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Length of longest subset consisting of A 0s and B 1s from an array of strings
  • Last Updated : 08 Apr, 2021

Given an array arr[] consisting of N binary strings, and two integers A and B, the task is to find the length of the longest subset consisting of at most A 0s and B 1s.

Examples:

Input: arr[] = {“1”, “0”, “0001”, “10”, “111001”}, A = 5, B = 3
Output: 4
Explanation: 
One possible way is to select the subset {arr[0], arr[1], arr[2], arr[3]}.
Total number of 0s and 1s in all these strings are 5 and 3 respectively.
Also, 4 is the length of the longest subset possible.

Input: arr[] = {“0”, “1”, “10”}, A = 1, B = 1
Output: 2
Explanation: 
One possible way is to select the subset {arr[0], arr[1]}.
Total number of 0s and 1s in all these strings is 1 and 1 respectively.
Also, 2 is the length of the longest subset possible.

Naive Approach: The simplest approach to solve the above problem is to use recursion. At every recursive call, the idea is to either include or exclude the current string. Once, all possibilities are considered, print the maximum length of subset obtained.



Below is the implementation of the above approach:

C++14




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to count 0's in a string
int count0(string s)
{
    // Stores count of 0s
    int count = 0;
  
    // Iterate over characters of string
    for (int i = 0; i < s.size(); i++) {
  
        // If current character is '0'
        if (s[i] == '0') {
            count++;
        }
    }
    return count;
}
  
// Recursive function to find the length of
// longest subset from an array of strings
// with at most A 0's and B 1's
int solve(vector<string> vec,
          int A, int B, int idx)
{
    // If idx is equal to N
    // or A + B is equal to 0
    if (idx == vec.size() || A + B == 0) {
        return 0;
    }
  
    // Stores the count of 0's in arr[idx]
    int zero = count0(vec[idx]);
  
    // Stores the count of 1's in arr[idx]
    int one = vec[idx].size() - zero;
  
    // Stores the length of the
    // subset if arr[i] is included
    int inc = 0;
  
    // If zero is less than or equal to A
    // and one is less than or equal to B
    if (zero <= A && one <= B) {
        inc = 1 + solve(vec, A - zero,
                        B - one, idx + 1);
    }
  
    // Stores the length of the subset
    // if arr[i] is excluded
    int exc = solve(vec, A, B, idx + 1);
  
    // Returns max of inc and exc
    return max(inc, exc);
}
  
// Function to find the length of the
// longest subset from an array of
// strings with at most A 0's and B 1's
int MaxSubsetlength(vector<string> arr,
                    int A, int B)
{
    // Return
    return solve(arr, A, B, 0);
}
  
// Driver Code
int main()
{
    vector<string> arr = { "1", "0", "10" };
    int A = 1, B = 1;
  
    cout << MaxSubsetlength(arr, A, B);
    return 0;
}
Output:
2

Time Complexity: O(2N
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized using Memoization, based on the following observations: 

Recursion Tree

  • It can be observed that there exists a overlapping subproblem and optimal substructure property.
    Therefore, the idea is to use dynamic programming to optimize the above approach.
  • The idea is to use the 3D-state dynamic programming.
  • Suppose Dp(i, A, B) represents the maximum length of the subset with at most A 0s and B 1s.
  • Then the transition of states can be defined by either selecting a string at ith index or not,
    • Dp(i, A, B) = Max(1+Dp(i+1, A- Z, B-O), Dp(i+1, A, B)),  
      where Z is the count of Os and O is the count of 0s.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to count number
// of 0s present in the string
int count0(string s)
{
    // Stores the count of 0s
    int count = 0;
  
    // Iterate over characters of string
    for (int i = 0; i < s.size(); i++) {
  
        // If current character is '0'
        if (s[i] == '0') {
            count++;
        }
    }
    return count;
}
// Recursive Function to find the length of
// longest subset from given array of strings
// with at most A 0s and B 1s
int solve(vector<string> vec, int A,
          int B, int idx,
          vector<vector<vector<int> > >& dp)
{
    // If idx is equal to N or
    // A + B is equal to 0
    if (idx == vec.size() || A + B == 0) {
        return 0;
    }
    // If the state is already calculated
    if (dp[A][B][idx] > 0) {
        return dp[A][B][idx];
    }
  
    // Stores the count of 0's
    int zero = count0(vec[idx]);
  
    // Stores the count of 1's
    int one = vec[idx].size() - zero;
  
    // Stores the length of longest
    // by including arr[idx]
    int inc = 0;
  
    // If zero is less than A
    // and one is less than B
    if (zero <= A && one <= B) {
        inc = 1
              + solve(vec, A - zero,
                      B - one, idx + 1, dp);
    }
  
    // Stores the length of longest subset
    // by excluding arr[idx]
    int exc = solve(vec, A, B, idx + 1, dp);
  
    // Assign
    dp[A][B][idx] = max(inc, exc);
  
    // Return
    return dp[A][B][idx];
}
  
// Function to find the length of the
// longest subset of an array of strings
// with at most A 0s and B 1s
int MaxSubsetlength(vector<string> arr,
                    int A, int B)
{
    // Stores all Dp-states
    vector<vector<vector<int> > > dp(
        A + 1,
        vector<vector<int> >(B + 1,
                             vector<int>(arr.size() + 1,
                                         0)));
    // Return
    return solve(arr, A, B, 0, dp);
}
  
// Driver Code
int main()
{
    vector<string> arr = { "1", "0", "10" };
    int A = 1, B = 1;
  
    cout << MaxSubsetlength(arr, A, B);
    return 0;
}
Output:
2

Time Complexity: O(N*A*B)
Auxiliary Space: O(N*A*B)

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