LeetCode - 565 Array Nesting


https://leetcode.com/problems/array-nesting
A zero-indexed array A consisting of N different integers is given. The array contains all integers in the range [0, N - 1].
Sets S[K] for 0 <= K < N are defined as follows:
S[K] = { A[K], A[A[K]], A[A[A[K]]], ... }.
Sets S[K] are finite for each K and should NOT contain duplicates.
Write a function that given an array A consisting of N integers, return the size of the largest set S[K] for this array.
Example 1:
Input: A = [5,4,0,3,1,6,2]
Output: 4
Explanation: 
A[0] = 5, A[1] = 4, A[2] = 0, A[3] = 3, A[4] = 1, A[5] = 6, A[6] = 2.

One of the longest S[K]:
S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0}
Note:
  1. N is an integer within the range [1, 20,000].
  2. The elements of A are all distinct.
  3. Each element of array A is an integer within the range [0, N-1].
https://discuss.leetcode.com/topic/90538/c-java-clean-code-o-n
The idea is to, start from every number, find circles in those index-pointer-chains, every time you find a set (a circle) mark every number as visited (-1) so that next time you won't step on it again.
    public int arrayNesting(int[] a) {
        int maxsize = 0;
        for (int i = 0; i < a.length; i++) {
            int size = 0;
            for (int k = i; a[k] >= 0; size++) {
                int ak = a[k];
                a[k] = -1; // mark a[k] as visited;
                k = ak;
            }
            maxsize = Integer.max(maxsize, size);
        }

        return maxsize;
    }
https://discuss.leetcode.com/topic/90527/java-o-n-time-o-1-space
public int arrayNesting(int[] nums) {
    int res = 0;
    for (int i=0;i<nums.length;i++) {
        if (nums[i] < 0) continue;//
        int length = 1, val = nums[i];
        while (Math.abs(val) != i) {
            length++;
            val = nums[Math.abs(val)];
            nums[Math.abs(val)] *= -1;//
        }
        res = Math.max(res, length);
    }
    return res;
}
X.
https://www.nowtoshare.com/zh/Article/Index/70565
public int arrayNesting(int[] nums) { HashSet<Integer> visited = new HashSet<>(); int maxLength=0; for(int i=0;i<nums.length;i++){ if(visited.contains(i)) continue; HashSet<Integer> loop =getLoopSize(nums, i); maxLength= Math.max(loop.size(), maxLength); for(int index: loop){ visited.add(index); } } return maxLength; } public HashSet<Integer> getLoopSize(int[] nums,int i){ HashSet<Integer> loop = new HashSet<>(); while(!loop.contains(i)){ loop.add(i); i=nums[i]; } return loop; }

https://discuss.leetcode.com/topic/90527/java-o-n-time-o-1-space
public int arrayNesting(int[] nums) {
    int res = 0;
    for (int i=0;i<nums.length;i++) {
        if (nums[i] < 0) continue;
        int length = 1, val = nums[i];
        while (Math.abs(val) != i) {
            length++;
            val = nums[Math.abs(val)];
            nums[Math.abs(val)] *= -1;
        }
        res = Math.max(res, length);
    }
    return res;
}
https://discuss.leetcode.com/topic/90537/this-is-actually-dfs
This is actually a DFS. Use a visited map to keep track of visited node. If a number is visited before, then the set that starts at this number must be smaller then previous max. So we can safely skip this number. In total it's O(n) complexity.
public int arrayNesting(int[] nums) {
        int max = Integer.MIN_VALUE;
        boolean[] visited = new boolean[nums.length];
        for (int i = 0; i < nums.length; i++) {
         if (visited[i]) 
          continue;
         max = Math.max(max, calcLength(nums, i, visited));
        }
        return max;
    }
 
 private int calcLength(int[] nums, int start, boolean[] visited) {
  int i = start, count = 0;
  while (count == 0 || i != start) {
   visited[i] = true;
   i = nums[i];
   count++;
  }
  return count;
 }

http://bookshadow.com/weblog/2017/05/28/leetcode-array-nesting/
DFS / 并查集
由于A是[0 .. N - 1]的排列,因此输入可以看做顶点集合V = [0 .. N - 1],边集合E = [[i, A[i]] (i ∈   [0 .. N - 1])的有向图
图的形态是一个或者多个O型的环(可以是自环),而不会出现ρ型的环
def arrayNesting(self, nums): """ :type nums: List[int] :rtype: int """ def search(idx): cnt = 0 while nums[idx] >= 0: cnt += 1 next = nums[idx] nums[idx] = -1 idx = next return cnt ans = 0 for x in range(len(nums)): if nums[x] >= 0: ans = max(ans, search(x)) return ans
X. Brute Force
https://discuss.leetcode.com/topic/90542/java-solution-union-find
    public int ArrayNesting(int[] nums) 
    {
        if (nums.Length == 0) return 0;
        
