Solution public class CacheMiss { public int missCount(int[] array, int size) { if (array == null) return 0; List<Integer> cache = new LinkedList<>(); int count = 0; for (int i = 0; i < array.length; i++) { if (cache.contains(array[i])) { cache.remove(new Integer(array[i])); } else { count++; if (size == cache.size()) cache.remove(0); } cache.add(array[i]); } return count; } }
Given a node from a cyclic linked list which is sorted in ascending order, write a function to insert a value into the list such that it remains a cyclic sorted list. The given node can be a reference to any single node in the list, and may not be necessarily the smallest value in the cyclic list. If there are multiple suitable places for insertion, you may choose any place to insert the new value.
There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolling up, down, left or right, but it won’t stop rolling until hitting a wall. When the ball stops, it could choose the next direction. Given the ball’s start position, the destination and the maze, find the shortest distance for the ball to stop at the destination. The distance is defined by the number of empty spaces traveled by the ball from the start position (excluded) to the destination (included).
There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolling up, down, left or right, but it won’t stop rolling until hitting a wall. When the ball stops, it could choose the next direction. Given the ball’s start position, the destination and the maze, determine whether the ball could stop at the destination. The maze is represented by a binary 2D array.
Given an array nums of n integers, find two integers in nums such that the sum is closest to a given number, target. Return the difference between the sum of the two integers and the target. Example Given array nums = [-1, 2, 1, -4], and target = 4. The minimum difference is 1. (4 - (2 + 1) = 1). Challenge: Do it in O(nlogn) time complexity. Solution public class Solution { /* * @param nums: an integer array * @param target: An integer * @return: the difference between the sum and the target */ public int twoSumClosest(int[] nums, int target) { if (nums == null && nums.
Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n). Example: Input: S = "ADOBECODEBANC", T = "ABC" Output: "BANC" Note If there is no such window in S that covers all characters in T, return the empty string “". If there is such window, you are guaranteed that there will always be only one unique minimum window in S.
Given a string s, find the length of the longest substring t that contains at most 2 distinct characters. Example 1 Input: "eceba" Output: 3 Explanation: t is "ece" which its length is 3. Example 2 Input: "ccaabbb" Output: 5 Explanation: t is "aabbb" which its length is 5. Solution class Solution { public int lengthOfLongestSubstringTwoDistinct(String s) { HashMap<Character, Integer> countMap = new HashMap<Character, Integer>(); int left = 0; int max = 0; for (int i = 0; i < s.
Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note You can only move either down or right at any point in time. Example: Input: [ [1,3,1], [1,5,1], [4,2,1] ] Output: 7 Explanation: Because the path 1→3→1→1→1 minimizes the sum. Solution class Solution { public int minPathSum(int[][] grid) { if (grid == null || grid.
Given a sorted array, two integers k and x, find the k closest elements to x in the array. The result should also be sorted in ascending order. If there is a tie, the smaller elements are always preferred. Example 1 Input: [1,2,3,4,5], k=4, x=3 Output: [1,2,3,4] Example 2 Input: [1,2,3,4,5], k=4, x=-1 Output: [1,2,3,4] Note The value k is positive and will always be smaller than the length of the sorted array.
Given a string, find the length of the longest substring T that contains at most k distinct characters. Example 1 Input: s = "eceba", k = 2 Output: 3 Explanation: T is "ece" which its length is 3. Example 2 Input: s = "aa", k = 1 Output: 2 Explanation: T is "aa" which its length is 2. Solution class Solution { public int lengthOfLongestSubstringKDistinct(String s, int k) { Map<Character, Integer> countMap = new HashMap<>(); int left = 0; int max = 0; for(int i = 0; i < s.