Length of longest subset consisting of A 0s and B 1s from an array of strings | Set 2
Last Updated :
13 Apr, 2023
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: Refer to the previous post of this article for the simplest approach to solve the problem.
Time Complexity: O(2N)
Auxiliary Space: O(1)
Dynamic Programming Approach: Refer to the previous post of this article for the Dynamic Programming approach.
Time Complexity: O(N*A*B)
Auxiliary Space: O(N * A * B)
Space-Optimized Dynamic Programming Approach: The space complexity in the above approach can be optimized based on the following observations:
- Initialize a 2D array, dp[A][B], where dp[i][j] represents the length of the longest subset consisting of at most i number of 0s and j number of 1s.
- To optimize the auxiliary space from the 3D table to the 2D table, the idea is to traverse the inner loops in reverse order.
- This ensures that whenever a value in dp[][] is changed, it will not be needed anymore in the current iteration.
- Therefore, the recurrence relation looks like this:
dp[i][j] = max (dp[i][j], dp[i – zeros][j – ones] + 1)
where zeros is the count of 0s and ones is the count of 1s in the current iteration.
Follow the steps below to solve the problem:
- Initialize a 2D array, say dp[A][B] and initialize all its entries as 0.
- Traverse the given array arr[] and for each binary string perform the following steps:
- After completing the above steps, print the value of dp[A][B] as the result.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int MaxSubsetlength(vector<string> arr,
int A, int B)
{
int dp[A + 1][B + 1];
memset(dp, 0, sizeof(dp));
for (auto& str : arr) {
int zeros = count(str.begin(),
str.end(), '0');
int ones = count(str.begin(),
str.end(), '1');
for (int i = A; i >= zeros; i--)
for (int j = B; j >= ones; j--)
dp[i][j] = max(
dp[i][j],
dp[i - zeros][j - ones] + 1);
}
return dp[A][B];
}
int main()
{
vector<string> arr
= { "1", "0", "0001",
"10", "111001" };
int A = 5, B = 3;
cout << MaxSubsetlength(arr, A, B);
return 0;
}
|
Java
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{
static int MaxSubsetlength(String arr[],
int A, int B)
{
int dp[][] = new int[A + 1][B + 1];
for(String str : arr)
{
int zeros = 0, ones = 0;
for(char ch : str.toCharArray())
{
if (ch == '0')
zeros++;
else
ones++;
}
for(int i = A; i >= zeros; i--)
for(int j = B; j >= ones; j--)
dp[i][j] = Math.max(
dp[i][j],
dp[i - zeros][j - ones] + 1);
}
return dp[A][B];
}
public static void main(String[] args)
{
String arr[] = { "1", "0", "0001",
"10", "111001" };
int A = 5, B = 3;
System.out.println(MaxSubsetlength(arr, A, B));
}
}
|
Python3
def MaxSubsetlength(arr, A, B):
dp = [[0 for i in range(B + 1)]
for i in range(A + 1)]
for str in arr:
zeros = str.count('0')
ones = str.count('1')
for i in range(A, zeros - 1, -1):
for j in range(B, ones - 1, -1):
dp[i][j] = max(dp[i][j],
dp[i - zeros][j - ones] + 1)
return dp[A][B]
if __name__ == '__main__':
arr = [ "1", "0", "0001", "10", "111001" ]
A, B = 5, 3
print (MaxSubsetlength(arr, A, B))
|
C#
using System;
class GFG {
static int MaxSubsetlength(string[] arr, int A, int B)
{
int[, ] dp = new int[A + 1, B + 1];
foreach(string str in arr)
{
int zeros = 0, ones = 0;
foreach(char ch in str.ToCharArray())
{
if (ch == '0')
zeros++;
else
ones++;
}
for (int i = A; i >= zeros; i--)
for (int j = B; j >= ones; j--)
dp[i, j] = Math.Max(
dp[i, j],
dp[i - zeros, j - ones] + 1);
}
return dp[A, B];
}
public static void Main(string[] args)
{
string[] arr = { "1", "0", "0001", "10", "111001" };
int A = 5, B = 3;
Console.WriteLine(MaxSubsetlength(arr, A, B));
}
}
|
Javascript
<script>
function MaxSubsetlength(arr, A, B)
{
var dp = Array.from(Array(A + 1),
()=>Array(B + 1).fill(0));
arr.forEach(str => {
var zeros = [...str].filter(x => x == '0').length;
var ones = [...str].filter(x => x == '1').length;
for(var i = A; i >= zeros; i--)
for(var j = B; j >= ones; j--)
dp[i][j] = Math.max(dp[i][j],
dp[i - zeros][j - ones] + 1);
});
return dp[A][B];
}
var arr = [ "1", "0", "0001",
"10", "111001" ];
var A = 5, B = 3;
document.write(MaxSubsetlength(arr, A, B));
</script>
|
Time Complexity: O(N * A * B)
Auxiliary Space: O(A * B)
Efficient Approach : using array instead of 2d matrix to optimize space complexity
The optimization comes from the fact that the in this approach we use a 1D array, which requires less memory compared to a 2D array. However, in this approach code requires an additional condition to check if the number of zeros is less than or equal to the allowed limit.
