Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2). Example: Given matrix = [ [3, 0, 1, 4, 2], [5, 6, 3, 2, 1], [1, 2, 0, 1, 5], [4, 1, 0, 1, 7], [1, 0, 3, 0, 5] ] sumRegion(2, 1, 4, 3) -> 8 sumRegion(1, 1, 2, 2) -> 11 sumRegion(1, 2, 2, 4) -> 12 Note: You may assume that the matrix does not change.

Given two arrays, write a function to compute their intersection. Example: Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2]. Note: Each element in the result should appear as many times as it shows in both arrays. The result can be in any order. Follow up: What if the given array is already sorted? How would you optimize your algorithm? What if nums1's size is small compared to nums2's size?

Given two arrays, write a function to compute their intersection. Example: Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2]. Note: Each element in the result must be unique. The result can be in any order. Solution: public class Solution { public int[] intersection(int[] nums1, int[] nums2) { Set<Integer> set1 = new HashSet<>(); for(int n : nums1) set1.add(n); Set<Integer> set2 = new HashSet<>(); for(int n :

Given a list of non negative integers, arrange them such that they form the largest number. For example, given [3, 30, 34, 5, 9], the largest formed number is 9534330. Note: The result may be very large, so you need to return a string instead of an integer. Solution: public class Solution { public String largestNumber(int[] nums) { String[] strs = new String[nums.length]; for(int i=0; i<nums.length; i++) strs[i] = ""+nums[i]; Arrays.sort(strs, new Comparator<String>(){ @Override public int compare(String a, String b){ String sA = a + b; String sB = b + a; return sA.compareTo(sB); } }); StringBuilder sb = new StringBuilder(); for(int i=strs.length-1; i>=0; i--){ sb.append(strs[i]); } return strs[strs.length-1].equals("0") ?

Implement a trie with insert, search, and startsWith methods. Solution: class TrieNode { boolean isLeaf; TrieNode[] next; // Initialize your data structure here. public TrieNode() { isLeaf = false; next = new TrieNode[26]; } } public class Trie { private TrieNode root; public Trie() { root = new TrieNode(); } // Inserts a word into the trie. public void insert(String word) { TrieNode cur = root; for(char c : word.toCharArray()){ if(cur.next[c-'a']==null){ cur.next[c-'a'] = new TrieNode(); } cur = cur.next[c-'a']; } cur.isLeaf = true; } // Returns if the word is in the trie.

Design a data structure that supports the following two operations: void addWord(word) bool search(word) search(word) can search a literal word or a regular expression string containing only letters a-z or .. A . means it can represent any one letter. For example: addWord("bad") addWord("dad") addWord("mad") search("pad") -> false search("bad") -> true search(".ad") -> true search("b..") -> true Note: You may assume that all words are consist of lowercase letters a-z. click to show hint.

Given a 2d grid map of ‘1’s (land) and ‘0’s (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water. Example 1: 11110 11010 11000 00000 Answer: 1 Example 2: 11000 11000 00100 00011 Answer: 3 Solution: public class Solution { public int numIslands(char[][]

Given two words (beginWord and endWord), and a dictionary’s word list, find the length of shortest transformation sequence from beginWord to endWord, such that: Only one letter can be changed at a time Each intermediate word must exist in the word list For example, Given: beginWord = “hit” endWord = “cog” wordList = [“hot”,“dot”,“dog”,“lot”,“log”] As one shortest transformation is “hit” -> “hot” -> “dot” -> “dog” -> “cog”, return its length 5.

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. The update(i, val) function modifies nums by updating the element at index i to val. Example: Given nums = [1, 3, 5] sumRange(0, 2) -> 9 update(1, 2) sumRange(0, 2) -> 8 Note: The array is only modifiable by the update function. You may assume the number of calls to update and sumRange function is distributed evenly.

Given two integers representing the numerator and denominator of a fraction, return the fraction in string format. If the fractional part is repeating, enclose the repeating part in parentheses. For example, Given numerator = 1, denominator = 2, return "0.5". Given numerator = 2, denominator = 1, return "2". Given numerator = 2, denominator = 3, return "0.(6)". Hint: No scary math, just apply elementary math knowledge. Still remember how to perform a long division?