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Commit0207b47

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Added tasks 2523, 2525, 2526, 2527, 2528
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‎README.md

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@@ -1848,6 +1848,11 @@ implementation 'com.github.javadev:leetcode-in-java:1.20'
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| # | Title | Difficulty | Tag | Time, ms | Time, %
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|------|----------------|-------------|-------------|----------|---------
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| 2528 |[Maximize the Minimum Powered City](src/main/java/g2501_2600/s2528_maximize_the_minimum_powered_city/Solution.java)| Hard | Array, Greedy, Binary_Search, Prefix_Sum, Sliding_Window, Queue | 51 | 77.59
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| 2527 |[Find Xor-Beauty of Array](src/main/java/g2501_2600/s2527_find_xor_beauty_of_array/Solution.java)| Medium | Array, Math, Bit_Manipulation | 1 | 100.00
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| 2526 |[Find Consecutive Integers from a Data Stream](src/main/java/g2501_2600/s2526_find_consecutive_integers_from_a_data_stream/DataStream.java)| Medium | Hash_Table, Design, Counting, Queue, Data_Stream | 32 | 94.65
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| 2525 |[Categorize Box According to Criteria](src/main/java/g2501_2600/s2525_categorize_box_according_to_criteria/Solution.java)| Easy | Math | 0 | 100.00
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| 2523 |[Closest Prime Numbers in Range](src/main/java/g2501_2600/s2523_closest_prime_numbers_in_range/Solution.java)| Medium | Math, Number_Theory | 1 | 100.00
18511856
| 2522 |[Partition String Into Substrings With Values at Most K](src/main/java/g2501_2600/s2522_partition_string_into_substrings_with_values_at_most_k/Solution.java)| Medium | String, Dynamic_Programming, Greedy | 6 | 84.66
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| 2521 |[Distinct Prime Factors of Product of Array](src/main/java/g2501_2600/s2521_distinct_prime_factors_of_product_of_array/Solution.java)| Medium | Array, Hash_Table, Math, Number_Theory | 2 | 100.00
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| 2520 |[Count the Digits That Divide a Number](src/main/java/g2501_2600/s2520_count_the_digits_that_divide_a_number/Solution.java)| Easy | Math | 0 | 100.00
@@ -1860,7 +1865,7 @@ implementation 'com.github.javadev:leetcode-in-java:1.20'
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| 2512 |[Reward Top K Students](src/main/java/g2501_2600/s2512_reward_top_k_students/Solution.java)| Medium | Array, String, Hash_Table, Sorting, Heap_Priority_Queue | 72 | 94.76
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| 2511 |[Maximum Enemy Forts That Can Be Captured](src/main/java/g2501_2600/s2511_maximum_enemy_forts_that_can_be_captured/Solution.java)| Easy | Array, Two_Pointers | 0 | 100.00
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| 2509 |[Cycle Length Queries in a Tree](src/main/java/g2501_2600/s2509_cycle_length_queries_in_a_tree/Solution.java)| Hard | Tree, Binary_Tree | 12 | 99.26
1863-
| 2508 |[Add Edges to Make Degrees of All Nodes Even](src/main/java/g2501_2600/s2508_add_edges_to_make_degrees_of_all_nodes_even/Solution.java)| Hard | Hash_Table, Graph |33 |100.00
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| 2508 |[Add Edges to Make Degrees of All Nodes Even](src/main/java/g2501_2600/s2508_add_edges_to_make_degrees_of_all_nodes_even/Solution.java)| Hard | Hash_Table, Graph |36 |95.00
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| 2507 |[Smallest Value After Replacing With Sum of Prime Factors](src/main/java/g2501_2600/s2507_smallest_value_after_replacing_with_sum_of_prime_factors/Solution.java)| Medium | Math, Number_Theory | 1 | 84.27
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| 2506 |[Count Pairs Of Similar Strings](src/main/java/g2501_2600/s2506_count_pairs_of_similar_strings/Solution.java)| Easy | Array, String, Hash_Table | 6 | 85.15
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| 2503 |[Maximum Number of Points From Grid Queries](src/main/java/g2501_2600/s2503_maximum_number_of_points_from_grid_queries/Solution.java)| Hard | Array, Sorting, Breadth_First_Search, Heap_Priority_Queue, Union_Find | 47 | 96.38
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packageg2501_2600.s2523_closest_prime_numbers_in_range;
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// #Medium #Math #Number_Theory #2023_04_19_Time_1_ms_(100.00%)_Space_39.9_MB_(97.74%)
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publicclassSolution {
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publicint[]closestPrimes(intleft,intright) {
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intdiff = -1;
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intx = -1;
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inty = -1;
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intprev = -1;
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for (inti =left;i <=right;i++) {
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booleanisPrime =isPrime(i);
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if (isPrime) {
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if (prev != -1) {
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intd =i -prev;
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if (diff == -1 ||d <diff) {
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diff =d;
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x =prev;
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y =i;
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if (diff <=2) {
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returnnewint[] {x,y};
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}
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}
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}
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prev =i;
