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Length of longest subset consisting of A 0s and B 1s from an array of strings

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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++




// 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;
}


Java




// Java program for the above approach
import java.util.*;
 
class GFG {
// Function to count 0's in a string
static int count0(String s)
{
    // Stores count of 0s
    int count = 0;
 
    // Iterate over characters of string
    for (int i = 0; i < s.length(); i++) {
 
        // If current character is '0'
        if (s.charAt(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
static int solve(String[] vec,
          int A, int B, int idx)
{
    // If idx is equal to N
    // or A + B is equal to 0
    if (idx == vec.length || 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].length() - 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 Math.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
static int MaxSubsetlength(String[] arr,
                    int A, int B)
{
    // Return
    return solve(arr, A, B, 0);
}
 
 
    public static void main (String[] args) {
    String[] arr = { "1", "0", "10" };
    int A = 1, B = 1;
 
    System.out.print(MaxSubsetlength(arr, A, B));
    }
}
// This code is contributed by offbeat


Python3




# Python3 program for the above approach
 
# Function to count 0's in a string
def count0(s):
     
    # Stores count of 0s
    count = 0
 
    # Iterate over characters of string
    for i in range(len(s)):
         
        # If current character is '0'
        if (s[i] == '0'):
            count += 1
             
    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
def solve(vec, A, B, idx):
     
    # If idx is equal to N
    # or A + B is equal to 0
    if (idx == len(vec) or A + B == 0):
        return 0
 
    # Stores the count of 0's in arr[idx]
    zero = count0(vec[idx])
 
    # Stores the count of 1's in arr[idx]
    one = len(vec[idx]) - zero
 
    # Stores the length of the
    # subset if arr[i] is included
    inc = 0
 
    # If zero is less than or equal to A
    # and one is less than or equal to B
    if (zero <= A and one <= B):
        inc = 1 + solve(vec, A - zero,
                             B - one, idx + 1)
 
    # Stores the length of the subset
    # if arr[i] is excluded
    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
def MaxSubsetlength(arr, A, B):
     
    # Return
    return solve(arr, A, B, 0)
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ "1", "0", "10" ]
    A = 1
    B = 1
     
    print(MaxSubsetlength(arr, A, B))
 
# This code is contributed by SURENDRA_GANGWAR


C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to count 0's in a string
static int count0(string s)
{
     
    // Stores count of 0s
    int count = 0;
 
    // Iterate over characters of string
    for(int i = 0; i < s.Length; 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
static int solve(List<string> vec, int A, int B,
                 int idx)
{
     
    // If idx is equal to N
    // or A + B is equal to 0
    if (idx == vec.Count || 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].Length - 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 Math.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
static int MaxSubsetlength(List<string> arr, int A,
                           int B)
{
     
    // Return
    return solve(arr, A, B, 0);
}
 
// Driver Code
public static void Main()
{
    List<string> arr = new List<string>{ "1", "0", "10" };
    int A = 1, B = 1;
 
    Console.WriteLine(MaxSubsetlength(arr, A, B));
}
}
 
// This code is contributed by ukasp


Javascript




<script>
 
// JavaScript program to implement
// the above approach
 
// Function to count 0's in a string
function count0(s)
{
    // Stores count of 0s
    let count = 0;
  
    // Iterate over characters of string
    for (let i = 0; i < s.length; 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
function solve(vec, A, B, idx)
{
    // If idx is equal to N
    // or A + B is equal to 0
    if (idx == vec.length || A + B == 0) {
        return 0;
    }
  
    // Stores the count of 0's in arr[idx]
    let zero = count0(vec[idx]);
  
    // Stores the count of 1's in arr[idx]
    let one = vec[idx].length - zero;
  
    // Stores the length of the
    // subset if arr[i] is included
    let 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
    let exc = solve(vec, A, B, idx + 1);
  
    // Returns max of inc and exc
    return Math.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
function MaxSubsetlength(arr, A, B)
{
    // Return
    return solve(arr, A, B, 0);
}
 
 
// Driver code
 
    let arr = [ "1", "0", "10" ];
    let A = 1, B = 1;
  
    document.write(MaxSubsetlength(arr, A, B));
           
</script>


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;
}


Java




// Java program for the above approach
 
import java.io.*;
 
class GFG {
   
// Function to count number
// of 0s present in the string
static int count0(String s)
{
    // Stores the count of 0s
    int count = 0;
 
    // Iterate over characters of string
    for (int i = 0; i < s.length(); i++) {
 
        // If current character is '0'
        if (s.charAt(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
static int solve(String []vec, int A,
          int B, int idx,
          int dp[][][])
{
    // If idx is equal to N or
    // A + B is equal to 0
    if (idx == vec.length || 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].length() - 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] = Math.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
static int MaxSubsetlength(String []arr,
                    int A, int B)
{
    // Stores all Dp-states
      int dp[][][] = new int[A+1][B+1][arr.length+1];
     
    // Return
    return solve(arr, A, B, 0, dp);
}
 
// Driver Code
public static void main (String[] args) {
    String arr[] = { "1", "0", "10" };
    int A = 1, B = 1;
 
    System.out.println(MaxSubsetlength(arr, A, B));
}
}
 
// This code is contributed by Dharanendra L V.


