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added tasks 350, 352, 354 #111

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javadev merged 13 commits into javadev:main from ThanhNIT:tasks-350-352-354
Dec 13, 2021
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View file Open in desktop
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package g0301_0400.s0350_intersection_of_two_arrays_ii;

// #Easy #Top_Interview_Questions #Array #Hash_Table #Sorting #Binary_Search #Two_Pointers

import java.util.Arrays;

public class Solution {
public int[] intersect(int[] nums1, int[] nums2) {
Arrays.sort(nums1);
Arrays.sort(nums2);
int i = 0;
int j = 0;
int k = 0;
while (i < nums1.length && j < nums2.length) {
if (nums1[i] < nums2[j]) {
i++;
} else if (nums1[i] > nums2[j]) {
j++;
} else {
nums1[k++] = nums1[i++];
j++;
}
}
return Arrays.copyOfRange(nums1, 0, k);
}
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You may keep comments in solutions.

Trailing comment may be moved to separate line.

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like that ?
image

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Yes, we may keep comments. We only need to move them, in separate line above.

}
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350\. Intersection of Two Arrays II

Easy

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**.

**Example 1:**

**Input:** nums1 = \[1,2,2,1\], nums2 = \[2,2\]

**Output:** \[2,2\]

**Example 2:**

**Input:** nums1 = \[4,9,5\], nums2 = \[9,4,9,8,4\]

**Output:** \[4,9\]

**Explanation:** \[9,4\] is also accepted.

**Constraints:**

* `1 <= nums1.length, nums2.length <= 1000`
* `0 <= nums1[i], nums2[i] <= 1000`

**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?
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package g0301_0400.s0352_data_stream_as_disjoint_intervals;

// #Hard #Binary_Search #Design #Ordered_Set


import java.util.Comparator;
import java.util.HashSet;
import java.util.NavigableSet;
import java.util.TreeSet;
import java.util.Set;


public class SummaryRanges {

private final NavigableSet<int[]> intervals;
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This solution is not very fast

image

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@javadev javadev Dec 13, 2021 •
edited
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I found better implemetation

public class SummaryRanges {
private static class Node {
 int min;
 int max;
 public Node(int min, int max) {
 this.min = min;
 this.max = max;
 }
 
 public Node(int val) {
 min = val;
 max = val;
 }
}
 List<Node> list;
 /** Initialize your data structure here. */
 public SummaryRanges() {
 list = new ArrayList<>();
 }
 
 public void addNum(int val) {
 int left = 0;
 int right = list.size()-1;
 int index = list.size();
 while (left <= right) {
 int mid = left + (right - left) / 2;
 Node node = list.get(mid);
 if (node.min <= val && val <= node.max) {
 return;
 } else if (val < node.min) {
 index = mid;
 right = mid - 1;
 } else if (val > node.max) {
 left = mid + 1;
 }
 }
 list.add(index, new Node(val));
 }
 
 public int[][] getIntervals() {
 for (int i = 1; i < list.size(); i++) {
 Node prev = list.get(i-1);
 Node curr = list.get(i);
 if (curr.min == prev.max+1) {
 prev.max = curr.max;
 list.remove(i--);
 }
 }
 int len = list.size();
 int[][] res = new int[len][2];
 for (int i = 0; i < len; i++) {
 Node node = list.get(i);
 res[i][0] = node.min;
 res[i][1] = node.max;
 }
 return res;
 }
}

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ok

private final Set<Integer> seen = new HashSet<>();

public SummaryRanges() {
this.intervals = new TreeSet<>(Comparator.comparingInt(i -> i[0]));
}

public void addNum(int val) {
if (seen.contains(val)) {
return;
}
seen.add(val);
int[] interval = new int[] {val, val};
this.intervals.add(interval);
this.merge(interval);
}

public int[][] getIntervals() {
int[][] res = new int[intervals.size()][2];
int i = 0;
for (int[] in : intervals) {
res[i] = in;
i++;
}
return res;
}

private void merge(int[] interval) {
// base case
if (interval == null) {
return;
}
int[] lower = intervals.lower(interval);

int[] higher;

if (lower != null && lower[1] == interval[0] - 1) {
lower[1] = interval[1];
intervals.remove(interval);
higher = intervals.higher(lower);
} else {
higher = intervals.higher(interval);
}
merge(higher);
}
}
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352\. Data Stream as Disjoint Intervals

Hard

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.

Implement the `SummaryRanges` class:

* `SummaryRanges()` Initializes the object with an empty stream.
* `void addNum(int val)` Adds the integer `val` to the stream.
* `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>.

**Example 1:**

**Input** \["SummaryRanges", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals"\] \[\[\], \[1\], \[\], \[3\], \[\], \[7\], \[\], \[2\], \[\], \[6\], \[\]\]
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It looks like two lines. It may be changes with code block.


