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‎1293_Shortest Path in a Grid with Obstacles Elimination.java

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// 1.2 O(mnk)/O(mn)
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// https://leetcode.com/problems/shortest-path-in-a-grid-with-obstacles-elimination/discuss/453652/Manhattan-distance-instead-of-normal-goal-check
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class Solution {
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public int shortestPath(int[][] grid, int k) {
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final int m = grid.length, n = grid[0].length;
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final int[][] dir = new int[][]{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
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int[][] dic = new int[m][n];
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int res = Integer.MAX_VALUE;
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Queue<Integer> qx = new LinkedList<>(), qy = new LinkedList<>(), qk = new LinkedList<>(), qd = new LinkedList<>();
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qx.add(0); qy.add(0); qd.add(0); qk.add(k);
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while (!qx.isEmpty()) {
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int x = qx.poll(), y = qy.poll(), rem = qk.poll(), d = qd.poll();
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System.out.println(x + " " + y + " " + rem + " " + d + " ");
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if ((m + n - 2 - (x + y)) <= rem)
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if (d + m + n - 2 - (x + y) < res)
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res = d + m + n - 2 - (x + y);
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for (int i = 0; i < dir.length; i++) {
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int nx = x + dir[i][0], ny = y + dir[i][1];
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if (nx < 0 || nx >= m || ny < 0 || ny >= n) continue;
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int nk = (grid[nx][ny] == 1 ? rem - 1 : rem), nd = d + 1;
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if (nk < 0) continue;
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if (dic[nx][ny] < nk) {
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dic[nx][ny] = nk;
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qx.add(nx); qy.add(ny); qd.add(nd); qk.add(nk);
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}
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}
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}
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return (res == Integer.MAX_VALUE ? -1 : res);
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}
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}
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// 1.0
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// https://leetcode.com/problems/shortest-path-in-a-grid-with-obstacles-elimination/discuss/451796/Java-Straightforward-BFS-O(MNK)-time-or-O(MNK)-space
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class Solution {
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public int shortestPath(int[][] grid, int k) {
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int step = 0, m = grid.length, n = grid[0].length;
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int[][] DIRS = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
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int[][] seen = new int[m][n]; // min obstacles elimination from (0,0) to (x, y)
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for (int i = 0; i < m; i++) {
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Arrays.fill(seen[i], Integer.MAX_VALUE);
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}
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Queue<int[]> q = new LinkedList<>();
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q.offer(new int[]{0, 0, 0});
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seen[0][0] = 0;
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while (!q.isEmpty()) {
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int size = q.size();
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while (size-- > 0) {
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int[] cur = q.poll();
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if (cur[0] == m - 1 && cur[1] == n - 1) {
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return step;
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}
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for (int[] dir : DIRS) {
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int x = dir[0] + cur[0];
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int y = dir[1] + cur[1];
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if (x < 0 || x >= m || y < 0 || y >= n) {
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continue;
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}
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int o = grid[x][y] + cur[2];
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if (o >= seen[x][y] || o > k) {
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continue;
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}
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seen[x][y] = o;
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q.offer(new int[]{x, y, o});
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}
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}
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++step;
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}
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return -1;
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}
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}
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// 1.1
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class Solution {
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public int shortestPath(int[][] grid, int k) {
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int step = 0, m = grid.length, n = grid[0].length;
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int[][] DIRS = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
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int[][] seen = new int[m][n]; // min obstacles elimination from (0,0) to (x, y)
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// for (int i = 0; i < m; i++) {
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// Arrays.fill(seen[i], Integer.MAX_VALUE);
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// }
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Queue<int[]> q = new LinkedList<>();
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q.offer(new int[]{0, 0, k + 1});
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seen[0][0] = 0;
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while (!q.isEmpty()) {
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int size = q.size();
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while (size-- > 0) {
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int[] cur = q.poll();
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// System.out.println(Arrays.toString(cur));
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if (cur[0] == m - 1 && cur[1] == n - 1) {
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return step;
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}
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for (int[] dir : DIRS) {
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int x = dir[0] + cur[0];
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int y = dir[1] + cur[1];
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if (x < 0 || x >= m || y < 0 || y >= n) {
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continue;
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}
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int o = - grid[x][y] + cur[2];
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if (o <= seen[x][y] || o < 0) {
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continue;
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}
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seen[x][y] = o;
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q.offer(new int[]{x, y, o});
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}
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}
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++step;
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}
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return -1;
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}
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}

