Leetcode Blog Post Algorithms

package leetcode.blog;import java.util.ArrayList;import java.util.Arrays;public class LeetcodeBlog {// Q1 Find Intersection of two Sorted Arrays// if either A or B 's length is too big, try to use Binary Search should be// a good idea, which is O(mlgn)// this method is O(m + n)public static ArrayList<Integer> findIntersection(int[] A, int[] B) {ArrayList<Integer> res = new ArrayList<Integer>();if (A.length == 0 || B.length == 0)return res;for (int i = 0, j = 0; i < A.length && j < B.length;) {if (A[i] == B[j]) {res.add(A[i]);i++;j++;} else if (A[i] < B[j])i++;elsej++;}return res;}// Q2 Reverse Linked List both Iteratively and Recursivelypublic static LinkedListNode reverseLinkedListIteratively(LinkedListNode head) {if (head == null)return head;LinkedListNode res = new LinkedListNode(0);LinkedListNode next;while (head != null) {next = head.next;head.next = res.next;res.next = head;head = next;}return res.next;}public static LinkedListNode reverseLinkedListRecursively(LinkedListNode head) {if (head == null)return head;if (head.next == null)return head;LinkedListNode newHead = reverseLinkedListRecursively(head.next);head.next.next = head;head.next = null;return newHead;}// Q3 Determine if a point is inside a rectangle/irregular polygon// Rectangle ABCD and Point P// 1. Calculate the sum of areas of all triangles. ABP, BCP, CDP, DAP.// if the sum is larger than the area of the rectangle, P is outside the// rectangle// if the sum is smaller than the area of the rectangle, P is inside the// rectangle// if equal, it is on one side line.// 2. Calculate the perpendicular distances of P from all the 4 lines// Q4 Rotating an array in placepublic static void rotateArray(int[] array, int k) {reverse(array, 0, array.length - 1);reverse(array, 0, k - 1);reverse(array, k, array.length - 1);}public static void reverse(int[] array, int left, int right) {int temp;while (left < right) {temp = array[left];array[left] = array[right];array[right] = temp;left++;right--;}}// Q5 Multiplication of Numbers/* * There is an array A[N] of N numbers. You have to compose an array * Output[N] such that Output[i] will be equal to multiplication of all the * elements of A[N] except A[i]. Solve it without division operator and in * O(n). For example Output[0] will be multiplication of A[1] to A[N-1] and * Output[1] will be multiplication of A[0] and from A[2] to A[N-1]. * Example: A: {4, 3, 2, 1, 2} OUTPUT: {12, 16, 24, 48, 24} */public static int[] multiplyNumbers(int[] num) {int n = num.length;int[] output = new int[n];if (n == 0)return output;Arrays.fill(output, 1);int left = 1, right = 1;for (int i = 0; i < n; i++) {output[i] *= left;output[n - i - 1] *= right;left *= num[i];right *= num[n - 1 - i];}return output;}/* * Q6 Generate a prime list from 0 up to n, using The Sieve of Erantosthenes * param n The upper bound of the prime list (including n) param prime[] An * array of truth value whether a number is prime */public static ArrayList<Integer> generatePrime(int n) {boolean[] prime = new boolean[n + 1];if (n < 2)return new ArrayList<Integer>();Arrays.fill(prime, true);prime[0] = false;prime[1] = false;int limit = (int) Math.sqrt((double) n);for (int i = 2; i <= limit; i++) {if (prime[i]) {for (int j = i * i; j <= n; j += i)prime[j] = false;}}ArrayList<Integer> res = new ArrayList<Integer>();for (int i = 0; i <= n; i++)if (prime[i])res.add(i);return res;}/* * Q7 Given a string of lowercase characters, reorder them such that the * same characters are at least distance d from each other. Input: { a, b, b * }, distance = 2 Output: { b, a, b } */public static int findMax(int freqeuncy[], boolean excep[]) {int max_i = -1;int max = -1;for (int i = 0; i < 26; i++) {if (!excep[i] && freqeuncy[i] > 0 && freqeuncy[i] > max) {max = freqeuncy[i];max_i = i;}}return max_i;}public static char[] keepDis(char[] str, int d) {int n = str.length;char[] res = new char[n];int freqeuncy[] = new int[26];for (int i = 0; i < n; i++)freqeuncy[str[i] - 'a']++;int distance[] = new int[26];boolean excep[] = new boolean[26];for (int i = 0; i < n; i++) {Arrays.fill(excep, false);boolean