How do you solve Ones and Zeroes on LeetCode?

Ones and Zeroes (LeetCode #474) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n * k) time complexity and O(m * n) space complexity. Key insight: Using dynamic programming reduces redundant calculations and improves efficiency.

What is the brute force solution for Ones and Zeroes?

The brute force approach for Ones and Zeroes is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible subsets of the input strings and checking each subset to see if it meets the criteria of having at most m 0's and n 1's. This method is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(m * n * k).

What algorithm or data structure is used to solve Ones and Zeroes?

Ones and Zeroes uses the following concepts: Array, String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Combinatorial Optimization.

What are the interview tips for Ones and Zeroes?

Always clarify the constraints and edge cases before diving into the solution. Explain your thought process clearly while coding, as it helps interviewers understand your approach. Practice writing both brute force and optimal solutions to solidify your understanding of the problem.

What are common mistakes when solving Ones and Zeroes?

Forgetting to initialize the DP array correctly. Updating the DP array in the wrong order, which can lead to incorrect results.

#474

Ones and Zeroes

Medium

ArrayStringDynamic ProgrammingDynamic ProgrammingCombinatorial Optimization

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(m * n * k)
Space	O(1)	O(m * n)

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Intuition

Time O(m * n * k)Space O(m * n)

The optimal approach uses dynamic programming to keep track of the maximum size of subsets that can be formed with given limits of 0's and 1's. This significantly reduces the number of redundant calculations compared to the brute force method.

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Algorithm

3 steps

1Step 1: Initialize a 2D DP array where dp[i][j] represents the maximum size of the subset with i 0's and j 1's.
2Step 2: Iterate through each string in the input array and count its 0's and 1's.
3Step 3: Update the DP array in reverse order to avoid overwriting values that are needed for calculations in the same iteration.

solution.py9 lines

1def findMaxForm(strs, m, n):
2    dp = [[0] * (n + 1) for _ in range(m + 1)]
3    for s in strs:
4        count_0 = s.count('0')
5        count_1 = s.count('1')
6        for i in range(m, count_0 - 1, -1):
7            for j in range(n, count_1 - 1, -1):
8                dp[i][j] = max(dp[i][j], dp[i - count_0][j - count_1] + 1)
9    return dp[m][n]

ℹ

Complexity note: The time complexity is O(m * n * k) where k is the number of strings. This is because we iterate through each string and update the DP table, which has dimensions m and n.

1Using dynamic programming reduces redundant calculations and improves efficiency.
2Understanding how to manage state in a DP table is crucial for solving similar problems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.