【Leetcode】474. Ones and Zeroes
Today, Leet weekly contest was hold on time. However, i was late about 15 minutes for checking out of the hotel.
It seems like every thing gone well. First problem was accepted by my first try. Second problem is not a complex task but has many coding work to solve it.
What makes me feel fool is the third problem which is absolutely a dp problem using two dimensions array. I was solved it by greedy. WTF, the reason of that I think may be I look down upon this problem.
I hope to solve it in easy stpes but I didn't think over it's weakness.
474. Ones and Zeroes
- User Accepted: 171
- User Tried: 359
- Total Accepted: 175
- Total Submissions: 974
- Difficulty: Medium
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
and1s
will both not exceed100
- 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 Solution {
public:
int findMaxForm(vector<string>& strs, int m, int n) {
vector<vector<int>>dp(m + , vector<int>(n + , ));
int len = strs.size();
for(int i = ; i < len; i ++){
int z = , o = ;
for(auto &ch : strs[i]){
if(ch == '') z ++;
else o ++;
}
for(int j = m; j >= z; j --){
for(int k = n; k >= o; k --){
dp[j][k] = max(dp[j][k], dp[j - z][k - o] + );
}
}
}
int ret = ;
for(int i = ; i <= m; i ++)
for(int j = ; j <= n; j ++)
ret = max(ret, dp[i][j]);
return ret;
}
};
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