done = false;while (!done) {int j = findMax(freqeuncy, excep);if (j == -1) {System.out.println("Invalid inputs.");return res;}excep[j] = true;if (distance[j] <= 0) {res[i] = (char) (j + 'a');freqeuncy[j]--;distance[j] = d;done = true;}}for (int k = 0; k < 26; k++)distance[k]--;}return res;}/* * Q8 Check whether an integer is a power of two. */boolean mystery(int x) {// bug is when x == 0return (x & (x - 1)) == 0;}boolean mysteryZero(int x) {return x != 0 && (x & (x - 1)) == 0;}/* * Q9 Number of 1 bits */public static int countOneBitsCommon(int x) {int counter = 0;while ((x & 0x1) != 0) {counter++;x = x >> 1;}return counter;}public static int countOneBits(int x) {int counter = 0;while (x != 0) {counter++;x = x & (x - 1);}return counter;}/* * Q10 * Pre-order BST constructor */public static int index = 0;public static TreeNode preorderConstructBST(int[] array, int min, int max){if(index < array.length && array[index] < max && array[index] > min){TreeNode root = new TreeNode(array[index++]);root.left = preorderConstructBST(array, min, root.val);root.right = preorderConstructBST(array, root.val, max);return root;}return null;}/* * Q11 * Deserialize a Binary Tree (any tree has an identical binary tree representation) */public static void serialize(TreeNode root, ArrayList<Integer> array){if(root == null)array.add(-1);else{array.add(root.val);serialize(root.left, array);serialize(root.right, array);}}/* * Q12 * Excel sheet row numbers */// start from 0public static String convert(int row){StringBuffer sb = new StringBuffer("");sb.append((char)('a' + row % 26));row /= 26;while(row != 0){int index = (row - 1) % 26;row = (row - 1) / 26;sb.insert(0, (char)(index + 'a'));}return sb.toString();}// start from 1public static String convert2(int row){StringBuffer sb = new StringBuffer("");while(row != 0){int index = row % 26;row /= 26;sb.insert(0, (char)(index + 'a' - 1));}return sb.toString();}// Q13 Search Young Tableau// O(n)public static boolean stepwise(int[][] table, int target){if(table.length == 0)return false;if(table[0].length == 0)return false;int row = table.length;int col = table[0].length;if(target < table[0][0] || target > table[row - 1][col - 1])return false;for(int i = 0, j = 0; i < row && j >= 0;){if(table[i][j] == target)return true;else if(table[i][j] > target)j--;elsei++;}return false;}// O(n^1.58)// row number ranges from top to bottom, column number ranges from left to rightpublic static boolean quadPartition(int[][] table, int target, int left, int top, int right, int bottom){if(left > right || top > bottom)return false;if(target < table[top][left] || target > table[bottom][right])return false;int col = (right + left) / 2;int row = (top + bottom) / 2;if(table[row][col] == target)return true;else if (left == right && top == bottom)return false;if(table[row][col] > target){// upper left || upper right || bottom leftreturn quadPartition(table, target, left, top, col, row) ||quadPartition(table, target, col + 1, top, right, row) ||quadPartition(table, target, left, row + 1, col, bottom);} else { // upper right || bottom right || bottom leftreturn quadPartition(table, target, col + 1, top, right, bottom) ||quadPartition(table, target, col + 1, row + 1, right, bottom) ||quadPartition(table, target, left, row + 1, col, bottom);}}// do search on mid column, and then search on upper right side and bottom left side// if two sub-matrices are equal sizes, then O(n)public static boolean binaryPartition(int[][] table, int target, int left, int top, int right, int bottom){if(left > right || top > bottom)return false;if(target < table[top][left] || target > table[bottom][right])return false;int mid = (left + right) / 2;int row = top;// Could use binary search here which reduce complexity to O(lgn * lgn)while(row <= bottom && table[row][mid] <= target){if(table[row][mid] == target)return true;row++;}// upper right || bottom leftreturn binaryPartition(table, target, mid + 1, top, top, row - 1) ||binaryPartition(table, target, left, row, mid - 1, bottom);}// Q14 Print boundary edges of a binary treepublic static