Han Zhu's Study Notes

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? Which algorithm is better? What if elements of  nums2  are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once? Thought process: Iterate through both arrays. Count the number of each number and put the counts into two maps. Iterate through both maps. For numbers that appear in both maps, put min(count1, count2) of that number into result. Solution: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 class Solution { public int [] inters

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. Thought process: Iterate through nums1. Put the elements into a hash set. Iterate through nums2. The elements that the hash set contains are the intersection. Solution: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 class Solution { public int [] intersection ( int [] nums1 , int [] nums2 ) { Set < Integer > set = new HashSet <>(); for ( int num : nums1 ) { set . add ( num ); } Set < Integer > intersection = new HashSet <>(); for ( int num : nums2 ) { if ( set . contains ( num )) { intersection . add ( num ); } } int [] array = new

Given two strings  s  and  t , write a function to determine if  t  is an anagram of  s . For example, s  = "anagram",  t  = "nagaram", return true. s  = "rat",  t  = "car", return false. Note: You may assume the string contains only lowercase alphabets. Follow up: What if the inputs contain unicode characters? How would you adapt your solution to such case? Thought process: Iterate through both strings, count the number of each character, and compare the counts. Solution: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 class Solution { public boolean isAnagram ( String s , String t ) { int [] count = new int [ 26 ]; for ( char c : s . toCharArray ()) { count [ c - 'a' ]++; } for ( char c : t . toCharArray ()) { count [ c - 'a' ]--; } for ( int i : count ) { if ( i != 0