|
| 1 | +packagemedium; |
| 2 | + |
| 3 | +importjava.util.TreeSet; |
| 4 | + |
| 5 | +/** |
| 6 | + * 220. Contains Duplicate III QuestionEditorial Solution My Submissions Total Accepted: 33823 Total |
| 7 | + * Submissions: 176491 Difficulty: Medium Given an array of integers, find out whether there are two |
| 8 | + * distinct indices i and j in the array such that the difference between nums[i] and nums[j] is at |
| 9 | + * most t and the difference between i and j is at most k. |
| 10 | + */ |
| 11 | +publicclassContainsDuplicateIII { |
| 12 | +/** |
| 13 | + * TreeSet: per Java doc, is a NavigableSet implementation based on a TreeMap. The elements are ordered |
| 14 | + * using their natural ordering, or by a Comparator provided at set creation time, depending on |
| 15 | + * which constructor is used. This implementation provides guaranteed log(n) time cost for the |
| 16 | + * basic operations (add, remove and contains). |
| 17 | + */ |
| 18 | + |
| 19 | +/** |
| 20 | + * TreeSet turns out to be a super handy data structure for this problem. It implements Set |
| 21 | + * interface and keeps elements in sorted order, plus it has two very handy APIs: |
| 22 | + * |
| 23 | + * https://docs.oracle.com/javase/7/docs/api/java/util/TreeSet.html#ceiling(E): Returns the |
| 24 | + * least element in this set greater than or equal to the given element, or null if there is no |
| 25 | + * such element. |
| 26 | + * |
| 27 | + * https://docs.oracle.com/javase/7/docs/api/java/util/TreeSet.html#floor(E): Returns the |
| 28 | + * greatest element in this set less than or equal to the given element, or null if there is no |
| 29 | + * such element. |
| 30 | + * |
| 31 | + * Idea: loop through this array, keep adding each element into the TreeSet, also keep an eye on |
| 32 | + * the size of the TreeSet, if it's greater than the required distance of k, then we remove the |
| 33 | + * left-most or oldest one from the set. For each element, we get the current floor and the |
| 34 | + * current ceiling and see if it works, if it does, then we return true immediately, otherwise, |
| 35 | + * we continue. Return false when we finished the loop. |
| 36 | + */ |
| 37 | + |
| 38 | +//Really, computing is all about data structures! Cool! Learned the gist again! |
| 39 | +publicbooleancontainsNearbyAlmostDuplicate(int[]nums,intk,intt) { |
| 40 | +TreeSet<Integer>set =newTreeSet<Integer>(); |
| 41 | +for(inti =0;i <nums.length;i++){ |
| 42 | +Integers =set.ceiling(nums[i]);//take the smallest greater number than nums[i] is there exists |
| 43 | +if(s !=null &&s -nums[i] <=t)returntrue;//it must be s - nums[i] here, otherwise, cases like [4,2] 2, 1 wont' pass, because we'll get 2-4 = -2 which is a negative number that for sure will be smaller than t |
| 44 | +Integerg =set.floor(nums[i]);//take the biggest smaller number than nums[i] if there exists |
| 45 | +if(g !=null && (long)nums[i] -g <=t)returntrue; |
| 46 | +set.add(nums[i]); |
| 47 | +if(set.size() >k)set.remove(nums[i-k]);//set doesn't have indices and it's not ordered, we could only specify the element |
| 48 | +//that we want to remove, this element is nums[i-k] |
| 49 | + } |
| 50 | +returnfalse; |
| 51 | + } |
| 52 | + |
| 53 | +//My naive approach prior to looking on Discuss: |
| 54 | +//it must be a O(n^2), I cannot think of any algorithms that do better than this |
| 55 | +// as expected, this following algorithm made it to 29/31 test cases and failed by the last |
| 56 | +// extreme test case due to TLE. |
| 57 | +publicbooleancontainsNearbyAlmostDuplicate_TLE(int[]nums,intk,intt) { |
| 58 | +for (inti =0;i <nums.length;i++) { |
| 59 | +for (intj =i +1;j <nums.length;j++) { |
| 60 | +longres =nums[j] -nums[i];// use long type to avoid overflow |
| 61 | +if (Math.abs(res) <=t &&j -i <=k) |
| 62 | +returntrue; |
| 63 | + } |
| 64 | + } |
| 65 | +returnfalse; |
| 66 | + } |
| 67 | + |
| 68 | +/** |
| 69 | + * converting to (long) is essential, otherwise cases like this: |
| 70 | + * |
| 71 | + * [-1,2147483647] |
| 72 | + * |
| 73 | + * 1 |
| 74 | + * |
| 75 | + * 2147483647 |
| 76 | + * |
| 77 | + * will overflow, i.e. Integer in Java is 32 bit which means Integer.MAX_VALUE =2147483647 and |
| 78 | + * Integer.MIN_VALUE = -2147483648, thus 2147483647 -(-1) = 2147483647+1 =-2147483648 instead of |
| 79 | + * 2147483648 (Java Integer won’t have this number), causing this test case to fail. |
| 80 | + */ |
| 81 | + |
| 82 | +publicstaticvoidmain(String...strings) { |
| 83 | +ContainsDuplicateIIItest =newContainsDuplicateIII(); |
| 84 | +// int[] nums = new int[]{-1, -1}; |
| 85 | +// int k = 1; |
| 86 | +// int t = 0; |
| 87 | + |
| 88 | +// int[] nums = new int[] { 1, 2 }; |
| 89 | +// int k = 0; |
| 90 | +// int t = 1; |
| 91 | + |
| 92 | +int[]nums =newint[] {4,2 }; |
| 93 | +intk =2; |
| 94 | +intt =1; |
| 95 | + |
| 96 | +// int[] nums = new int[]{2, 2}; |
| 97 | +// int k = 3; |
| 98 | +// int t = 0; |
| 99 | + |
| 100 | +// int[] nums = new int[]{1}; |
| 101 | +// int k = 1; |
| 102 | +// int t = 1; |
| 103 | +System.out.println(test.containsNearbyAlmostDuplicate(nums,k,t)); |
| 104 | + } |
| 105 | +} |