LeetCode 0474. Ones and Zeroes Solution in Java, Python, C++, JavaScript, Go & Rust | Explanation + Code

Description

You are 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 0's and n 1's in the subset.

A set x is a subset of a set y if all elements of x are also elements of y.

 

Example 1:

Input: strs = ["10","0001","111001","1","0"], m = 5, n = 3
Output: 4
Explanation: 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
Explanation: 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

Solutions

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.

For f[i][j][k], we have two choices:

• Do not select the i-th string, in which case f[i][j][k]=f[i-1][j][k];
• Select the i-th string, in which case f[i][j][k]=f[i-1][j-a][k-b]+1, where a and b are the number of zeros and ones in the i-th string, respectively.

We take the maximum of these two choices to obtain the value of f[i][j][k].

The final answer is f[sz][m][n].

The time complexity is O(sz × m × n), and the space complexity is O(sz × m × n), where sz is the length of the array strs, and m and n are the upper limits on the number of zeros and ones, respectively.

PythonJavaC++GoTypeScriptRust

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](code-box)

class Solution { public int findMaxForm(String[] strs, int m, int n) { int sz = strs.length; int[][][] f = new int[sz + 1][m + 1][n + 1]; for (int i = 1; i <= sz; ++i) { int[] cnt = count(strs[i - 1]); for (int j = 0; j <= m; ++j) { for (int k = 0; k <= n; ++k) { f[i][j][k] = f[i - 1][j][k]; if (j >= cnt[0] && k >= cnt[1]) { f[i][j][k] = Math.max(f[i][j][k], f[i - 1][j - cnt[0]][k - cnt[1]] + 1); } } } } return f[sz][m][n]; } private int[] count(String s) { int[] cnt = new int[2]; for (int i = 0; i < s.length(); ++i) { ++cnt[s.charAt(i) - '0']; } return cnt; } }(code-box)

class Solution { public: int findMaxForm(vector<string>& strs, int m, int n) { int sz = strs.size(); int f[sz + 1][m + 1][n + 1]; memset(f, 0, sizeof(f)); for (int i = 1; i <= sz; ++i) { auto [a, b] = count(strs[i - 1]); for (int j = 0; j <= m; ++j) { for (int k = 0; k <= n; ++k) { f[i][j][k] = f[i - 1][j][k]; if (j >= a && k >= b) { f[i][j][k] = max(f[i][j][k], f[i - 1][j - a][k - b] + 1); } } } } return f[sz][m][n]; } pair<int, int> count(string& s) { int a = count_if(s.begin(), s.end(), [](char c) { return c == '0'; }); return {a, s.size() - a}; } };(code-box)

func findMaxForm(strs []string, m int, n int) int { sz := len(strs) f := make([][][]int, sz+1) for i := range f { f[i] = make([][]int, m+1) for j := range f[i] { f[i][j] = make([]int, n+1) } } for i := 1; i <= sz; i++ { a, b := count(strs[i-1]) for j := 0; j <= m; j++ { for k := 0; k <= n; k++ { f[i][j][k] = f[i-1][j][k] if j >= a && k >= b { f[i][j][k] = max(f[i][j][k], f[i-1][j-a][k-b]+1) } } } } return f[sz][m][n] } func count(s string) (int, int) { a := strings.Count(s, "0") return a, len(s) - a }(code-box)

function findMaxForm(strs: string[], m: number, n: number): number { const sz = strs.length; const f = Array.from({ length: sz + 1 }, () => Array.from({ length: m + 1 }, () => Array.from({ length: n + 1 }, () => 0)), ); const count = (s: string): [number, number] => { let a = 0; for (const c of s) { a += c === '0' ? 1 : 0; } return [a, s.length - a]; }; for (let i = 1; i <= sz; ++i) { const [a, b] = count(strs[i - 1]); for (let j = 0; j <= m; ++j) { for (let k = 0; k <= n; ++k) { f[i][j][k] = f[i - 1][j][k]; if (j >= a && k >= b) { f[i][j][k] = Math.max(f[i][j][k], f[i - 1][j - a][k - b] + 1); } } } } return f[sz][m][n]; }(code-box)

impl Solution { pub fn find_max_form(strs: Vec<String>, m: i32, n: i32) -> i32 { let sz = strs.len(); let m = m as usize; let n = n as usize; let mut f = vec![vec![vec![0; n + 1]; m + 1]; sz + 1]; for i in 1..=sz { let a = strs[i - 1].chars().filter(|&c| c == '0').count(); let b = strs[i - 1].len() - a; for j in 0..=m { for k in 0..=n { f[i][j][k] = f[i - 1][j][k]; if j >= a && k >= b { f[i][j][k] = f[i][j][k].max(f[i - 1][j - a][k - b] + 1); } } } } f[sz][m][n] as i32 } }(code-box)

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 × n).

PythonJavaC++GoTypeScriptRust

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

class Solution { public int findMaxForm(String[] strs, int m, int n) { int[][] f = new int[m + 1][n + 1]; for (String s : strs) { int[] cnt = count(s); for (int i = m; i >= cnt[0]; --i) { for (int j = n; j >= cnt[1]; --j) { f[i][j] = Math.max(f[i][j], f[i - cnt[0]][j - cnt[1]] + 1); } } } return f[m][n]; } private int[] count(String s) { int[] cnt = new int[2]; for (int i = 0; i < s.length(); ++i) { ++cnt[s.charAt(i) - '0']; } return cnt; } }(code-box)

class Solution { public: int findMaxForm(vector<string>& strs, int m, int n) { int f[m + 1][n + 1]; memset(f, 0, sizeof(f)); for (auto& s : strs) { auto [a, b] = count(s); for (int i = m; i >= a; --i) { for (int j = n; j >= b; --j) { f[i][j] = max(f[i][j], f[i - a][j - b] + 1); } } } return f[m][n]; } pair<int, int> count(string& s) { int a = count_if(s.begin(), s.end(), [](char c) { return c == '0'; }); return {a, s.size() - a}; } };(code-box)

func findMaxForm(strs []string, m int, n int) int { f := make([][]int, m+1) for i := range f { f[i] = make([]int, n+1) } for _, s := range strs { a, b := count(s) for j := m; j >= a; j-- { for k := n; k >= b; k-- { f[j][k] = max(f[j][k], f[j-a][k-b]+1) } } } return f[m][n] } func count(s string) (int, int) { a := strings.Count(s, "0") return a, len(s) - a }(code-box)

function findMaxForm(strs: string[], m: number, n: number): number { const f = Array.from({ length: m + 1 }, () => Array.from({ length: n + 1 }, () => 0)); const count = (s: string): [number, number] => { let a = 0; for (const c of s) { a += c === '0' ? 1 : 0; } return [a, s.length - a]; }; for (const s of strs) { const [a, b] = count(s); for (let i = m; i >= a; --i) { for (let j = n; j >= b; --j) { f[i][j] = Math.max(f[i][j], f[i - a][j - b] + 1); } } } return f[m][n]; }(code-box)

impl Solution { pub fn find_max_form(strs: Vec<String>, m: i32, n: i32) -> i32 { let m = m as usize; let n = n as usize; let mut f = vec![vec![0; n + 1]; m + 1]; for s in strs { let a = s.chars().filter(|&c| c == '0').count(); let b = s.len() - a; for i in (a..=m).rev() { for j in (b..=n).rev() { f[i][j] = f[i][j].max(f[i - a][j - b] + 1); } } } f[m][n] } }(code-box)