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}
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}
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returnnewint[] {x,y};
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}
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privatebooleanisPrime(intx) {
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if (x ==1) {
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returnfalse;
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}
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doublesqrt =Math.sqrt(x);
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for (inti =2;i <=sqrt;i++) {
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if (x %i ==0) {
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returnfalse;
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}
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}
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returntrue;
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}
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}
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2523\. Closest Prime Numbers in Range
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Medium
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Given two positive integers`left` and`right`, find the two integers`num1` and`num2` such that:
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*`left <= nums1 < nums2 <= right` .
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*`nums1` and`nums2` are both**prime** numbers.
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*`nums2 - nums1` is the**minimum** amongst all other pairs satisfying the above conditions.
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Return_the positive integer array_`ans = [nums1, nums2]`._If there are multiple pairs satisfying these conditions, return the one with the minimum_`nums1`_value or_`[-1, -1]`_if such numbers do not exist._
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A number greater than`1` is called**prime** if it is only divisible by`1` and itself.
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**Example 1:**
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**Input:** left = 10, right = 19
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**Output:**[11,13]
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**Explanation:** The prime numbers between 10 and 19 are 11, 13, 17, and 19.
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The closest gap between any pair is 2, which can be achieved by[11,13] or[17,19].
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Since 11 is smaller than 17, we return the first pair.
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**Example 2:**
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**Input:** left = 4, right = 6
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**Output:**[-1,-1]
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**Explanation:** There exists only one prime number in the given range, so the conditions cannot be satisfied.
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**Constraints:**
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* <code>1 <= left <= right <= 10<sup>6</sup></code>
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packageg2501_2600.s2525_categorize_box_according_to_criteria;
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// #Easy #Math #2023_04_19_Time_0_ms_(100.00%)_Space_40.1_MB_(36.65%)
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publicclassSolution {
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publicStringcategorizeBox(intlength,intwidth,intheight,intmass) {
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longvol = (long)length *width *height;
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booleanb =false;
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booleanh =false;
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if (length >=10000 ||width >=10000 ||height >=10000 ||vol >=1000000000) {
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b =true;
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}
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if (mass >=100) {
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h =true;
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}
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if (b &&h) {
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return"Both";
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}elseif (!b && !h) {
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return"Neither";
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}elseif (b) {
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return"Bulky";
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}else {
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return"Heavy";
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}
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}
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}
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2525\. Categorize Box According to Criteria
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Easy
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Given four integers`length`,`width`,`height`, and`mass`, representing the dimensions and mass of a box, respectively, return_a string representing the**category** of the box_.
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* The box is`"Bulky"` if:
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***Any** of the dimensions of the box is greater or equal to <code>10<sup>4</sup></code>.
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* Or, the**volume** of the box is greater or equal to <code>10<sup>9</sup></code>.
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* If the mass of the box is greater or equal to`100`, it is`"Heavy".`
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* If the box is both`"Bulky"` and`"Heavy"`, then its category is`"Both"`.
