[Algorithm] 474. Ones and Zeroes

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:

  1. The given numbers of 0s and 1s will both not exceed 100
  2. 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".
/**
 * @param {string[]} strs
 * @param {number} m
 * @param {number} n
 * @return {number}
 */
var findMaxForm = function(strs, m, n) {
    if(!strs || strs.length === 0) return 0;

    let dp = [...Array(m+1)].map(_ => Array(n+1).fill(0));

    strs.forEach(str=>{
        const [zero, one] = calculateBinary(str);

        for(let i = m; i >= zero; i--){
            for(let j = n ; j >= one; j--){
                dp[i][j] = Math.max(dp[i][j],dp[i - zero][j - one] + 1);
            }
        }
    })

    return dp[m][n]
};

var calculateBinary = function(str){
    let zero = 0, one = 0;
    for(let i = 0 ; i < str.length ; i++){
        if(str[i] === "1")  one++;
        else    zero++;
    }

    return [zero, one];
}

原文地址:https://www.cnblogs.com/Answer1215/p/12334209.html

时间: 2025-01-15 08:00:49

[Algorithm] 474. Ones and Zeroes的相关文章

474. Ones and Zeroes

本周继续练习动态规划的相关题目. 题目: 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 s

过中等难度题目.0310

  .   8  String to Integer (atoi)    13.9% Medium   . 151 Reverse Words in a String      15.7% Medium     . 288 Unique Word Abbreviation      15.8% Medium     . 29 Divide Two Integers      16.0% Medium     . 166 Fraction to Recurring Decimal      17.

继续过中等难度.0309

  .   8  String to Integer (atoi)    13.9% Medium   . 151 Reverse Words in a String      15.7% Medium     . 288 Unique Word Abbreviation      15.8% Medium     . 29 Divide Two Integers      16.0% Medium     . 166 Fraction to Recurring Decimal      17.

LeetCode Problems List 题目汇总

No. Title Level Rate 1 Two Sum Medium 17.70% 2 Add Two Numbers Medium 21.10% 3 Longest Substring Without Repeating Characters Medium 20.60% 4 Median of Two Sorted Arrays Hard 17.40% 5 Longest Palindromic Substring Medium 20.70% 6 ZigZag Conversion Ea

Leetcode problems classified by company 题目按公司分类(Last updated: October 2, 2017)

Sorted by frequency of problems that appear in real interviews.Last updated: October 2, 2017Google (214)534 Design TinyURL388 Longest Absolute File Path683 K Empty Slots340 Longest Substring with At Most K Distinct Characters681 Next Closest Time482

[Algorithm] 283. Move Zeroes

Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements. Example: Input: [0,1,0,3,12] Output: [1,3,12,0,0] Note: You must do this in-place without making a copy of the array

Leetcode之动态规划(DP)专题-474. 一和零(Ones and Zeroes)

在计算机界中,我们总是追求用有限的资源获取最大的收益. 现在,假设你分别支配着 m 个 0 和 n 个 1.另外,还有一个仅包含 0 和 1 字符串的数组. 你的任务是使用给定的 m 个 0 和 n 个 1 ,找到能拼出存在于数组中的字符串的最大数量.每个 0 和 1 至多被使用一次. 注意: 给定 0 和 1 的数量都不会超过 100. 给定字符串数组的长度不会超过 600. 示例 1: 输入: Array = {"10", "0001", "11100

Strassen algorithm(O(n^lg7))

Let A, B be two square matrices over a ring R. We want to calculate the matrix product C as {\displaystyle \mathbf {C} =\mathbf {A} \mathbf {B} \qquad \mathbf {A} ,\mathbf {B} ,\mathbf {C} \in R^{2^{n}\times 2^{n}}} If the matrices A, B are not of ty

Light oj 1138 - Trailing Zeroes (III) 【二分查找 &amp;&amp; N!中末尾连续0的个数】

1138 - Trailing Zeroes (III) PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB You task is to find minimal natural number N, so that N! contains exactly Q zeroes on the trail in decimal notation. As you know N! = 1*2*...*N. F