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".
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]
