#474

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Ones and Zeroes

Solve the Ones and Zeroes problem using dynamic programming with state transition, focusing on array and string manipulation.

array string dynamic programming

Problem Statement

You are given an array of binary strings strs and two integers m and n. The goal is to determine the size of the largest subset of strs such that the subset contains at most m 0's and n 1's in total.

A set x is a subset of a set y if all elements of x are also present in y. For each binary string in the array, count the number of 0's and 1's, and choose a subset where the total number of 0's does not exceed m and the total number of 1's does not exceed n.

Examples

Example 1

Input: strs = ["10","0001","111001","1","0"], m = 5, n = 3

Output: 4

The largest subset with at most 5 0's and 3 1's is {"10", "0001", "1", "0"}, so the answer is 4. Other valid but smaller subsets include {"0001", "1"} and {"10", "1", "0"}. {"111001"} is an invalid subset because it contains 4 1's, greater than the maximum of 3.

Example 2

Input: strs = ["10","0","1"], m = 1, n = 1

Output: 2

The largest subset is {"0", "1"}, so the answer is 2.

Constraints

1 <= strs.length <= 600
1 <= strs[i].length <= 100
strs[i] consists only of digits '0' and '1'.
1 <= m, n <= 100

Solution Approach

Dynamic Programming State Transition

Use dynamic programming with a state transition approach. Create a 2D DP table where each entry dp[i][j] represents the maximum subset size with i 0's and j 1's. Transition through each binary string and update the DP table by considering the number of 0's and 1's in the string. This allows you to evaluate the largest subset with the constraints on 0's and 1's.

Space Optimization

Since the current DP table only depends on the previous row, optimize the space complexity by using a 1D DP array. This reduces the space usage from O(m * n) to O(n) by updating the array in reverse order to avoid overwriting values prematurely.

Subset Consideration

Iterate through all the binary strings, and for each, determine how it can contribute to the current DP state. This process involves checking if including the string violates the constraints on 0's and 1's and, if valid, updating the DP table to maximize the subset size.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this problem depends on the final approach but typically falls within O(m * n * k), where k is the number of strings and m and n are the given constraints for 0's and 1's. The space complexity can be optimized to O(m * n) using dynamic programming. If space optimization is used, it reduces to O(n).

What Interviewers Usually Probe

Look for clear understanding of dynamic programming concepts.
Check for efficient space optimization techniques in DP.
Evaluate ability to break down a problem into manageable subproblems.

Common Pitfalls or Variants

Common pitfalls

Overcomplicating the DP state transitions or failing to properly update the table.
Ignoring the space complexity aspect and using an overly large DP table.
Misunderstanding the problem constraints, especially in counting 0's and 1's.

Follow-up variants

Consider variations with additional constraints, like limiting the number of binary strings that can be selected.
Modify the problem by changing the allowed counts of 0's and 1's or adding more characters to the binary strings.
Test how the solution handles larger inputs and evaluate performance scalability.

FAQ

How can dynamic programming help solve the Ones and Zeroes problem?

Dynamic programming helps by breaking the problem down into smaller subproblems, where the solution to each subproblem depends on the previous state. This is useful in optimizing the subset selection process while keeping track of the constraints on 0's and 1's.

What is the space optimization strategy for Ones and Zeroes?

Space optimization can be achieved by using a 1D DP array instead of a 2D table. This allows you to reduce the space complexity from O(m * n) to O(n) by updating the array in reverse order.

Can this problem be solved without dynamic programming?

While a brute force solution is possible, it is inefficient due to its high time complexity. Dynamic programming offers a much more efficient approach for solving the problem within the given constraints.

What is the time complexity of the Ones and Zeroes problem?

The time complexity of the solution is O(m * n * k), where m and n are the constraints on 0's and 1's, and k is the number of binary strings. This complexity comes from the need to iterate through each string and update the DP table.

How do I handle the edge cases for this problem?

Edge cases involve situations like empty strings, strings with only 0's or only 1's, or constraints where m or n is small. Make sure to properly handle such cases by checking the current DP state before updating it.

terminal

Solution

Solution 1: Dynamic Programming

We define $f[i][j][k]$ as the maximum number of strings that can be obtained from the first $i$ strings using $j$ zeros and $k$ ones. Initially, $f[i][j][k]=0$, and the answer is $f[sz][m][n]$, where $sz$ is the length of the array $strs$.

class Solution:
    def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
        sz = len(strs)
        f = [[[0] * (n + 1) for _ in range(m + 1)] for _ in range(sz + 1)]
        for i, s in enumerate(strs, 1):
            a, b = s.count("0"), s.count("1")
            for j in range(m + 1):
                for k in range(n + 1):
                    f[i][j][k] = f[i - 1][j][k]
                    if j >= a and k >= b:
                        f[i][j][k] = max(f[i][j][k], f[i - 1][j - a][k - b] + 1)
        return f[sz][m][n]

Solution 2: Dynamic Programming (Space Optimization)

We notice that the calculation of $f[i][j][k]$ only depends on $f[i-1][j][k]$ and $f[i-1][j-a][k-b]$. Therefore, we can eliminate the first dimension and optimize the space complexity to $O(m \times n)$.

class Solution:
    def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
        sz = len(strs)
        f = [[[0] * (n + 1) for _ in range(m + 1)] for _ in range(sz + 1)]
        for i, s in enumerate(strs, 1):
            a, b = s.count("0"), s.count("1")
            for j in range(m + 1):
                for k in range(n + 1):
                    f[i][j][k] = f[i - 1][j][k]
                    if j >= a and k >= b:
                        f[i][j][k] = max(f[i][j][k], f[i - 1][j - a][k - b] + 1)
        return f[sz][m][n]

Continue Practicing

#472 concatenated words #691 stickers to spell word #792 number of matching subsequences