The Fellow Programmer

Given a pattern and a string s , find if s  follows the same pattern. Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in s . Example 1: Input: pattern = "abba", s = "dog cat cat dog" Output: true Example 2: Input: pattern = "abba", s = "dog cat cat fish" Output: false Example 3: Input: pattern = "aaaa", s = "dog cat cat dog" Output: false Example 4: Input: pattern = "abba", s = "dog dog dog dog" Output: false   Constraints: 1 <= pattern.length <= 300 pattern contains only lower-case English letters. 1 <= s.length <= 3000 s contains only lower-case English letters and spaces ' ' . s does not contain any leading or trailing spaces. All the words in s are separated by a single space .   class Solution {     public boolean wordPattern(String pattern, String s) {         H

Given an array of size n , find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. You may assume that the array is non-empty and the majority element always exist in the array. Example 1: Input: [3,2,3] Output: 3 Example 2: Input: [2,2,1,1,1,2,2] Output: 2   class Solution { public int majorityElement(int[] nums) { Arrays.sort(nums); return nums[nums.length/2]; } } Here, we are sorting the entire array using sort() of Arrays class. Then, number at index position n/2 would be a majority element. There is no chance of wrong values in odd or even length.   Below is another method using HashMap. METHOD 2: class Solution { public int majorityElement(int[] nums) { HashMap<Integer, Integer>list = new HashMap<Integer, Integer>(); int repeatLen