‎1295_Find Numbers with Even Number of Digits.java

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class Solution {
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public int findNumbers(int[] nums) {
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int cnt = 0;
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for (int num : nums) {
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// System.out.println((Math.log(num) / Math.log(10)) + " " + (int)(Math.log(num) / Math.log(10)) % 2);
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if ((int)(Math.log(num + 0.5) / Math.log(10)) % 2 == 1)
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cnt++;
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}
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return cnt;
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}
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}

‎1305_All Elements in Two Binary Search Trees.java

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// 1.0 in-order traversal both trees and add to result list, sort it. O(nlogn)/O(n)
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// 2.0 two pointers, like 2-way merge sort, add smaller element to result in each step, and in-order traversal.
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// O(n)/O(n)
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public List<Integer> getAllElements(TreeNode root1, TreeNode root2) {
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Deque<TreeNode> stack1 = new LinkedList<>(), stack2 = new LinkedList<>();
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List<Integer> res = new LinkedList<>();
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TreeNode p1 = root1, p2 = root2;
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if (p1 != null) {
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while (p1.left != null) {
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stack1.push(p1);
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p1 = p1.left;
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}
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}
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if (p2 != null) {
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while (p2.left != null) {
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stack2.push(p2);
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p2 = p2.left;
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}
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}
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while (p1 != null || p2 != null) {
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int temp1 = (p1 == null) ? (int)1e9 : p1.val, temp2 = (p2 == null) ? (int)1e9 : p2.val;
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// System.out.println(temp1 + " " + temp2);
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// System.out.println(res);
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if (temp1 > temp2) {
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res.add(p2.val);
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if (p2.right != null) {
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p2 = p2.right;
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while (p2.left != null) {
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stack2.push(p2);
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p2 = p2.left;
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}
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} else {
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p2 = stack2.isEmpty() ? null : stack2.pop();
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}
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} else {
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res.add(p1.val);
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if (p1.right != null) {
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p1 = p1.right;
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while (p1.left != null) {
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stack1.push(p1);
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p1 = p1.left;
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}
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} else {
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p1 = stack1.isEmpty() ? null : stack1.pop();
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}
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}
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}
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return res;
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}
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}
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// 2.01
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class Solution {
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private TreeNode getNext(TreeNode p, Deque<TreeNode> s) {
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while (p.left != null) {
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s.push(p);
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p = p.left;
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}
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return p;
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}
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public List<Integer> getAllElements(TreeNode root1, TreeNode root2) {
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Deque<TreeNode> stack1 = new LinkedList<>(), stack2 = new LinkedList<>();
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List<Integer> res = new LinkedList<>();
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TreeNode p1 = root1, p2 = root2;
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if (p1 != null) p1 = getNext(p1, stack1);
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if (p2 != null) p2 = getNext(p2, stack2);
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while (p1 != null || p2 != null) {
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int temp1 = (p1 == null) ? (int)1e9 : p1.val, temp2 = (p2 == null) ? (int)1e9 : p2.val;
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if (temp1 > temp2) {
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res.add(p2.val);
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if (p2.right != null) {
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p2 = p2.right;
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p2 = getNext(p2, stack2);
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} else {
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p2 = stack2.isEmpty() ? null : stack2.pop();
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}
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} else {
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res.add(p1.val);
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if (p1.right != null) {
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p1 = p1.right;
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p1 = getNext(p1, stack1);
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} else {
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p1 = stack1.isEmpty() ? null : stack1.pop();
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}
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}
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}
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return res;
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}
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}
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//2.02
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class Solution {
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private TreeNode getNext(TreeNode p, Deque<TreeNode> s) {
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while (p.left != null) {
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s.push(p);
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p = p.left;
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}
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return p;
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}
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public List<Integer> getAllElements(TreeNode root1, TreeNode root2) {
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Deque<TreeNode> stack1 = new LinkedList<>(), stack2 = new LinkedList<>();
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List<Integer> res = new LinkedList<>();
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TreeNode p1 = root1, p2 = root2;
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if (p1 != null) p1 = getNext(p1, stack1);
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if (p2 != null) p2 = getNext(p2, stack2);
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while (p1 != null || p2 != null) {
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int temp1 = (p1 == null) ? (int)1e9 : p1.val, temp2 = (p2 == null) ? (int)1e9 : p2.val;
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if (temp1 > temp2) {
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Deque<TreeNode> tmpStack = stack1; stack1 = stack2; stack2 = tmpStack;
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TreeNode tmpp = p1; p1 = p2; p2 = tmpp;
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}
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res.add(p1.val);
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if (p1.right != null) {
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p1 = p1.right;
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p1 = getNext(p1, stack1);
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} else {
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p1 = stack1.isEmpty() ? null : stack1.pop();
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}
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}
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return res;
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}
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}
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//2.1
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// https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/469952/Java-Inorder-traversal-and-store-to-list

