[GangBean] Week 8

design-add-and-search-words-data-structure/GangBean.java

@@ -0,0 +1,47 @@
+class WordDictionary {
+    private Map<Character, WordDictionary> children;
+    private boolean isEnd;
+
+    public WordDictionary() {
+        this.children = new HashMap<>();
+        this.isEnd = false;
+    }
+
+    public void addWord(String word) {
+        char[] arr = word.toCharArray();
+        WordDictionary next = this;
+        for (int i = 0; i < arr.length; i++) {
+            char c = arr[i];
+            next.children.putIfAbsent(c, new WordDictionary());
+            next = next.children.get(c);
+        }
+        next.isEnd = true;
+    }
+
+    public boolean search(String word) {
+        // System.out.println(this);
+        return search(word, 0);
+    }
+
+    private boolean search(String word, int idx) {
+        if (idx == word.length()) return this.isEnd;
+        char c = word.charAt(idx);
+        if (c == '.') {
+            return this.children.values().stream().anyMatch(child -> child.search(word, idx+1));
+        }
+        if (!this.children.containsKey(c)) return false;
+        return this.children.get(c).search(word, idx+1);
+    }
+
+    public String toString() {
+        return String.format("%s -> %s", isEnd, this.children);
+    }
+}
+
+/**
+ * Your WordDictionary object will be instantiated and called as such:
+ * WordDictionary obj = new WordDictionary();
+ * obj.addWord(word);
+ * boolean param_2 = obj.search(word);
+ */
+

longest-increasing-subsequence/GangBean.java

@@ -0,0 +1,22 @@
+class Solution {
+    /**
+    1. understanding
+    - dp[n] : length of longest increasing subsequence in 0...n
+    2. space
+    - time: O(N^2)
+    - space: O(N)
+     */
+    public int lengthOfLIS(int[] nums) {
+        int[] dp = new int[nums.length];
+        Arrays.fill(dp, 1);
+        for (int i = 0; i < nums.length; i++) { // O(N)
+            for (int j = 0; j <= i; j++) { // O(N)
+                if (nums[j] < nums[i]) {
+                    dp[i] = Math.max(dp[j]+1, dp[i]);
+                }
+            }
+        }
+        return Arrays.stream(dp).max().orElse(1);
+    }
+}
+

longest-substring-without-repeating-characters/GangBean.java

@@ -0,0 +1,38 @@
+class Solution {
+    /**
+    1. understanding
+    - [a] -> [a,b] -> [a,b,c] -> [a] -> [a,b] -> [a,b,c] -> [b] -> [b]
+    2. complexity
+    - time: O(N)
+    - space: O(N)
+     */
+    public int lengthOfLongestSubstring(String s) {
+        int left = 0;
+        int right = 0;
+        int len = s.length();
+        HashSet charSet = new HashSet();
+        int ret = (len==0)?0:1;
+        while (left<len && right<len) {
+            if (left == right) {
+                right++;
+                charSet = new HashSet();
+                charSet.add(s.charAt(left));
+            } else {
+                if (!charSet.contains(s.charAt(right))) {
+                    charSet.add(s.charAt(right));
+                    int tmpLen = right - left + 1;
+                    if (tmpLen > ret) {
+                        ret = tmpLen;
+                    }
+                    right++;
+                } else {
+                    left++;
+                    right = left;
+                }
+            }
+        }
+
+        return ret;
+    }
+}
+

number-of-1-bits/GangBean.java

@@ -0,0 +1,11 @@
+class Solution {
+    public int hammingWeight(int n) {
+        int count = 0;
+        while (n >= 1) {
+            count += n % 2;
+            n /= 2;
+        }
+        return count;
+    }
+}
+

number-of-islands/GangBean.java

@@ -0,0 +1,38 @@
+class Solution {
+    /**
+    1. understanding
+    - for each grid cell, if it is land, then move horizontally and vertically, to find connecting lands
+    2. complexity
+    - time: O(N * M) where grid is N * M matrix
+    - space: O(1)
+     */
+    public int numIslands(char[][] grid) {
+        int ret = 0;
+        for (int i=0; i<grid.length; i++) {
+            for (int j=0; j<grid[0].length; j++) {
+                if (grid[i][j] == '1') {
+                    findIsland(grid,i,j,'.');
+                    ret++;
+                }
+            }
+        }
+        return ret;
+    }
+
+    private void findIsland(char[][] grid, int i, int j, char color) {
+        if (grid[i][j] == '1') {
+            grid[i][j] = color;
+        }
+        int[] dx = {0,1,0,-1};
+        int[] dy = {1,0,-1,0};
+
+        for (int dir=0; dir<4; dir++) {
+            int nx = i + dx[dir];
+            int ny = j + dy[dir];
+            if (0<= nx&&nx<grid.length && 0<=ny&&ny<grid[0].length && grid[nx][ny]=='1') {
+                findIsland(grid,nx,ny,color);
+            }
+        }
+    }
+}
+

