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++ 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 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 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# 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 |
<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> |
2
Time Complexity: O(2N)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized using Memoization, based on the following observations:
- 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++ 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 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 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. |
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 |
<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> |
2
Time Complexity: O(N*A*B)
Auxiliary Space: O(N*A*B)