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Commitfd1cee4

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packageg2801_2900.s2874_maximum_value_of_an_ordered_triplet_ii;
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// #Medium #Array #2023_12_22_Time_2_ms_(99.67%)_Space_57.5_MB_(41.20%)
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publicclassSolution {
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publiclongmaximumTripletValue(int[]nums) {
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int[]diff =newint[nums.length];
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inttempMax =nums[0];
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for (inti =1;i <diff.length -1;i++) {
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diff[i] =tempMax -nums[i];
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tempMax =Math.max(tempMax,nums[i]);
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}
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longmax =Long.MIN_VALUE;
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tempMax =nums[nums.length -1];
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for (inti =nums.length -2;i >0;i--) {
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max =Math.max(max, (long)tempMax *diff[i]);
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tempMax =Math.max(tempMax,nums[i]);
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}
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returnMath.max(max,0);
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}
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}
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2874\. Maximum Value of an Ordered Triplet II
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Medium
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You are given a**0-indexed** integer array`nums`.
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Return_**the maximum value over all triplets of indices**_`(i, j, k)`_such that_`i < j < k`_._ If all such triplets have a negative value, return`0`.
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The**value of a triplet of indices**`(i, j, k)` is equal to`(nums[i] - nums[j]) * nums[k]`.
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**Example 1:**
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**Input:** nums =[12,6,1,2,7]
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**Output:** 77
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**Explanation:** The value of the triplet (0, 2, 4) is (nums[0] - nums[2])\* nums[4] = 77. It can be shown that there are no ordered triplets of indices with a value greater than 77.
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**Example 2:**
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**Input:** nums =[1,10,3,4,19]
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**Output:** 133
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**Explanation:** The value of the triplet (1, 2, 4) is (nums[1] - nums[2])\* nums[4] = 133. It can be shown that there are no ordered triplets of indices with a value greater than 133.
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**Example 3:**
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**Input:** nums =[1,2,3]
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**Output:** 0
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**Explanation:** The only ordered triplet of indices (0, 1, 2) has a negative value of (nums[0] - nums[1])\* nums[2] = -3. Hence, the answer would be 0.
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**Constraints:**
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* <code>3 <= nums.length <= 10<sup>5</sup></code>
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* <code>1 <= nums[i] <= 10<sup>6</sup></code>
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packageg2801_2900.s2875_minimum_size_subarray_in_infinite_array;
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// #Medium #Array #Hash_Table #Prefix_Sum #Sliding_Window
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// #2023_12_22_Time_6_ms_(87.63%)_Space_55.1_MB_(54.84%)
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publicclassSolution {
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publicintminSizeSubarray(int[]nums,inttarget) {
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intsum =0;
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for (intnum :nums) {
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sum +=num;
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}
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if (sum ==0) {
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return -1;
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}
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intresult = (target /sum) *nums.length;
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sum =target %sum;
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intcurrentSum =0;
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intmin =nums.length;
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intstart =0;
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for (inti =0;i <nums.length *2;i++) {
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currentSum +=nums[i %nums.length];
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while (currentSum >sum) {
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currentSum -=nums[start %nums.length];
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start++;
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}
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if (currentSum ==sum) {
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min =Math.min(min,i -start +1);
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}
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}
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if (min ==nums.length) {
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return -1;
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}
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returnresult +min;
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}
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}
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2875\. Minimum Size Subarray in Infinite Array
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Medium
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You are given a**0-indexed** array`nums` and an integer`target`.
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A**0-indexed** array`infinite_nums` is generated by infinitely appending the elements of`nums` to itself.
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Return_the length of the**shortest** subarray of the array_`infinite_nums`_with a sum equal to_`target`_._ If there is no such subarray return`-1`.
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**Example 1:**
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**Input:** nums =[1,2,3], target = 5
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**Output:** 2
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**Explanation:** In this example infinite\_nums =[1,2,3,1,2,3,1,2,...]. The subarray in the range[1,2], has the sum equal to target = 5 and length = 2. It can be proven that 2 is the shortest length of a subarray with sum equal to target = 5.