**Output:** \[null, null, \[\[1, 1\]\], null, \[\[1, 1\], \[3, 3\]\], null, \[\[1, 1\], \[3, 3\], \[7, 7\]\], null, \[\[1, 3\], \[7, 7\]\], null, \[\[1, 3\], \[6, 7\]\]\]

**Explanation:**
```
SummaryRanges summaryRanges = new SummaryRanges();
summaryRanges.addNum(1); // arr = \[1\]
summaryRanges.getIntervals(); // return \[\[1, 1\]\]
summaryRanges.addNum(3); // arr = \[1, 3\]
summaryRanges.getIntervals(); // return \[\[1, 1\], \[3, 3\]\]
summaryRanges.addNum(7); // arr = \[1, 3, 7\]
summaryRanges.getIntervals(); // return \[\[1, 1\], \[3, 3\], \[7, 7\]\]
summaryRanges.addNum(2); // arr = \[1, 2, 3, 7\]
summaryRanges.getIntervals(); // return \[\[1, 3\], \[7, 7\]\]
summaryRanges.addNum(6); // arr = \[1, 2, 3, 6, 7\]
summaryRanges.getIntervals(); // return \[\[1, 3\], \[6, 7\]\]
```
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Please remove escapes and replace code block with 4 spaces ident.

image

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We may use four space ident for code blocks.


**Constraints:**

* <code>0 <= val <= 10<sup>4</sup></code>
* At most <code>3 * 10<sup>4</sup></code> calls will be made to `addNum` and `getIntervals`.

**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;

// #Hard #Array #Dynamic_Programming #Sorting #Binary_Search

import java.util.Arrays;

public class Solution {
public int maxEnvelopes(int[][] envelopes) {
Arrays.sort(envelopes, (a, b) -> a[0] != b[0] ? a[0] - b[0] : b[1] - a[1]);
int[] tails = new int[envelopes.length];
int size = 0;
for (int[] enve : envelopes) {
int i = 0;
int j = size;

while (i != j) {
int mid = i + ((j - i) >> 1);

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Please remove empty lines

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I usually code spaces like that to be easier to see than just writing a block

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The idea to use plugin to format codes. If it allows to remove empty line - remove it.

if (tails[mid] < enve[1]) {
i = mid + 1;
} else {
j = mid;
}
}
tails[i] = enve[1];
if (i == size) {
size++;
}
}
return size;
}
}
31 changes: 31 additions & 0 deletions src/main/java/g0301_0400/s0354_russian_doll_envelopes/readme.md
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354\. Russian Doll Envelopes

Hard

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.

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.

Return _the maximum number of envelopes you can Russian doll (i.e., put one inside the other)_.

**Note:** You cannot rotate an envelope.

**Example 1:**

**Input:** envelopes = \[\[5,4\],\[6,4\],\[6,7\],\[2,3\]\]

**Output:** 3

**Explanation:** The maximum number of envelopes you can Russian doll is `3` (\[2,3\] => \[5,4\] => \[6,7\]).

**Example 2:**

**Input:** envelopes = \[\[1,1\],\[1,1\],\[1,1\]\]

**Output:** 1

**Constraints:**

* `1 <= envelopes.length <= 5000`
* `envelopes[i].length == 2`
* <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;

import static org.hamcrest.CoreMatchers.equalTo;
import static org.hamcrest.MatcherAssert.assertThat;

import org.junit.jupiter.api.Test;

class SolutionTest {
@Test
void intersect() {
assertThat(
new Solution().intersect(new int[] {1, 2, 2, 1}, new int[] {2, 2}),
equalTo(new int[] {2, 2}));
}

@Test
void intersect2() {
assertThat(
new Solution().intersect(new int[] {4, 9, 5}, new int[] {9, 4, 9, 8, 4}),
equalTo(new int[] {4, 9}));
}
}
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package g0301_0400.s0352_data_stream_as_disjoint_intervals;

import static org.hamcrest.CoreMatchers.equalTo;
import static org.hamcrest.MatcherAssert.assertThat;
import org.junit.jupiter.api.Test;

class SummaryRangesTest {
SummaryRanges summaryRanges = new SummaryRanges();

@Test
void testGetInterval() {
summaryRanges.addNum(1);
assertThat(summaryRanges.getIntervals(), equalTo(new int[][] {{1, 1}}));
}

@Test
void testGetInterval2() {
summaryRanges.addNum(1);
summaryRanges.addNum(3);
assertThat(summaryRanges.getIntervals(), equalTo(new int[][] {{1, 1}, {3, 3}}));
}

@Test
void testGetInterval3() {
summaryRanges.addNum(1);
summaryRanges.addNum(3);
summaryRanges.addNum(7);
assertThat(summaryRanges.getIntervals(), equalTo(new int[][] {{1, 1}, {3, 3}, {7, 7}}));
}

@Test
void testGetInterval4() {
summaryRanges.addNum(1);
summaryRanges.addNum(2);
summaryRanges.addNum(3);
summaryRanges.addNum(7);
assertThat(summaryRanges.getIntervals(), equalTo(new int[][] {{1, 3}, {7, 7}}));
}

@Test
void testGetInterval5() {
summaryRanges.addNum(1);
summaryRanges.addNum(2);
summaryRanges.addNum(3);
summaryRanges.addNum(6);
summaryRanges.addNum(7);
assertThat(summaryRanges.getIntervals(), equalTo(new int[][] {{1, 3}, {6, 7}}));
}
}
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package g0301_0400.s0354_russian_doll_envelopes;

import static org.hamcrest.CoreMatchers.equalTo;
import static org.hamcrest.MatcherAssert.assertThat;

import org.junit.jupiter.api.Test;

class SolutionTest {

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Please remove empty line.

@Test
void testMaxEnvelopes() {
assertThat(
new Solution().maxEnvelopes(new int[][] {{5, 4}, {6, 4}, {6, 7}, {2, 3}}),
equalTo(3));
}

@Test
void testMaxEnvelopes2() {
assertThat(new Solution().maxEnvelopes(new int[][] {{1, 1}, {1, 1}, {1, 1}}), equalTo(1));
}
}

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