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Commit fbcbf2f

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[Function add]
1. Add leetcode solutions with tag amazon.
1 parent 03a2113 commit fbcbf2f

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‎leetcode/212. Word Search II.md

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@@ -198,3 +198,54 @@ class Solution {
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}
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}
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```
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### Amazon session
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* Method 1: dfs
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```Java
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class Solution {
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private char[][] board;
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private int height, width;
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private static final int[][] dir = new int[][]{{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
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public List<String> findWords(char[][] board, String[] words) {
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List<String> result = new ArrayList<>();
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if(board == null || board.length == 0 || board[0].length == 0) return result;
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this.board = board;
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this.height = board.length;
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this.width = board[0].length;
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LABEL:
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for(String word: words){
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for(int i = 0; i < height; i++){
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for(int j = 0; j < width; j++){
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if(word.charAt(0) == board[i][j]){
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Set<Integer> visited = new HashSet<>();
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visited.add(i * width + j);
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if(dfs(i, j, word, 1, visited)){
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result.add(word);
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continue LABEL;
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}
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}
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}
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}
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}
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return result;
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}
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private boolean dfs(int x, int y, String word, int index, Set<Integer> visited){
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if(index == word.length()) return true;
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else if(index < word.length()){
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int tx = 0, ty = 0;
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for(int d = 0; d < 4; d++){
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tx = x + dir[d][0];
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ty = y + dir[d][1];
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if(tx >= 0 && tx < height && ty >= 0 && ty < width
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&& !visited.contains(tx * width + ty)
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&& board[tx][ty] == word.charAt(index)){
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visited.add(tx * width + ty);
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if(dfs(tx, ty, word, index + 1, visited)) return true;
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visited.remove(tx * width + ty);
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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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}
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```

‎leetcode/269. Alien Dictionary.md

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@@ -94,4 +94,51 @@ Note:
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return result.toString();
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}
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}
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```
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```
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### Amazon session
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* Method 1: Topological sort
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```Java
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class Solution {
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public String alienOrder(String[] words) {
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Map<Character, Integer> indegree = new HashMap<>(); // key: letter, value: its indegree
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Map<Character, Set<Character>> map = new HashMap<>(); // key: request, value: its childrens
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for(String word: words){
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for(char c : word.toCharArray()){
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indegree.put(c, 0);
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}
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}
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for(int i = 1; i < words.length; i++){
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int minLen = Math.min(words[i].length(), words[i - 1].length());
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char[] arr1 = words[i - 1].toCharArray(), arr2 = words[i].toCharArray();
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for(int j = 0; j < minLen; j++){
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if(arr1[j] != arr2[j]){ //arr1[j] is ahead of arr2[j]
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Set<Character> set = map.getOrDefault(arr1[j], new HashSet<>());
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if(!set.contains(arr2[j])) indegree.put(arr2[j], indegree.get(arr2[j]) + 1);
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set.add(arr2[j]);
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map.put(arr1[j], set);
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break;
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}
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}
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}
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Queue<Character> q = new LinkedList<>();
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for(Map.Entry<Character, Integer> entry: indegree.entrySet()){
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if(entry.getValue() == 0){
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q.offer(entry.getKey());
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}
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}
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StringBuilder sb = new StringBuilder();
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while(!q.isEmpty()){
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char cur = q.poll();
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sb.append(cur);
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if(!map.containsKey(cur)) continue; // current character doesn't have neighbour.
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Set<Character> neighbours = map.get(cur);
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for(char c : neighbours){
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indegree.put(c, indegree.get(c) - 1);
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if(indegree.get(c) == 0) q.offer(c);
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}
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}
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return sb.length() == indegree.size() ? sb.toString(): "";
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}
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}
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```

‎leetcode/472. Concatenated Words.md

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@@ -62,4 +62,42 @@ Note:
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return dp[len];
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}
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}
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```
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### Amazon session
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* Method 1: DP + Set
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```Java
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class Solution {
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private Set<String> set;
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public List<String> findAllConcatenatedWordsInADict(String[] words) {
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List<String> result = new ArrayList<>();
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if(words == null || words.length <= 1) return result;
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Arrays.sort(words, (a, b)->{
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return a.length() - b.length();
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});
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this.set = new HashSet<>();
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set.add(words[0]);
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for(int i = 1; i < words.length; i++){
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if(dfs(words[i])) result.add(words[i]);
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set.add(words[i]);
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}
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return result;
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}
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private boolean dfs(String cur){
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if(set.isEmpty()) return false;
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int len = cur.length();
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boolean[] dp = new boolean[len + 1];
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dp[0] = true;
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for(int end = 1; end <= len; end++){
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for(int start = 0; start < end; start++){
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String sub = cur.substring(start, end);
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if(set.contains(sub)){
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dp[end] |= dp[start];
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}
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if(dp[end]) break;
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}
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}
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return dp[len];
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}
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}
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```

‎leetcode/895. Maximum Frequency Stack.md

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@@ -87,6 +87,45 @@ Note:
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*/
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```
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### Amazon Session
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* Method 1: two maps + deque
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* Fix some bugs in first try, since all frequencies are countinuous, we don't have to use while loop to find next maxCount.
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```Java
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class FreqStack {
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private int maxCount;
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private Map<Integer, Integer> countMap;
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private Map<Integer, Deque<Integer>> freqMap;
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public FreqStack() {
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this.countMap = new HashMap<>();
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this.freqMap = new HashMap<>();
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}
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public void push(int x) {
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countMap.put(x, countMap.getOrDefault(x, 0) + 1);
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int freq = countMap.get(x);
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maxCount = Math.max(maxCount, freq);
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Deque<Integer> deque = freqMap.getOrDefault(freq, new ArrayDeque<>());
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deque.push(x);
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freqMap.put(freq, deque);
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}
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public int pop() {
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int res = freqMap.get(maxCount).pop();
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countMap.put(res, countMap.get(res) - 1);
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if(freqMap.get(maxCount).size() == 0){
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freqMap.remove(maxCount);
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maxCount--;
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}
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return res;
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}
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}
90122

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/**
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* Your FreqStack object will be instantiated and called as such:
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* FreqStack obj = new FreqStack();
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* obj.push(x);
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* int param_2 = obj.pop();
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*/
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```
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### Reference
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1. [Java - intuition and thought process](https://leetcode.com/problems/maximum-frequency-stack/discuss/289916/Java-intuition-and-thought-process)

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