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* If the box is neither`"Bulky"` nor`"Heavy"`, then its category is`"Neither"`.
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* If the box is`"Bulky"` but not`"Heavy"`, then its category is`"Bulky"`.
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* If the box is`"Heavy"` but not`"Bulky"`, then its category is`"Heavy"`.
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**Note** that the volume of the box is the product of its length, width and height.
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**Example 1:**
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**Input:** length = 1000, width = 35, height = 700, mass = 300
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**Output:** "Heavy"
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**Explanation:**
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None of the dimensions of the box is greater or equal to 10<sup>4</sup>.
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Its volume = 24500000 <= 10<sup>9</sup>. So it cannot be categorized as "Bulky".
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However mass >= 100, so the box is "Heavy".
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Since the box is not "Bulky" but "Heavy", we return "Heavy".
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**Example 2:**
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**Input:** length = 200, width = 50, height = 800, mass = 50
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**Output:** "Neither"
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**Explanation:**
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None of the dimensions of the box is greater or equal to 10<sup>4</sup>.
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Its volume = 8\* 10<sup>6</sup> <= 10<sup>9</sup>. So it cannot be categorized as "Bulky".
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Its mass is also less than 100, so it cannot be categorized as "Heavy" either.
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Since its neither of the two above categories, we return "Neither".
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**Constraints:**
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* <code>1 <= length, width, height <= 10<sup>5</sup></code>
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* <code>1 <= mass <= 10<sup>3</sup></code>
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packageg2501_2600.s2526_find_consecutive_integers_from_a_data_stream;
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// #Medium #Hash_Table #Design #Counting #Queue #Data_Stream
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// #2023_04_19_Time_32_ms_(94.65%)_Space_80.2_MB_(39.13%)
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publicclassDataStream {
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privatefinalintvalue;
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privatefinalintk;
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privateintcount;
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publicDataStream(intvalue,intk) {
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this.value =value;
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this.k =k;
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}
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publicbooleanconsec(intnum) {
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count =num ==value ?count +1 :0;
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returncount >=k;
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}
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}
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/*
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* Your DataStream object will be instantiated and called as such:
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* DataStream obj = new DataStream(value, k);
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* boolean param_1 = obj.consec(num);
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*/
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2526\. Find Consecutive Integers from a Data Stream
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Medium
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For a stream of integers, implement a data structure that checks if the last`k` integers parsed in the stream are**equal** to`value`.
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Implement the**DataStream** class:
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*`DataStream(int value, int k)` Initializes the object with an empty integer stream and the two integers`value` and`k`.
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*`boolean consec(int num)` Adds`num` to the stream of integers. Returns`true` if the last`k` integers are equal to`value`, and`false` otherwise. If there are less than`k` integers, the condition does not hold true, so returns`false`.
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**Example 1:**
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**Input**["DataStream", "consec", "consec", "consec", "consec"]
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[[4, 3],[4],[4],[4],[3]]
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**Output:**[null, false, false, true, false]
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**Explanation:**
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DataStream dataStream = new DataStream(4, 3); //value = 4, k = 3
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dataStream.consec(4); // Only 1 integer is parsed, so returns False.
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dataStream.consec(4); // Only 2 integers are parsed.
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// Since 2 is less than k, returns False.
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dataStream.consec(4); // The 3 integers parsed are all equal to value, so returns True.
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dataStream.consec(3); // The last k integers parsed in the stream are [4,4,3].
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// Since 3 is not equal to value, it returns False.
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**Constraints:**
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* <code>1 <= value, num <= 10<sup>9</sup></code>
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* <code>1 <= k <= 10<sup>5</sup></code>
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* At most <code>10<sup>5</sup></code> calls will be made to`consec`.
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packageg2501_2600.s2527_find_xor_beauty_of_array;
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// #Medium #Array #Math #Bit_Manipulation #2023_04_19_Time_1_ms_(100.00%)_Space_59.1_MB_(62.11%)
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publicclassSolution {
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publicintxorBeauty(int[]arr) {
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inti =1;
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while (i <arr.length) {
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arr[0] ^=arr[i];
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i++;
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}
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returnarr[0];
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}
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}
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2527\. Find Xor-Beauty of Array
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Medium
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You are given a**0-indexed** integer array`nums`.