        var global = 1;
        for (int i = 0; i < nums.Length; i++)
        {
            var local = 1;
            while(nums[i] != i)
            {
                var temp = nums[i];
                nums[i] = nums[temp];
                nums[temp] = temp;

                local++;
            }

            global = Math.Max(local, global);
        }

        return global;
    }

LeetCode 599 - Minimum Index Sum of Two Lists


https://leetcode.com/problems/minimum-index-sum-of-two-lists
Suppose Andy and Doris want to choose a restaurant for dinner, and they both have a list of favorite restaurants represented by strings.
You need to help them find out their common interest with the least list index sum. If there is a choice tie between answers, output all of them with no order requirement. You could assume there always exists an answer.
Example 1:
Input:
["Shogun", "Tapioca Express", "Burger King", "KFC"]
["Piatti", "The Grill at Torrey Pines", "Hungry Hunter Steakhouse", "Shogun"]
Output: ["Shogun"]
Explanation: The only restaurant they both like is "Shogun".
Example 2:
Input:
["Shogun", "Tapioca Express", "Burger King", "KFC"]
["KFC", "Shogun", "Burger King"]
Output: ["Shogun"]
Explanation: The restaurant they both like and have the least index sum is "Shogun" with index sum 1 (0+1).
Note:
  1. The length of both lists will be in the range of [1, 1000].
  2. The length of strings in both lists will be in the range of [1, 30].
  3. The index is starting from 0 to the list length minus 1.
  4. No duplicates in both lists.
https://www.nowtoshare.com/zh/Article/Index/70564
public String[] findRestaurant(String[] list1, String[] list2) { HashMap<String, Integer> memo = new HashMap<>(); for(int i=0;i<list1.length; i++){ memo.put(list1[i],i); } List<String> commons = new ArrayList<>(); int minIndex=Integer.MAX_VALUE; for(int i=0; i<list2.length; i++){ if(memo.containsKey(list2[i])){ int idx= memo.get(list2[i])+i; if(idx <minIndex){ commons = new ArrayList<>(); commons.add(list2[i]); minIndex =idx; } else if(idx == minIndex){ commons.add(list2[i]); } } } String[] ret = new String[commons.size()]; for(int i=0;i<ret.length;i++){ ret[i] =commons.get(i); } return ret; }

https://discuss.leetcode.com/topic/90534/java-o-n-m-time-o-n-space
public String[] findRestaurant(String[] list1, String[] list2) {
    Map<String, Integer> map = new HashMap<>();
    List<String> res = new LinkedList<>();
    int minSum = Integer.MAX_VALUE;
    for (int i=0;i<list1.length;i++) map.put(list1[i], i);
    for (int i=0;i<list2.length;i++) {
        Integer j = map.get(list2[i]);
        if (j != null && i + j <= minSum) {
            if (i + j < minSum) { res = new LinkedList<>(); minSum = i+j; }
            res.add(list2[i]);
        }
    }
    return res.toArray(new String[res.size()]);
}
https://discuss.leetcode.com/topic/90545/java-solution-hashmap-s
    public String[] findRestaurant(String[] list1, String[] list2) {
        List<String> result = null;
        Map<String, Integer> map1 = new HashMap<>();
        Map<String, Integer> map2 = new HashMap<>();
        int min = Integer.MAX_VALUE;
        
        for (int i = 0; i < list1.length; i++) {
            map1.put(list1[i], i);
        }
        for (int i = 0; i < list2.length; i++) {
            map2.put(list2[i], i);
        }
        
        for (int i = 0; i < list1.length; i++) {
            if (map2.containsKey(list1[i])) {
                int sum = map1.get(list1[i]) + map2.get(list1[i]);
                if (sum < min) {
                    min = sum;
                    result = new ArrayList<String>();
                    result.add(list1[i]);
                }
                else if (sum == min) {
                    result.add(list1[i]);
                }
            }
        }
        