‎1306_Jump Game III.java

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// 1.0 bfs, without judge 0 : 0ms, with : 3ms, O(n)/O(n)
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class Solution {
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public boolean canReach(int[] arr, int start) {
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// boolean hasZero = false;
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// for (Integer idx : arr) {
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// if (idx == 0) {
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// hasZero = true;
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// break;
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// }
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// }
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// if (!hasZero) return false;
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boolean[] visited = new boolean[arr.length];
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Queue<Integer> q = new LinkedList<>();
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visited[start] = true;
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q.add(start);
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while (!q.isEmpty()) {
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int cur = q.poll();
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if (arr[cur] == 0) return true;
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int[] next = new int[]{cur + arr[cur], cur - arr[cur]};
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for (int idx : next) {
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if (idx < arr.length && idx >= 0 && !visited[idx]) {
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visited[idx] = true;
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q.add(idx);
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}
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}
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}
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return false;
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}
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}

‎1309_Decrypt String from Alphabet to Integer Mapping.java

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class Solution {
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public String freqAlphabets(String s) {
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int p = 0;
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StringBuilder res = new StringBuilder();
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char[] ch = s.toCharArray();
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while (p < s.length()) {
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if ((ch[p] == '1' || ch[p] == '2') && (p < s.length() - 2) && (ch[p + 2] == '#')) {
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res.append(Character.valueOf((char)('a' + (ch[p] - '0') * 10 + (ch[p + 1] - '0') - 1)));
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p += 3;
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} else {
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res.append(Character.valueOf((char)('a' + ch[p] - '0' - 1)));
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p++;
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}
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}
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return res.toString();
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}
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}

‎1310_XOR Queries of a Subarray.java

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// https://leetcode.com/problems/xor-queries-of-a-subarray/discuss/470787/JavaC%2B%2BPython-Straight-Forward-Solution
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// https://leetcode.com/problems/xor-queries-of-a-subarray/discuss/474351/pythonc%2B%2Bjava-prefix-xor
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class Solution {
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public int[] xorQueries(int[] arr, int[][] queries) {
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for (int i = 1; i < arr.length; i++) {
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arr[i] ^= arr[i - 1];
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}
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int[] res = new int[queries.length];
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for (int i = 0; i < queries.length; i++) {
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int[] q = queries[i];
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res[i] = q[0] > 0 ? arr[q[1]] ^ arr[q[0] - 1] : arr[q[1]];
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}
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return res;
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}
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}
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s[i] = a[0] + a[1] .. a[i]
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s[i] - s[j] = a[j + 1] + a[j + 2] ... a[i]

‎1313_Decompress Run-Length Encoded List.java

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// O(n)/O(n)
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class Solution {
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public int[] decompressRLElist(int[] nums) {
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int len = 0, pt = 0;
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for (int i = 0; i < nums.length; i += 2) len += nums[i];
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int[] res = new int[len];
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for (int i = 0; i < nums.length; i += 2) {
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for (int j = 0; j < nums[i]; j++) {
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res[pt++] = nums[i + 1];
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}
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}
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return res;
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}
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}
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// wrong, can't change orders
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class Solution {
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public int[] decompressRLElist(int[] nums) {
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int[] map = new int[101];
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int len = 0, pt = 0;
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for (int i = 0; i < nums.length; i += 2) {
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map[nums[i + 1]] += nums[i];
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len += nums[i];
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}
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int[] res = new int[len];
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for (int i = 1; i <= 100; i++) {
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for (int j = 0; j < map[i]; j++) {
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res[pt++] = i;
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}
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}
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return res;
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}
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}

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