Python3




# Python3 program for the above approach
 
# Function to count number
# of 0s present in the string
def count0(s):
   
    # Stores the count of 0s
    count = 0
   
    # Iterate over characters of string
    for i in range(len(s)):
       
        # If current character is '0'
        if (s[i] == '0'):
            count+=1
    return count
  
# Recursive Function to find the length of
# longest subset from given array of strings
# with at most A 0s and B 1s
def solve(vec, A, B, idx, dp):
   
    # If idx is equal to N or
    # A + B is equal to 0
    if (idx == len(vec) or (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
    zero = count0(vec[idx])
   
    # Stores the count of 1's
    one = len(vec[idx]) - zero
   
    # Stores the length of longest
    # by including arr[idx]
    inc = 0
   
    # If zero is less than A
    # and one is less than B
    if (zero <= A and one <= B):
        inc = 1 + solve(vec, A - zero, B - one, idx + 1, dp)
   
    # Stores the length of longest subset
    # by excluding arr[idx]
    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
def MaxSubsetlength(arr, A, B):
    # Stores all Dp-states
    dp = [[ [0 for col in range(len(arr) + 1)] for col in range(B + 1)] for row in range(A + 1)]
       
    # Return
    return solve(arr, A, B, 0, dp)
 
arr = [ "1", "0", "10" ]
A, B = 1, 1
print(MaxSubsetlength(arr, A, B))
 
# This code is contributed by suresh07.


C#




using System;
 
public class GFG{
     
    // Function to count number
// of 0s present in the string
static int count0(string s)
{
    // Stores the count of 0s
    int count = 0;
  
    // Iterate over characters of string
    for (int i = 0; i < s.Length; 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
static int solve(string []vec, int A,
          int B, int idx,
          int[,,] dp)
{
    // If idx is equal to N or
    // A + B is equal to 0
    if (idx == vec.Length || 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].Length - 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] = Math.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
static int MaxSubsetlength(string []arr,
                    int A, int B)
{
    // Stores all Dp-states
      int[,,] dp = new int[A+1,B+1,arr.Length+1];
      
    // Return
    return solve(arr, A, B, 0, dp);
}
  
// Driver Code   
    static public void Main ()
    {
         
        string[] arr = { "1", "0", "10" };
    int A = 1, B = 1;
  
    Console.WriteLine(MaxSubsetlength(arr, A, B));
         
    }
}
 
// This code is contributed by rag2127


Javascript




<script>
// Javascript program for the above approach
 
// Function to count number
// of 0s present in the string
function count0(s)
{
 
    // Stores the count of 0s
    let count = 0;
  
    // Iterate over characters of string
    for (let i = 0; i < s.length; 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
function solve(vec, A, B, idx, dp)
{
 
    // If idx is equal to N or
    // A + B is equal to 0
    if (idx == vec.length || 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
    let zero = count0(vec[idx]);
  
    // Stores the count of 1's
    let one = vec[idx].length - zero;
  
    // Stores the length of longest
    // by including arr[idx]
    let 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]
    let exc = solve(vec, A, B, idx + 1, dp);
  
    // Assign
    dp[A][B][idx] = Math.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
function MaxSubsetlength(arr, A, B)
{
 
    // Stores all Dp-states
      let dp = new Array(A+1);
    for(let i = 0; i < A + 1; i++)
    {
        dp[i] = new Array(B+1);
        for(let j = 0; j < B+1;j++)
        {
            dp[i][j] = new Array(arr.length+1);
            for(let k = 0; k < arr.length + 1; k++)
            {
                dp[i][j][k] = 0;
            }
        }
    }
      
    // Return
    return solve(arr, A, B, 0, dp);
}
 
// Driver Code
let arr = [ "1", "0", "10" ];
let  A = 1, B = 1;
document.write(MaxSubsetlength(arr, A, B));
 
// This code is contributed by avanitrachhadiya2155
</script>


Output: 

2

 

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

 



Last Updated : 04 Oct, 2021
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