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**Example 2:**
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**Input:** nums =[1,1,1,2,3], target = 4
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**Output:** 2
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**Explanation:** In this example infinite\_nums =[1,1,1,2,3,1,1,1,2,3,1,1,...]. The subarray in the range[4,5], has the sum equal to target = 4 and length = 2. It can be proven that 2 is the shortest length of a subarray with sum equal to target = 4.
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**Example 3:**
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**Input:** nums =[2,4,6,8], target = 3
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**Output:** -1
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**Explanation:** In this example infinite\_nums =[2,4,6,8,2,4,6,8,...]. It can be proven that there is no subarray with sum equal to target = 3.
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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>5</sup></code>
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* <code>1 <= target <= 10<sup>9</sup></code>
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packageg2801_2900.s2876_count_visited_nodes_in_a_directed_graph;
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// #Hard #Dynamic_Programming #Graph #Memoization
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// #2023_12_22_Time_36_ms_(93.33%)_Space_76.9_MB_(25.00%)
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importjava.util.List;
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publicclassSolution {
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publicint[]countVisitedNodes(List<Integer>edges) {
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intn =edges.size();
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boolean[]visited =newboolean[n];
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int[]ans =newint[n];
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int[]level =newint[n];
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for (inti =0;i <n;i++) {
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if (!visited[i]) {
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visit(edges,0,i,ans,visited,level);
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}
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}
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returnans;
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}
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privateint[]visit(
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List<Integer>edges,intcount,intcurr,int[]ans,boolean[]visited,int[]level) {
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if (ans[curr] !=0) {
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returnnewint[] {-1,ans[curr]};
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}
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if (visited[curr]) {
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returnnewint[] {level[curr],count -level[curr]};
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}
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level[curr] =count;
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visited[curr] =true;
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int[]ret =visit(edges,count +1,edges.get(curr),ans,visited,level);
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if (ret[0] == -1 ||count <ret[0]) {
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ret[1]++;
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}
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ans[curr] =ret[1];
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returnret;
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}
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}
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2876\. Count Visited Nodes in a Directed Graph
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Hard
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There is a**directed** graph consisting of`n` nodes numbered from`0` to`n - 1` and`n` directed edges.
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You are given a**0-indexed** array`edges` where`edges[i]` indicates that there is an edge from node`i` to node`edges[i]`.
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Consider the following process on the graph:
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* You start from a node`x` and keep visiting other nodes through edges until you reach a node that you have already visited before on this**same** process.
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Return_an array_`answer`_where_`answer[i]`_is the number of**different** nodes that you will visit if you perform the process starting from node_`i`.
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**Example 1:**
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![](https://assets.leetcode.com/uploads/2023/08/31/graaphdrawio-1.png)
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**Input:** edges =[1,2,0,0]
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**Output:**[3,3,3,4]
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**Explanation:** We perform the process starting from each node in the following way:
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- Starting from node 0, we visit the nodes 0 -> 1 -> 2 -> 0. The number of different nodes we visit is 3.
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- Starting from node 1, we visit the nodes 1 -> 2 -> 0 -> 1. The number of different nodes we visit is 3.
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- Starting from node 2, we visit the nodes 2 -> 0 -> 1 -> 2. The number of different nodes we visit is 3.
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- Starting from node 3, we visit the nodes 3 -> 0 -> 1 -> 2 -> 0. The number of different nodes we visit is 4.
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**Example 2:**
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![](https://assets.leetcode.com/uploads/2023/08/31/graaph2drawio.png)
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**Input:** edges =[1,2,3,4,0]
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**Output:**[5,5,5,5,5]
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**Explanation:** Starting from any node we can visit every node in the graph in the process.