Implementations Steps:
- Initialize an array dp of size B+1 with all entries as 0.
- Traverse through the given array of strings and for each string, count the number of 0’s and 1’s in it.
- For each string, iterate from B to the number of 1’s in the string and update the value of dp[j] as maximum of dp[j] and dp[j – ones] + (1 if (j >= ones && A >= zeros) else 0).
- Finally, return dp[B] as the length of the longest subset of strings with at most A 0’s and B 1’s.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
int MaxSubsetlength(vector<string> arr, int A, int B)
{
int dp[B + 1];
memset(dp, 0, sizeof(dp));
for (auto& str : arr) {
int zeros = count(str.begin(), str.end(), '0');
int ones = count(str.begin(), str.end(), '1');
for (int j = B; j >= ones; j--)
dp[j] = max(
dp[j],
dp[j - ones]
+ ((j >= ones && A >= zeros) ? 1 : 0));
}
return dp[B];
}
int main()
{
vector<string> arr
= { "1", "0", "0001", "10", "111001" };
int A = 5, B = 3;
cout << MaxSubsetlength(arr, A, B);
return 0;
}
|
Java
import java.util.*;
class Main {
static int MaxSubsetlength(List<String> arr, int A, int B) {
int[] dp = new int[B + 1];
Arrays.fill(dp, 0);
for (String str : arr) {
int zeros = 0, ones = 0;
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == '0') zeros++;
else ones++;
}
for (int j = B; j >= ones; j--) {
dp[j] = Math.max(dp[j], dp[j - ones] + ((j >= ones && A >= zeros) ? 1 : 0));
}
}
return dp[B];
}
public static void main(String[] args) {
List<String> arr = Arrays.asList("1", "0", "0001", "10", "111001");
int A = 5, B = 3;
System.out.println(MaxSubsetlength(arr, A, B));
}
}
|
Python3
def MaxSubsetlength(arr, A, B):
dp = [0] * (B + 1)
for str in arr:
zeros = str.count('0')
ones = str.count('1')
for j in range(B, ones - 1, -1):
dp[j] = max(
dp[j],
dp[j - ones]
+ ((j >= ones and A >= zeros) == True))
return dp[B]
arr = ["1", "0", "0001", "10", "111001"]
A, B = 5, 3
print(MaxSubsetlength(arr, A, B))
|
C#
using System;
using System.Collections.Generic;
class MainClass
{
static int MaxSubsetlength(List<string> arr, int A, int B)
{
int[] dp = new int[B + 1];
Array.Fill(dp, 0);
foreach (string str in arr)
{
int zeros = 0, ones = 0;
foreach (char c in str) {
if (c == '0') zeros++;
else ones++;
}
for (int j = B; j >= ones; j--)
{
dp[j] = Math.Max(dp[j], dp[j - ones] + ((j >= ones && A >= zeros) ? 1 : 0));
}
}
return dp[B];
}
public static void Main(string[] args)
{
List<string> arr = new List<string> {"1", "0", "0001", "10", "111001"};
int A = 5, B = 3;
Console.WriteLine(MaxSubsetlength(arr, A, B));
}
}
|
Javascript
function MaxSubsetlength(arr, A, B) {
let dp = new Array(B + 1).fill(0);
for (let i = 0; i < arr.length; i++) {
const str = arr[i];
let zeros = 0,
ones = 0;
for (let j = 0; j < str.length; j++) {
if (str.charAt(j) === '0') zeros++;
else ones++;
}
for (let j = B; j >= ones; j--) {
dp[j] = Math.max(dp[j], dp[j - ones] + ((j >= ones && A >= zeros) ? 1 : 0));
}
}
return dp[B];
}
const arr = ["1", "0", "0001", "10", "111001"];
const A = 5,
B = 3;
console.log(MaxSubsetlength(arr, A, B));
|
Output :
4
Time Complexity: O(N * A * B)
Auxiliary Space: O(B)
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...