        String[] res = new String[result.size()];
        for (int i = 0; i < result.size(); i++) {
            res[i] = result.get(i);
        }
        
        return res;
    }


LeetCode 598 - Range Addition II


https://leetcode.com/problems/range-addition-ii
Given an m * n matrix M initialized with all 0's and several update operations.
Operations are represented by a 2D array, and each operation is represented by an array with two positive integers a and b, which means M[i][j] should be added by one for all 0 <= i < a and 0 <= j < b.
You need to count and return the number of maximum integers in the matrix after performing all the operations.
Example 1:
Input: 
m = 3, n = 3
operations = [[2,2],[3,3]]
Output: 4
Explanation: 
Initially, M = 
[[0, 0, 0],
 [0, 0, 0],
 [0, 0, 0]]

After performing [2,2], M = 
[[1, 1, 0],
 [1, 1, 0],
 [0, 0, 0]]

After performing [3,3], M = 
[[2, 2, 1],
 [2, 2, 1],
 [1, 1, 1]]

So the maximum integer in M is 2, and there are four of it in M. So return 4.
Note:
  1. The range of m and n is [1,40000].
  2. The range of a is [1,m], and the range of b is [1,n].
  3. The range of operations size won't exceed 10,000.
https://discuss.leetcode.com/topic/90547/java-solution-find-min
思路1: brute force - O(m * n * k), k = number of ops
思路2: 从 ops 中找到被操作次数最多的 row 和 column,这可以通过遍历 ops 数组,分别找到 row 和 column 的最小值来完成
You don't need the first if statement if you init row = m and col = n :)
    public int maxCount(int m, int n, int[][] ops) {
        if (ops == null || ops.length == 0) {
            return m * n;
        }
        
        int row = Integer.MAX_VALUE, col = Integer.MAX_VALUE;
        for(int[] op : ops) {
            row = Math.min(row, op[0]);
            col = Math.min(col, op[1]);
        }
        
        return row * col;
    }
https://discuss.leetcode.com/topic/90540/c-java-clean-code
https://www.nowtoshare.com/zh/Article/Index/70563
    public int maxCount(int m, int n, int[][] ops) {
        for (int[] op : ops) {
            m = Math.min(op[0], m);
            n = Math.min(op[1], n);
        }
        return m * n;        
    }

LeetCode 588 - Design In-Memory File System


http://bookshadow.com/weblog/2017/05/21/leetcode-design-in-memory-file-system/
Design an in-memory file system to simulate the following functions:
ls: Given a path in string format. If it is a file path, return a list that only contains this file's name. If it is a directory path, return the list of file and directory names in this directory. Your output (file and directory names together) should in lexicographic order.
mkdir: Given a directory path that does not exist, you should make a new directory according to the path. If the middle directories in the path don't exist either, you should create them as well. This function has void return type.
addContentToFile: Given a file path and file content in string format. If the file doesn't exist, you need to create that file containing given content. If the file already exists, you need to append given content to original content. This function has void return type.
readContentFromFile: Given a file path, return its content in string format.
Example:
Input: 
["FileSystem","ls","mkdir","addContentToFile","ls","readContentFromFile"]
[[],["/"],["/a/b/c"],["/a/b/c/d","hello"],["/"],["/a/b/c/d"]]
Output:
[null,[],null,null,["a"],"hello"]
Explanation:
filesystem
Note:
  1. You can assume all file or directory paths are absolute paths which begin with / and do not end with / except that the path is just "/".
  2. You can assume that all operations will be passed valid parameters and users will not attempt to retrieve file content or list a directory or file that does not exist.
  3. You can assume that all directory names and file names only contain lower-case letters, and same names won't exist in the same directory.

题目大意:

设计内存文件系统模拟实现如下功能:
ls:给定路径字符串,若对应一个目录,则输出其中包含的目录和文件(字典序);若对应一个文件,则只输出该文件名
mkdir:创建目录,若目录不存在,则递归创建缺少的目录
addContentToFile:在文件中追加内容,若文件不存在,则新建
readContentFromFile:从文件中读取内容并返回
注意:
  1. 你可以假设文件和目录的路径均为绝对路径,以/开始,并且结尾不包含/,除非路径就是"/"本身
  2. 你可以假设所有操作均合法,用户不会常识读取一个不存在的文件,或者ls一个不存在的目录
  3. 你可以假设所有目录名称和文件名称都只包含小写字母,并且在同一目录下不会存在同名的目录或者文件
树形结构(Tree)
目录节点Node包含两个字段dirs和files,分别存储其中包含的子目录节点和文件内容
X. Use TreeMap to contain children nodes - so they are sorted.
https://discuss.leetcode.com/topic/90016/file-system-as-a-tree-java
    private class TreeNode implements Comparable<TreeNode> {
        String label;
        boolean isDir;
        Map<String, TreeNode> dirContents = new TreeMap<>();
        String fileContents;
        