reverse-linked-list/GangBean.java

@@ -0,0 +1,25 @@
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode() {}
+ *     ListNode(int val) { this.val = val; }
+ *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
+ * }
+ */
+class Solution {
+    public ListNode reverseList(ListNode head) {
+        ListNode curr = head;
+        ListNode prev = null;
+        while (curr != null) {
+            ListNode next = curr.next;
+            curr.next = prev;
+            prev = curr;
+            curr = next;
+        }
+
+        return prev;
+    }
+}
+

set-matrix-zeroes/GangBean.java

@@ -0,0 +1,40 @@
+class Solution {
+    /**
+    1. understanding
+    - iterate over all cells, if value is 0, then add row and col to sweep target set.
+    - for each row target set, and col target set, sweep all it's value to 0
+    2. complexity
+    - time: O(m * n)
+    - space: O(m + n)
+     */
+    public void setZeroes(int[][] matrix) {
+        Set<Integer> rows = new HashSet<>(); // O(m)
+        Set<Integer> cols = new HashSet<>(); // O(n)
+
+        for (int row = 0; row < matrix.length; row++) { // O(m)
+            for (int col = 0; col < matrix[row].length; col++) { // O(n)
+                if (matrix[row][col] == 0) { // O(m * n)
+                    rows.add(row);
+                    cols.add(col);
+                }
+            }
+        }
+
+        for (int row: rows) { // O(m)
+            int col = 0;
+            while (col < matrix[row].length) { // O(n)
+                matrix[row][col] = 0;
+                col++;
+            }
+        }
+
+        for (int col: cols) { // O(n)
+            int row = 0;
+            while (row < matrix.length) { // O(m)
+                matrix[row][col] = 0;
+                row++;
+            }
+        }
+    }
+}
+

spiral-matrix/GangBean.java

@@ -0,0 +1,37 @@
+class Solution {
+    /**
+    1. understanding
+    - 3 x 3 = N x M
+    - (1,2,3) -> (6,9) // (8,7) -> (4) // (5)
+    - upper: step M count, N -= 1 // right: step N count, M -= 1 // bottom: step M count, N -= 1 // left: step N count, M -= 1
+    2. complexity:
+    - time: O(N * M)
+    - space: O(1)
+    */
+    public List<Integer> spiralOrder(int[][] matrix) {
+        int r = 0;
+        int c = -1;
+        int dir = 1;
+        int N = matrix.length;
+        int M = matrix[0].length;
+        List<Integer> ret = new ArrayList<>();
+
+        while (0 < N && 0 < M) {
+            for (int i = 0; i < M; i++) {
+                c += dir;
+                ret.add(matrix[r][c]);
+            }
+            N -= 1;
+            for (int i = 0; i < N; i++) {
+                r += dir;
+                ret.add(matrix[r][c]);
+            }
+            M -= 1;
+
+            dir *= -1;
+        }
+
+        return ret;
+    }
+}
+

unique-paths/GangBean.java

@@ -0,0 +1,24 @@
+import java.math.BigDecimal;
+
+class Solution {
+    /**
+    1. understanding
+    - To reach destination, you have to move bottom direction in m-1 times, and move to right direction in n-1 times.
+    - in example 2, [(DDR), (DRD), (RDD)] is the paths.
+    - so, the number of paths are combinations of (m-1) D and (n-1) R
+    - (m+n-2)!/(m-1)!(n-1)!, where ! means factorial, n! = 1*2*...*n
+    - factorial[n]: n!
+    2. complexity
+    - time: O(m+n)
+    - space: O(m+n)
+     */
+    public int uniquePaths(int m, int n) {
+        BigDecimal[] dp = new BigDecimal[m+n];
+        Arrays.fill(dp, BigDecimal.ONE);
+        for (int num = 2; num < m+n; num++) {
+            dp[num] = dp[num-1].multiply(new BigDecimal(num));
+        }
+        return dp[m+n-2].divide(dp[m-1]).divide(dp[n-1]).intValue();
+    }
+}
+