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The**effective value** of three indices`i`,`j`, and`k` is defined as`((nums[i] | nums[j]) & nums[k])`.
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The**xor-beauty** of the array is the XORing of**the effective values of all the possible triplets** of indices`(i, j, k)` where`0 <= i, j, k < n`.
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Return_the xor-beauty of_`nums`.
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**Note** that:
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*`val1 | val2` is bitwise OR of`val1` and`val2`.
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*`val1 & val2` is bitwise AND of`val1` and`val2`.
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**Example 1:**
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**Input:** nums =[1,4]
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**Output:** 5
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**Explanation:**
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The triplets and their corresponding effective values are listed below:
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- (0,0,0) with effective value ((1 | 1) & 1) = 1
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- (0,0,1) with effective value ((1 | 1) & 4) = 0
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- (0,1,0) with effective value ((1 | 4) & 1) = 1
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- (0,1,1) with effective value ((1 | 4) & 4) = 4
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- (1,0,0) with effective value ((4 | 1) & 1) = 1
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- (1,0,1) with effective value ((4 | 1) & 4) = 4
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- (1,1,0) with effective value ((4 | 4) & 1) = 0
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- (1,1,1) with effective value ((4 | 4) & 4) = 4
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Xor-beauty of array will be bitwise XOR of all beauties = 1 ^ 0 ^ 1 ^ 4 ^ 1 ^ 4 ^ 0 ^ 4 = 5.
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**Example 2:**
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**Input:** nums =[15,45,20,2,34,35,5,44,32,30]
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**Output:** 34
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**Explanation:**`The xor-beauty of the given array is 34.`
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**Constraints:**
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* <code>1 <= nums.length <= 10<sup>5</sup></code>
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* <code>1 <= nums[i] <= 10<sup>9</sup></code>
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packageg2501_2600.s2528_maximize_the_minimum_powered_city;
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// #Hard #Array #Greedy #Binary_Search #Prefix_Sum #Sliding_Window #Queue
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// #2023_04_19_Time_51_ms_(77.59%)_Space_60_MB_(5.17%)
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publicclassSolution {
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privatebooleancanIBeTheMinimum(long[]power,longminimum,intk,intr) {
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intn =power.length;
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long[]extraPower =newlong[n];
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for (inti =0;i <n;i++) {
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if (i >0) {
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extraPower[i] +=extraPower[i -1];
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}
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longcurPower =power[i] +extraPower[i];
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longreq =minimum -curPower;
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if (req <=0) {
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continue;
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}
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if (req >k) {
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returnfalse;
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}
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k -=req;
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extraPower[i] += (req);
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if (i +2 *r +1 <n) {
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extraPower[i +2 *r +1] -= (req);
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}
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}
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returntrue;
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}
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privatelong[]calculatePowerArray(int[]stations,intr) {
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intn =stations.length;
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long[]preSum =newlong[n];
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for (inti =0;i <n;i++) {
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intst =i -r;
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intlast =i +r +1;
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if (st <0) {
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st =0;
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}
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preSum[st] +=stations[i];
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if (last <n) {
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preSum[last] -=stations[i];
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}
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}
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for (inti =1;i <n;i++) {
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preSum[i] +=preSum[i -1];
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}
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returnpreSum;
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}
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publiclongmaxPower(int[]stations,intr,intk) {
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longmin =0;
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longsum = (long)Math.pow(10,10) + (long)Math.pow(10,9);
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long[]power =calculatePowerArray(stations,r);
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longans = -1;
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while (min <=sum) {
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longmid = (min +sum) >>1;
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if (canIBeTheMinimum(power,mid,k,r)) {
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ans =mid;
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min =mid +1;
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}else {
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sum =mid -1;
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}
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}
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returnans;
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}
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}

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