 0): return -sys.maxsize - 1 # If index reaches out of bound if (i >= len(arr)): return 0 # Return the prestored result if (dp[m][n][i] != -1): return dp[m][n][i] # Intialise count of 0's and 1's # to 0 for the current state zero = 0 one = 0 # Calculate the number of 1's and # 0's in current string for c in arr[i]: if (c == '0'): zero += 1 else: one += 1 # Include the current string and # recurr for the next iteration x = 1 + countString(m - zero, n - one, arr, i + 1) # Exclude the current string and # recurr for the next iteration y = countString(m, n, arr, i + 1) dp[m][n][i] = max(x, y) # Update the maximum of the above # two states to the current dp state return dp[m][n][i] # Driver Code if __name__ == '__main__': arr = ["10", "0001", "1","111001", "0"] # N 0's and M 1's N = 3 M = 5 # Function call print(countString(M, N, arr, 0)) # This code is contributed by Surendra_Gangwar |
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C#
// C# program for the above approach using System; class GFG{ // 3D dp table to store the state value static int [,,]dp = new int[100, 100, 100]; // Function that count the combination // of 0's and 1's from the given list // of String static int countString(int m, int n, String []arr, int i) { // Base Case if count of 0's or 1's // becomes negative if (m < 0 || n < 0) { return int.MinValue; } // If index reaches out of bound if (i >= arr.Length) { return 0; } // Return the prestored result if (dp[m, n, i] != -1) { return dp[m, n, i]; } // Intialise count of 0's and 1's // to 0 for the current state int zero = 0, one = 0; // Calculate the number of 1's and // 0's in current String foreach (char c in arr[i].ToCharArray()) { if (c == '0') { zero++; } else { one++; } } // Include the current String and // recurr for the next iteration int x = 1 + countString(m - zero, n - one, arr, i + 1); // Exclude the current String and // recurr for the next iteration int y = countString(m, n, arr, i + 1); // Update the maximum of the above // two states to the current dp state return dp[m, n, i] = Math.Max(x, y); } // Driver Code public static void Main(String[] args) { String []arr = { "10", "0001", "1", "111001", "0" }; // N 0's and M 1's int N = 3, M = 5; // Intialise dp array to -1 for(int i = 0; i < 100; i++){ for(int j = 0; j < 100; j++){ for(int l = 0; l < 100; l++) dp[i, j, l] = -1; } } // Function call Console.Write(countString(M, N, arr, 0)); } } // This code is contributed by Rajput-Ji |
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Output:
4
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