# 474.ones-and-zeros-cn

## 题目地址

https://leetcode.com/problems/ones-and-zeroes/

## 题目描述

``````In the computer world, use restricted resource you have to generate maximum benefit is what we always want to pursue.

For now, suppose you are a dominator of m 0s and n 1s respectively. On the other hand, there is an array with strings consisting of only 0s and 1s.

Now your task is to find the maximum number of strings that you can form with given m 0s and n 1s. Each 0 and 1 can be used at most once.

Note:

The given numbers of 0s and 1s will both not exceed 100
The size of given string array won't exceed 600.

Example 1:

Input: Array = {"10", "0001", "111001", "1", "0"}, m = 5, n = 3
Output: 4

Explanation: This are totally 4 strings can be formed by the using of 5 0s and 3 1s, which are “10,”0001”,”1”,”0”

Example 2:

Input: Array = {"10", "0", "1"}, m = 1, n = 1
Output: 2

Explanation: You could form "10", but then you'd have nothing left. Better form "0" and "1".``````

## 思路

### 复杂度分析

• 时间复杂度: `O(2^len * s) - len is Strs length, s is the average string length`

• 空间复杂度: `O(1)`

### 解法二 - 记忆 + 递归

`memo[i][j][k] - 表示前i个字符串中能用j个0和k和1组成的最多的字符串个数。`

`helpe(strs, i, j, k, memo)` 递归求解： 1. 如果`memo[i][j][k] != 0`, 表示已经求解过了，所以直接返回值即可 2. 如果没有求解过，判断当前剩余的`0``1`的个数`j``k`是否大于等于当前字符串的`0``1`的个数， a. 满足条件，则可以取当前字符串， 3. 不取当前字符串。 4. 取最大值保存在`memo[i][j][k]`

### 复杂度分析

• 时间复杂度: `O(l * m * n) - l 是字符串数组的长度，m是0的个数，n是1的个数`

• 空间复杂度: `O(l * m * n) - memo 三维数组`

### 解法三 - 三维DP

`dp[i][j][k] - 表示前i个字符串中能用j个0和k和1组成的最多的字符串个数。`

• `j >= count0 && k >= count1`,

`dp[i][j][k] = max(dp[i - 1][j][k], dp[i - 1][j - count0][k - count1] + 1)`

• 不满条件，不取当前字符串

`dp[i][j][k] = dp[i - 1][j][k]`

### 复杂度分析

• 时间复杂度: `O(l * m * n) - l 是字符串数组的长度，m是0的个数，n是1的个数`

• 空间复杂度: `O(l * m * n) - dp 三维数组`

### 解法四 - 二维DP

`dp[i][j] = max(dp[i][j], dp[i - count0][j - count1] + 1)`

### 复杂度分析

• 时间复杂度: `O(l * m * n) - l 是字符串数组的长度，m是0的个数，n是1的个数`

• 空间复杂度: `O(m * n) - dp 三维数组`

## 代码 (`Java/Python3`)

### 解法一 - 递归 (超时)

Java code

``````class OnesAndZerosBFRecursive {
public int findMaxForm2(String[] strs, int m, int n) {
return helper(strs, 0, m, n);
}
private int helper(String[] strs, int idx, int j, int k) {
if (idx == strs.length) return 0;
// count current idx string number of zeros and ones
int[] counts = countZeroOnes(strs[idx]);
// if j >= count0 && k >= count1, take current index string
int takeCurrStr = j - counts[0] >= 0 && k - counts[1] >= 0
? 1 + helper(strs, idx + 1, j - counts[0], k - counts[1])
: -1;
// don't take current index string strs[idx], continue next string
int notTakeCurrStr = helper(strs, idx + 1, j, k);
return Math.max(takeCurrStr, notTakeCurrStr);
}
private int[] countZeroOnes(String s) {
int[] res = new int[2];
for (char ch : s.toCharArray()) {
res[ch - '0']++;
}
return res;
}
}``````

Python3 code

``````class Solution:
def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
return self.helper(strs, m, n, 0)

def helper(self, strs, m, n, idx):
if idx == len(strs):
return 0
take_curr_str = -1
count0, count1 = strs[idx].count('0'), strs[idx].count('1')
if m >= count0 and n >= count1:
take_curr_str = max(take_curr_str, self.helper(strs, m - count0, n - count1, idx + 1) + 1)
not_take_curr_str = self.helper(strs, m, n, idx + 1)
return max(take_curr_str, not_take_curr_str)``````

### 解法二 - 记忆 + 递归

Java code