        public int compareTo(TreeNode other) {
            return label.compareTo(other.label);
        }
        
        public int hashCode() {
            return Objects.hash(label);
        }
        
        public boolean equals(Object treeNode) {
            TreeNode other = (TreeNode) treeNode;
            return Objects.equals(label, other.label);
        }
    }    
    
    private TreeNode sentinel = new TreeNode();

    public FileSystem() {
        TreeNode root = new TreeNode();
        root.label = "";
        root.isDir = true;
        sentinel.dirContents.put("", root);
    }
    
    public List<String> ls(String path) {
        String[] labels = path.split("/");
        TreeNode curr = (labels.length != 0) ? sentinel : sentinel.dirContents.get("");
        List<String> result = new ArrayList<>();
        for(String label : labels) {
            curr = curr.dirContents.get(label);
            System.out.println(">>" + curr);
        }
        if(curr.isDir) {
            for(String label : curr.dirContents.keySet()) {
                result.add(label);
            }
        }
        else {
            result.add(curr.label);
        }
        
        return result;
    }
    
    public void mkdir(String path) {
        String[] labels = path.split("/");
        TreeNode curr = sentinel;
        int currIndex = 0;
        for(String label : labels) {
            TreeNode next = curr.dirContents.get(label);
            if(next == null) {
                break;
            }
            currIndex++;
            curr = next;
        }
        
        for(int i = currIndex; i < labels.length; i++) {
            TreeNode newDir = new TreeNode();
            newDir.isDir = true;
            newDir.label = labels[i];
            curr.dirContents.put(labels[i], newDir);
            
            curr = newDir;
        }
    }
    
    public void addContentToFile(String filePath, String content) {
        String[] labels = filePath.split("/");
        TreeNode curr = sentinel, prev = null;
        
        for(String label : labels) {
            prev = curr;
            if(curr == null) {
                break;
            }
            curr = curr.dirContents.get(label);
        }
        
        if(curr == null) {
            curr = new TreeNode();
            curr.isDir = false;
            curr.label = labels[labels.length - 1];
            curr.fileContents = content;
            prev.dirContents.put(curr.label, curr);
        }
        else {
            curr.fileContents = curr.fileContents == null ? content : curr.fileContents + content;
        }
    }
    
    public String readContentFromFile(String filePath) {
        String[] labels = filePath.split("/");
        TreeNode curr = sentinel;
        
        for(String label : labels) {
            curr = curr.dirContents.get(label);
        }
        
        return (curr.fileContents == null) ? "" : curr.fileContents;
    }

X. https://discuss.leetcode.com/topic/90000/java-solution-file-class
https://discuss.leetcode.com/topic/90004/file-system-as-graph
    class File {
        boolean isFile = false;
        Map<String, File> children = new HashMap<>();
        String content = "";
    }
    
    File root = null;
    
    public FileSystem() {
        root = new File();
    }
    
    public List<String> ls(String path) {
        String[] dirs = path.split("/");
        File node = root;
        List<String> result = new ArrayList<>();
        String name = "";
        for (String dir : dirs) {
            if (dir.length() == 0) continue;
            if (!node.children.containsKey(dir)) {
                return result;
            }
            node = node.children.get(dir);
            name = dir;
        }
        
        if (node.isFile) {
            result.add(name);
        }
        else {
            for (String key : node.children.keySet()) {
                result.add(key);
            }
        }
        
        Collections.sort(result);
        
        return result;
    }
    
    public void mkdir(String path) {
        String[] dirs = path.split("/");
        File node = root;
        for (String dir : dirs) {
            if (dir.length() == 0) continue;
            if (!node.children.containsKey(dir)) {
                File file = new File();
                node.children.put(dir, file);
            }
            node = node.children.get(dir);
        }
    }
    
    public void addContentToFile(String filePath, String content) {
        String[] dirs = filePath.split("/");
        File node = root;
        for (String dir : dirs) {
            if (dir.length() == 0) continue;
            if (!node.children.containsKey(dir)) {
                File file = new File();
                node.children.put(dir, file);
            }
            node = node.children.get(dir);
        }
        node.isFile = true;
        node.content += content;
    }
    
    public String readContentFromFile(String filePath) {
        String[] dirs = filePath.split("/");
        File node = root;
        for (String dir : dirs) {
            if (dir.length() == 0) continue;
            if (!node.children.containsKey(dir)) {
                File file = new File();
                node.children.put(dir, file);
            }
            node = node.children.get(dir);
        }

        return node.content;
    }

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