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Commit 53a3bed

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Added tasks 350, 352, 354.
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package g0301_0400.s0350_intersection_of_two_arrays_ii;
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// #Easy #Top_Interview_Questions #Array #Hash_Table #Sorting #Binary_Search #Two_Pointers
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import java.util.Arrays;
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public class Solution {
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public int[] intersect(int[] nums1, int[] nums2) {
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// First sort the array
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Arrays.sort(nums1);
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// Similar this one as well
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Arrays.sort(nums2);
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// "i" will point at nums1
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int i = 0;
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// "j" will point at nums2
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int j = 0;
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// "k" will point at nums1 and helps in getting the intersection answer;
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int k = 0;
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// Loop will run until "i" & "j" doesn't reach the array boundary;
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while (i < nums1.length && j < nums2.length) {
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if (nums1[i] < nums2[j]) {
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// Check if nums1 value is less then nums2 value;
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// Increment "i"
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i++;
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} else if (nums1[i] > nums2[j]) {
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// Check if nums2 value is less then nums1 value;
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// Increment "j"
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j++;
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} else {
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// Check if nums1 value is equals to nums2 value;
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// Dump into nums1 and increment k, increment i & increment j as well;
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nums1[k++] = nums1[i++];
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j++;
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}
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}
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// Only return nums1 0th index to kth index value, because that's will be our intersection;
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return Arrays.copyOfRange(nums1, 0, k);
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}
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}
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350\. Intersection of Two Arrays II
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Easy
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Given two integer arrays `nums1` and `nums2`, return _an array of their intersection_. Each element in the result must appear as many times as it shows in both arrays and you may return the result in **any order**.
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**Example 1:**
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**Input:** nums1 = \[1,2,2,1\], nums2 = \[2,2\]
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**Output:** \[2,2\]
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**Example 2:**
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**Input:** nums1 = \[4,9,5\], nums2 = \[9,4,9,8,4\]
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**Output:** \[4,9\]
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**Explanation:** \[9,4\] is also accepted.
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**Constraints:**
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* `1 <= nums1.length, nums2.length <= 1000`
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* `0 <= nums1[i], nums2[i] <= 1000`
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**Follow up:**
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* What if the given array is already sorted? How would you optimize your algorithm?
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* What if `nums1`'s size is small compared to `nums2`'s size? Which algorithm is better?
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* 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?
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package g0301_0400.s0352_data_stream_as_disjoint_intervals;
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// #Hard #Binary_Search #Design #Ordered_Set
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import java.util.ArrayList;
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import java.util.List;
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public class SummaryRanges {
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private static class Node {
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int min;
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int max;
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public Node(int min, int max) {
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this.min = min;
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this.max = max;
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}
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public Node(int val) {
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min = val;
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max = val;
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}
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}
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List<Node> list;
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// Initialize your data structure here.
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public SummaryRanges() {
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list = new ArrayList<>();
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}
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public void addNum(int val) {
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int left = 0;
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int right = list.size() - 1;
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int index = list.size();
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while (left <= right) {
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int mid = left + (right - left) / 2;
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Node node = list.get(mid);
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if (node.min <= val && val <= node.max) {
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return;
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} else if (val < node.min) {
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index = mid;
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right = mid - 1;
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} else if (val > node.max) {
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left = mid + 1;
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}
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}
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list.add(index, new Node(val));
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}
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public int[][] getIntervals() {
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for (int i = 1; i < list.size(); i++) {
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Node prev = list.get(i - 1);
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Node curr = list.get(i);
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if (curr.min == prev.max + 1) {
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prev.max = curr.max;
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list.remove(i--);
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}
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}
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int len = list.size();
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int[][] res = new int[len][2];
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for (int i = 0; i < len; i++) {
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Node node = list.get(i);
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res[i][0] = node.min;
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res[i][1] = node.max;
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}
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return res;
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}
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}
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352\. Data Stream as Disjoint Intervals
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Hard
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Given a data stream input of non-negative integers <code>a<sub>1</sub>, a<sub>2</sub>, ..., a<sub>n</sub></code>, summarize the numbers seen so far as a list of disjoint intervals.
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Implement the `SummaryRanges` class:
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* `SummaryRanges()` Initializes the object with an empty stream.
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* `void addNum(int val)` Adds the integer `val` to the stream.
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* `int[][] getIntervals()` Returns a summary of the integers in the stream currently as a list of disjoint intervals <code>[start<sub>i</sub>, end<sub>i</sub>]</code>.
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**Example 1:**
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**Input** \["SummaryRanges", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals"\] \[\[\], \[1\], \[\], \[3\], \[\], \[7\], \[\], \[2\], \[\], \[6\], \[\]\]
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**Output:** \[null, null, \[\[1, 1\]\], null, \[\[1, 1\], \[3, 3\]\], null, \[\[1, 1\], \[3, 3\], \[7, 7\]\], null, \[\[1, 3\], \[7, 7\]\], null, \[\[1, 3\], \[6, 7\]\]\]
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**Explanation:**
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```
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SummaryRanges summaryRanges = new SummaryRanges();
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summaryRanges.addNum(1); // arr = [1]
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summaryRanges.getIntervals(); // return [[1, 1]]
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summaryRanges.addNum(3); // arr = [1, 3]
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summaryRanges.getIntervals(); // return [[1, 1], [3, 3]]
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summaryRanges.addNum(7); // arr = [1, 3, 7]
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summaryRanges.getIntervals(); // return [[1, 1], [3, 3], [7, 7]]
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summaryRanges.addNum(2); // arr = [1, 2, 3, 7]
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summaryRanges.getIntervals(); // return [[1, 3], [7, 7]]
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summaryRanges.addNum(6); // arr = [1, 2, 3, 6, 7]
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summaryRanges.getIntervals(); // return [[1, 3], [6, 7]]
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```
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**Constraints:**
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* <code>0 <= val <= 10<sup>4</sup></code>
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* At most <code>3 * 10<sup>4</sup></code> calls will be made to `addNum` and `getIntervals`.
