grid47 Exploring patterns and algorithms
Sep 20, 2024
6 min read
Solution to LeetCode 474: Ones and Zeroes Problem
Given an array of binary strings strs and two integers m and n, return the size of the largest subset of strs such that there are at most m zeros and n ones in the subset.
Problem
Approach
Steps
Complexity
Input: The input consists of an array `strs` of binary strings and two integers `m` and `n`.
Example: strs = ['11', '0001', '101', '1', '0'], m = 4, n = 3
Constraints:
• 1 <= strs.length <= 600
• 1 <= strs[i].length <= 100
• strs[i] consists only of digits '0' and '1'.
• 1 <= m, n <= 100
Output: Return the size of the largest subset of strings where the total number of zeros is at most `m` and the total number of ones is at most `n`.
Example: Output: 4
Constraints:
• The output should be an integer representing the size of the largest valid subset.
Goal: Find the largest subset of strings that meets the criteria of having at most `m` zeros and `n` ones.
Steps:
• 1. Initialize a dynamic programming table `dp` to track the largest subset size for each combination of zeros and ones.
• 2. For each string in `strs`, count the number of zeros and ones.
• 3. Update the DP table by considering the current string and checking if adding it to any existing valid subset would exceed the limits of `m` zeros and `n` ones.
• 4. Return the maximum subset size found in the DP table.
Goal: Constraints for the problem are based on the size of the input array, the length of each binary string, and the values of `m` and `n`.
Steps:
• The array `strs` contains up to 600 binary strings.
• Each binary string has a length between 1 and 100 characters.
• The integers `m` and `n` are between 1 and 100.
Assumptions:
• The strings in `strs` only contain '0' and '1'.
• The constraints on `m` and `n` ensure that brute-force approaches are not feasible.
• Input: strs = ['11', '0001', '101', '1', '0'], m = 4, n = 3
• Explanation: The largest valid subset is {'11', '0001', '1', '0'}, which has 4 zeros and 3 ones, satisfying the conditions.
• Input: strs = ['10', '1', '0'], m = 2, n = 2
• Explanation: The largest valid subset is {'10', '1', '0'}, which contains 2 zeros and 2 ones.
Approach: The approach involves dynamic programming to keep track of the largest subset that satisfies the constraints on zeros and ones.
Observations:
• This is a variant of the knapsack problem, where we are bounded by both zeros and ones in the strings.
• We need to use dynamic programming to track the largest valid subset size.
• Using dynamic programming (DP), we can efficiently track the maximum number of strings we can select given the constraints.
Steps:
• 1. Initialize a DP table where dp[i][j] represents the largest subset with at most `i` zeros and `j` ones.
• 2. Loop through each string, calculate the number of zeros and ones in the string.
• 3. Update the DP table by checking if adding the current string to an existing subset remains valid (does not exceed the limits of zeros and ones).
• 4. Return the value in dp[m][n], which will be the largest subset size with at most `m` zeros and `n` ones.
Empty Inputs:
• Empty input array should return 0 as no strings can be selected.
Large Inputs:
• Handle cases where the number of strings is large (up to 600).
Special Values:
• Consider inputs where all strings are '0' or '1', or combinations of strings with very few zeros or ones.
Constraints:
• Ensure that the DP table is efficiently updated to handle all possible combinations of zeros and ones.
intfindMaxForm(vector<string>& strs, int zeros, int ones) {
int i, j, p = strs.size();
vector<vector<int>> dp(zeros +1, vector<int>(ones+1));
for(auto&s : strs) {
int currOne = count(s.begin(), s.end(), '1');
int currZero = s.size() - currOne;
for(int j = ones ; j >= currOne; j--)
for(int i = zeros; i >= currZero; i--) {
dp[i][j] = max(dp[i][j], 1+ dp[i- currZero ][j-currOne]);
}
}
return dp[zeros][ones];
}
1 : Function Definition
intfindMaxForm(vector<string>& strs, int zeros, int ones) {
Defines the function `findMaxForm`, which takes a list of binary strings and two integers `zeros` and `ones` as input and returns the maximum number of strings that can be formed using at most the given number of 0's and 1's.
2 : Variable Initialization
int i, j, p = strs.size();
Initializes the variables `i` and `j` for the loops, and `p` to store the size of the `strs` vector.
Initializes the DP table `dp` with dimensions `(zeros + 1) x (ones + 1)` to store the maximum number of strings that can be formed with a given number of zeros and ones.
4 : Outer Loop Over Strings
for(auto&s : strs) {
Iterates over each string `s` in the `strs` vector to process it one by one.
5 : Count Ones in Current String
int currOne = count(s.begin(), s.end(), '1');
Counts the number of 1's in the current string `s` using the `count()` function.
6 : Count Zeros in Current String
int currZero = s.size() - currOne;
Calculates the number of 0's in the current string `s` by subtracting the number of 1's from the total size of the string.
7 : Inner Loop Over Ones
for(int j = ones ; j >= currOne; j--)
Iterates over the number of ones in the DP table in reverse order (from `ones` down to `currOne`) to ensure we don't overwrite previously computed values.
8 : Inner Loop Over Zeros
for(int i = zeros; i >= currZero; i--) {
Iterates over the number of zeros in the DP table in reverse order (from `zeros` down to `currZero`) to avoid overwriting values during the DP calculation.
Updates the DP table at position `dp[i][j]` by considering whether to include the current string. If included, the value becomes 1 + the previous value at `dp[i - currZero][j - currOne]`.
10 : Return Result
return dp[zeros][ones];
Returns the value stored in `dp[zeros][ones]`, which represents the maximum number of strings that can be formed with `zeros` and `ones`.
Best Case: O(m * n * strs.length)
Average Case: O(m * n * strs.length)
Worst Case: O(m * n * strs.length)
Description: Time complexity is proportional to the product of `m`, `n`, and the length of the input array `strs`.
Best Case: O(m * n)
Worst Case: O(m * n)
Description: Space complexity is proportional to the size of the DP table, which is `O(m * n)`.
