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  • 474. Ones and Zeroes (Solution 2)

    /**
    474. Ones and Zeroes
    https://leetcode.com/problems/ones-and-zeroes/
    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. 1 <= strs.length <= 600
    2. 1 <= strs[i].length <= 100
    3. strs[i] consists only of digits '0' and '1'.
    4. 1 <= m, n <= 100
    */
    
    pub struct Solution {}
    
    impl Solution {
        /*
        Solution 2: Recursive + Memorization, Time:O(m*n*length), Space:O(m*n*length);
        */
        pub fn find_max_form(strs: Vec<String>, m: i32, n: i32) -> i32 {
            //memo help to save index,m,n
            let M = (m as usize) + 1;
            let N = (n as usize) + 1;
            //M rows * N cols, size of length
            let mut memo = vec![vec![vec![-1; N]; M]; strs.len()];
            Self::help(0, m, n, &strs, memo.as_mut())
            //3D array example: two 3 rows * 4 cols arrays
            //let mut memo = vec![vec![vec![0;4];3];2];
        }
    
        pub fn help(index: i32, m: i32, n: i32, strs: &Vec<String>, memo: &mut Vec<Vec<Vec<i32>>>) -> i32 {
            let length: i32 = strs.len() as i32;
            if index == length {
                return 0;
            }
            let currentValue = memo[index as usize][m as usize][n as usize];
            if (currentValue != -1) {
                return currentValue;
            }
            let mut currentString: String = String::from(&strs[index as usize]);
            let mut zeroCount = Self::countZero(&currentString);
            let mut oneCount = currentString.len() as i32 - zeroCount;
            //m for 0, n for 1
            let mut calculateCurStr = 0;
            if m >= zeroCount && n >= oneCount {
                calculateCurStr = 1 + Self::help(index + 1, m - zeroCount, n - oneCount, &strs, memo);
            }
            let mut calculateOtherStr = Self::help(index + 1, m, n, &strs, memo);
            let res = std::cmp::max(calculateCurStr, calculateOtherStr);
            memo[index as usize][m as usize][n as usize] = res;
            res
        }
    
        fn countZero(str: &String) -> i32 {
            let mut result = 0;
            for c in str.chars() {
                if (c == '0') {
                    result = result + 1;
                }
            }
            result
        }
    }
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  • 原文地址:https://www.cnblogs.com/johnnyzhao/p/15733236.html
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