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**Follow up:** What if there are lots of merges and the number of disjoint intervals is small compared to the size of the data stream?
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package g0301_0400.s0354_russian_doll_envelopes;
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// #Hard #Array #Dynamic_Programming #Sorting #Binary_Search
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import java.util.Arrays;
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public class Solution {
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public int maxEnvelopes(int[][] envelopes) {
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Arrays.sort(envelopes, (a, b) -> a[0] != b[0] ? a[0] - b[0] : b[1] - a[1]);
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int[] tails = new int[envelopes.length];
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int size = 0;
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for (int[] enve : envelopes) {
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int i = 0;
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int j = size;
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while (i != j) {
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int mid = i + ((j - i) >> 1);
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if (tails[mid] < enve[1]) {
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i = mid + 1;
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} else {
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j = mid;
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}
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}
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tails[i] = enve[1];
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if (i == size) {
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size++;
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}
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}
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return size;
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}
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}
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354\. Russian Doll Envelopes
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Hard
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You are given a 2D array of integers `envelopes` where <code>envelopes[i] = [w<sub>i</sub>, h<sub>i</sub>]</code> represents the width and the height of an envelope.
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One envelope can fit into another if and only if both the width and height of one envelope are greater than the other envelope's width and height.
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Return _the maximum number of envelopes you can Russian doll (i.e., put one inside the other)_.
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**Note:** You cannot rotate an envelope.
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**Example 1:**
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**Input:** envelopes = \[\[5,4\],\[6,4\],\[6,7\],\[2,3\]\]
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**Output:** 3
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**Explanation:** The maximum number of envelopes you can Russian doll is `3` (\[2,3\] => \[5,4\] => \[6,7\]).
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**Example 2:**
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**Input:** envelopes = \[\[1,1\],\[1,1\],\[1,1\]\]
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**Output:** 1
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**Constraints:**
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* `1 <= envelopes.length <= 5000`
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* `envelopes[i].length == 2`
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* <code>1 <= w<sub>i</sub>, h<sub>i</sub> <= 10<sup>4</sup></code>
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package g0301_0400.s0350_intersection_of_two_arrays_ii;
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import static org.hamcrest.CoreMatchers.equalTo;
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import static org.hamcrest.MatcherAssert.assertThat;
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import org.junit.jupiter.api.Test;
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class SolutionTest {
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@Test
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void intersect() {
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assertThat(
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new Solution().intersect(new int[] {1, 2, 2, 1}, new int[] {2, 2}),
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equalTo(new int[] {2, 2}));
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}
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@Test
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void intersect2() {
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assertThat(
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new Solution().intersect(new int[] {4, 9, 5}, new int[] {9, 4, 9, 8, 4}),
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equalTo(new int[] {4, 9}));
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}
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}
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package g0301_0400.s0352_data_stream_as_disjoint_intervals;
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import static org.hamcrest.CoreMatchers.equalTo;
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import static org.hamcrest.MatcherAssert.assertThat;
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import org.junit.jupiter.api.Test;
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class SummaryRangesTest {
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SummaryRanges summaryRanges = new SummaryRanges();
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@Test
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void testGetInterval() {
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summaryRanges.addNum(1);
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assertThat(summaryRanges.getIntervals(), equalTo(new int[][] {{1, 1}}));
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}
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@Test
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void testGetInterval2() {
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summaryRanges.addNum(1);
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summaryRanges.addNum(3);
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assertThat(summaryRanges.getIntervals(), equalTo(new int[][] {{1, 1}, {3, 3}}));
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}
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@Test
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void testGetInterval3() {
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summaryRanges.addNum(1);
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summaryRanges.addNum(3);
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summaryRanges.addNum(7);
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assertThat(summaryRanges.getIntervals(), equalTo(new int[][] {{1, 1}, {3, 3}, {7, 7}}));
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}
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@Test
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void testGetInterval4() {
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summaryRanges.addNum(1);
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summaryRanges.addNum(2);
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summaryRanges.addNum(3);
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summaryRanges.addNum(7);
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assertThat(summaryRanges.getIntervals(), equalTo(new int[][] {{1, 3}, {7, 7}}));
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}
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@Test
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void testGetInterval5() {
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summaryRanges.addNum(1);
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summaryRanges.addNum(2);
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summaryRanges.addNum(3);
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summaryRanges.addNum(6);
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summaryRanges.addNum(7);
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assertThat(summaryRanges.getIntervals(), equalTo(new int[][] {{1, 3}, {6, 7}}));
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}
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}
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package g0301_0400.s0354_russian_doll_envelopes;
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import static org.hamcrest.CoreMatchers.equalTo;
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import static org.hamcrest.MatcherAssert.assertThat;
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import org.junit.jupiter.api.Test;
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class SolutionTest {
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@Test
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void testMaxEnvelopes() {
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assertThat(
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new Solution().maxEnvelopes(new int[][] {{5, 4}, {6, 4}, {6, 7}, {2, 3}}),
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equalTo(3));
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}
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@Test
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void testMaxEnvelopes2() {
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assertThat(new Solution().maxEnvelopes(new int[][] {{1, 1}, {1, 1}, {1, 1}}), equalTo(1));
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}
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}

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