``````class OnesAndZerosMemoRecur {
public int findMaxForm4(String[] strs, int m, int n) {
return helper(strs, 0, m, n, new int[strs.length][m + 1][n + 1]);
}
private int helper(String[] strs, int idx, int j, int k, int[][][] memo) {
if (idx == strs.length) return 0;
// check if already calculated, return value
if (memo[idx][j][k] != 0) {
return memo[idx][j][k];
}
int[] counts = countZeroOnes(strs[idx]);
// if satisfy condition, take current string, strs[idx], update count0 and count1
int takeCurrStr = j - counts[0] >= 0 && k - counts[1] >= 0
? 1 + helper(strs, idx + 1, j - counts[0], k - counts[1], memo)
: -1;
// not take current string
int notTakeCurrStr = helper(strs, idx + 1, j, k, memo);
// always keep track the max value into memory
memo[idx][j][k] = Math.max(takeCurrStr, notTakeCurrStr);
return memo[idx][j][k];
}
private int[] countZeroOnes(String s) {
int[] res = new int[2];
for (char ch : s.toCharArray()) {
res[ch - '0']++;
}
return res;
}
}``````

Python3 code - (TLE)

``````class Solution:
def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
memo = {k:[[0]*(n+1) for _ in range(m+1)] for k in range(len(strs)+1)}
return self.helper(strs, 0, m, n, memo)

def helper(self, strs, idx, m, n, memo):
if idx == len(strs):
return 0
if memo[idx][m][n] != 0:
return memo[idx][m][n]
take_curr_str = -1
count0, count1 = strs[idx].count('0'), strs[idx].count('1')
if m >= count0 and n >= count1:
take_curr_str = max(take_curr_str, self.helper(strs, idx + 1, m - count0, n - count1, memo) + 1)
not_take_curr_str = self.helper(strs, idx + 1, m, n, memo)
memo[idx][m][n] = max(take_curr_str, not_take_curr_str)
return memo[idx][m][n]``````

### 解法三 - 3D DP

Java code

``````class OnesAndZeros3DDP {
public int findMaxForm(String[] strs, int m, int n) {
int l = strs.length;
int [][][] d = new int[l + 1][m + 1][n + 1];
for (int i = 0; i <= l; i ++){
int [] nums = new int[]{0,0};
if (i > 0){
nums = countZeroOnes(strs[i - 1]);
}
for (int j = 0; j <= m; j ++){
for (int k = 0; k <= n; k ++){
if (i == 0) {
d[i][j][k] = 0;
} else if (j >= nums[0] && k >= nums[1]){
d[i][j][k] = Math.max(d[i - 1][j][k], d[i - 1][j - nums[0]][k - nums[1]] + 1);
} else {
d[i][j][k] = d[i - 1][j][k];
}
}
}
}
return d[l][m][n];
}
}``````

### 解法四 - 2D DP

Java code

``````class OnesAndZeros2DDP {
public int findMaxForm(String[] strs, int m, int n) {
int[][] dp = new int[m + 1][n + 1];
for (String s : strs) {
int[] counts = countZeroOnes(s);
for (int i = m; i >= counts[0]; i--) {
for (int j = n; j >= counts[1]; j--) {
dp[i][j] = Math.max(1 + dp[i - counts[0]][j - counts[1]], dp[i][j]);
}
}
}
return dp[m][n];
}
private int[] countZeroOnes(String s) {
int[] res = new int[2];
for (char ch : s.toCharArray()) {
res[ch - '0']++;
}
return res;
}
}``````

Python3 code

``````class Solution:
def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
l = len(strs)
dp = [[0]*(n+1) for _ in range(m+1)]
for i in range(1, l + 1):
count0, count1 = strs[i - 1].count('0'), strs[i - 1].count('1')
for i in reversed(range(count0, m + 1)):
for j in reversed(range(count1, n + 1)):
dp[i][j] = max(dp[i][j], 1 + dp[i - count0][j - count1])
return dp[m][n]``````

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