ArrayList<Integer> printEdges(TreeNode root){ArrayList<Integer> res = new ArrayList<Integer>();if(root == null)return res;res.add(root.val);printLeftBottomEdges(root.left, true, res);printBottomRightEdges(root.right, true, res);return res;}public static void printLeftBottomEdges(TreeNode root, boolean print, ArrayList<Integer> res){if(root == null)return;if(print || (root.left == null && root.right == null))res.add(root.val);printLeftBottomEdges(root.left, print, res);printLeftBottomEdges(root.right, print && root.left == null ? true : false, res);}public static void printBottomRightEdges(TreeNode root, boolean print, ArrayList<Integer> res){if(root == null)return;printBottomRightEdges(root.left, print && root.right == null ? true : false, res);printBottomRightEdges(root.right, print, res);if(print || (root.left == null && root.right == null))res.add(root.val);}// Q15public static String Q15replace(String str, String pattern){if(str == null || pattern == null || pattern.length() > str.length())return str;StringBuilder sb = new StringBuilder("");int lastStart = 0, curStart = 0, curEnd = 0;while(curStart != str.length()){if(curStart + pattern.length() <= str.length() && match(str.substring(curStart, curStart + pattern.length()), pattern)){sb.append(str.substring(lastStart, curStart));if(sb.length() == 0 || sb.charAt(sb.length() - 1) != 'X')sb.append('X');curEnd = curStart + pattern.length();lastStart = curEnd;curStart = curEnd;} else {curStart++;}}if(curEnd != str.length())sb.append(str.substring(lastStart, str.length()));return sb.toString();}public static boolean match(String s, String p){int i = 0;while(i != s.length() && i != p.length()){if(s.charAt(i) != p.charAt(i))return false;i++;}if(i == p.length())return true;return false;}// Q16 from random7() to random10()// E(# of call random7) is 2.45. We throw away idx from 41 to 49public static int random10(){int row, col, idx;do {row = random7();col = random7();idx = col + (row - 1) * 7;} while(idx > 40);return 1 + (idx - 1) % 10;}public static int random7(){return (int)(Math.random() * 7 + 1);}// E(# of call random7) is 2.21. We utilize integers from 41 to 49, and integers from 61 to 63public static int random10_better(){int row, col, idx;while(true){row = random7();col = random7();idx = col + (row - 1) * 7;if(idx <= 40)return 1 + (idx - 1) % 10;row = row - 40;col = random7();idx = col + (row - 1) * 7;if(idx <= 60)return 1 + (idx - 1) % 10;row = row - 60;col = random7();idx = col + (row - 1) * 7;if(idx <= 20)return 1 + (idx - 1) % 10;}}//Q 17// Largest Subtree which is a BSTstatic TreeNode largestSubRoot = null;static int min, max, maxNodes;// min and max are for subtree, so the root.val should be outside of [min, max]public static TreeNode largestSubtree(TreeNode root){findLargestSubtree(root);return largestSubRoot;}public static int findLargestSubtree(TreeNode root){if(root == null)return 0;boolean isBST = true;int leftNodes = findLargestSubtree(root.left);int curMin = (leftNodes == 0) ? root.val : min;if(leftNodes == -1 || (leftNodes != 0 && root.val <= max)) // root.val in the range of subtree values, thus not BSTisBST = false;int rightNodes = findLargestSubtree(root.right);int curMax = (rightNodes == 0) ? root.val : max;if(rightNodes == -1 || (rightNodes != 0 && root.val >= min)) // root.val in the range of subtree values, thus not BSTisBST = false;if(isBST){min = curMin;max = curMax;int totalNodes = leftNodes + rightNodes + 1;if(totalNodes > maxNodes){maxNodes = totalNodes;largestSubRoot = root;}return totalNodes;} elsereturn -1;}// Q18static int maxNodesBST;static TreeNode largestBST;static TreeNode child;public int findLargestBST(TreeNode root, int min, int max){if(root == null)return 0;if(min < root.val && root.val < max){int leftNodes = findLargestBST(root.left, min, root.val);TreeNode leftChild = (leftNodes == 0) ? null : child;int rightNodes = findLargestBST(root.right, root.val, max);TreeNode rightChild = (rightNodes == 0) ? null : child;TreeNode parent = new TreeNode(root.val);parent.left = leftChild;parent.right = rightChild;child = parent;int totalNodes = leftNodes + rightNodes + 1;if(totalNodes > maxNodes){maxNodes = totalNodes;largestBST = parent;}return totalNodes;} else {findLargestBST(root, Integer.MIN_VALUE, Integer.MAX_VALUE);return 0;}}// Q19public TreeNode constructFromList(LinkedListNode head, int start, int end){if(start > end)return null;int mid = (start + end) / 2;TreeNode left = constructFromList(head, start, mid - 1);TreeNode parent = new TreeNode(head.value);parent.left = left;head = head.next;parent.right = constructFromList(head, mid + 1, end);return parent;}public static void main(String[] args) {// Q1int[] A = { 1, 2, 3, 4, 5, 6, 7 };int[] B = { 4, 5, 6, 7, 8, 9, 10, 11, 12 };ArrayList<Integer> res = findIntersection(A, B);for (int i = 0; i < res.size(); i++)System.out.print(res.get(i) + " ");System.out.println();// Q2LinkedListNode head = new LinkedListNode(1);head.next = new LinkedListNode(2);head.next.next = new LinkedListNode(3);head = reverseLinkedListIteratively(head);while (head != null) {System.out.print(head.value + " ");head = head.next;}System.out.println();// Q3head = new LinkedListNode(1);head.next = new LinkedListNode(2);head.next.next = new LinkedListNode(3);head = reverseLinkedListRecursively(head);while (head != null) {System.out.print(head.value + " ");head = head.next;}System.out.println();// Q4rotateArray(A, 3);for (int i = 0; i < A.length; i++)System.out.print(A[i] + " ");System.out.println();// Q5A = new int[] { 4, 3, 2, 1, 2 };A = multiplyNumbers(A);for (int i = 0; i < A.length; i++)System.out.print(A[i] + " ");System.out.println();// Q6ArrayList<Integer> primes = generatePrime(103);for (int i = 0; i < primes.size(); i++)System.out.print(primes.get(i) + " ");System.out.println();// Q7char[] charList = keepDis(new char[] { 'a', 'b', 'b' }, 2);for (int i = 0; i < charList.length; i++)System.out.print(charList[i] + " ");System.out.println();// Q9System.out.println("Count 1 bits: " + countOneBitsCommon(7));System.out.println("Count 1 bits: " + countOneBits(7));// Q10int[] array = new int[]{7, 3, 1, 0, 2, 4, 6, 8};TreeNode  root = preorderConstructBST(array, Integer.MIN_VALUE, Integer.MAX_VALUE);System.out.println();//Q11ArrayList<Integer> selist = new ArrayList<Integer>();serialize(root, selist);for(int i = 0; i < selist.size(); i++){System.out.print(selist.get(i) + " ");}System.out.println();// Q12System.out.println(convert(730));System.out.println(convert(27));System.out.println(convert(3));System.out.println(convert2(730));System.out.println(convert2(27));System.out.println(convert2(3));//Q13System.out.println("Search Young Tableau");int[][] table = {{1, 4, 7, 11, 15},{2, 5, 8, 12, 19},{3, 6, 9, 16, 22},{10, 13, 14, 17, 24},{18, 21, 23, 26, 30}};System.out.println(stepwise(table, 10));System.out.println(quadPartition(table, 10, 0, 0, table[0].length - 1, table.length - 1));System.out.println(binaryPartition(table, 10, 0, 0, table[0].length - 1, table.length - 1));// Q14int[] sorted_array = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};TreeNode tree_root = TreeNode.construct(sorted_array, 0, sorted_array.length - 1);TreeNode.printLevel(tree_root);ArrayList<Integer> list = printEdges(tree_root);for(int i = 0; i < list.size(); i++)System.out.print(list.get(i) + " ");System.out.println();// Q15System.out.println(Q15replace("a", "a"));System.out.println(Q15replace("aa", "aa"));System.out.println(Q15replace("aa", "a"));System.out.println(Q15replace("aa", "aaa"));System.out.println(Q15replace("abc", "abc"));System.out.println(Q15replace("abcabc", "abc"));System.out.println(Q15replace("abcaabcaabc", "abc"));System.out.println(Q15replace("abcdeffdfegabcaaaabcab", "abc"));System.out.println(Q15replace("aabaaabaa", "aa"));// Q17TreeNode root17 = new TreeNode(10);root17.left = new TreeNode(5);root17.left.left = new TreeNode(1);root17.left.right = new TreeNode(8);root17.right = new TreeNode(15);root17.right.right = new TreeNode(7);largestSubtree(root17);// Test. Pass. test.left will remain 6 no matter how left got changedTreeNode test = new TreeNode(5);TreeNode left = new TreeNode(6);TreeNode left_next = new TreeNode(7);test.left = left;//left = new TreeNode(8);left = left_next;// Q18}}