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**Constraints:**
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*`n == edges.length`
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* <code>2 <= n <= 10<sup>5</sup></code>
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*`0 <= edges[i] <= n - 1`
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*`edges[i] != i`
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2877\. Create a DataFrame from List
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Easy
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Write a solution to**create** a DataFrame from a 2D list called`student_data`. This 2D list contains the IDs and ages of some students.
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The DataFrame should have two columns,`student_id` and`age`, and be in the same order as the original 2D list.
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The result format is in the following example.
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**Example 1:**
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**Input:** student\_data:`[ [1, 15], [2, 11], [3, 11], [4, 20] ]`
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**Output:**
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+------------+-----+
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| student_id | age |
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+------------+-----+
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| 1 | 15 |
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| 2 | 11 |
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| 3 | 11 |
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| 4 | 20 |
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+------------+-----+
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**Explanation:** A DataFrame was created on top of student\_data, with two columns named`student_id` and`age`.
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# #Easy #2023_12_22_Time_406_ms_(82.57%)_Space_59.2_MB_(81.15%)
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importpandasaspd
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defcreateDataframe(student_data:List[List[int]])->pd.DataFrame:
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column_name= ['student_id','age']
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result=pd.DataFrame(student_data,columns=column_name)
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returnresult
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2878\. Get the Size of a DataFrame
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Easy
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DataFrame`players:`
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+-------------+--------+
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| Column Name | Type |
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+-------------+--------+
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| player_id | int |
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| name | object |
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| age | int |
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| position | object |
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| ... | ... |
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+-------------+--------+
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Write a solution to calculate and display the**number of rows and columns** of`players`.
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Return the result as an array:
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`[number of rows, number of columns]`
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The result format is in the following example.
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**Example 1:**
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**Input:**
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+-----------+----------+-----+-------------+--------------------+
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| player_id | name | age | position | team |
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+-----------+----------+-----+-------------+--------------------+
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| 846 | Mason | 21 | Forward | RealMadrid |
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| 749 | Riley | 30 | Winger | Barcelona |
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| 155 | Bob | 28 | Striker | ManchesterUnited |
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| 583 | Isabella | 32 | Goalkeeper | Liverpool |
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| 388 | Zachary | 24 | Midfielder | BayernMunich |
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| 883 | Ava | 23 | Defender | Chelsea |
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| 355 | Violet | 18 | Striker | Juventus |
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| 247 | Thomas | 27 | Striker | ParisSaint-Germain |
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| 761 | Jack | 33 | Midfielder | ManchesterCity |
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| 642 | Charlie | 36 | Center-back | Arsenal |
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+-----------+----------+-----+-------------+--------------------+
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**Output:**[10, 5]
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**Explanation:** This DataFrame contains 10 rows and 5 columns.
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# #Easy #2023_12_22_Time_413_ms_(94.68%)_Space_59.9_MB_(74.79%)
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importpandasaspd
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defgetDataframeSize(players:pd.DataFrame)->List[int]:
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return[players.shape[0],players.shape[1]]
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packageg2801_2900.s2874_maximum_value_of_an_ordered_triplet_ii;
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importstaticorg.hamcrest.CoreMatchers.equalTo;
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importstaticorg.hamcrest.MatcherAssert.assertThat;
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importorg.junit.jupiter.api.Test;
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classSolutionTest {
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@Test
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voidmaximumTripletValue() {
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assertThat(newSolution().maximumTripletValue(newint[] {12,6,1,2,7}),equalTo(77L));
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}
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@Test
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voidmaximumTripletValue2() {
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assertThat(newSolution().maximumTripletValue(newint[] {1,10,3,4,19}),equalTo(133L));
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}
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@Test
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voidmaximumTripletValue3() {
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assertThat(newSolution().maximumTripletValue(newint[] {1,2,3}),equalTo(0L));
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}
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}
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packageg2801_2900.s2875_minimum_size_subarray_in_infinite_array;
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importstaticorg.hamcrest.CoreMatchers.equalTo;
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importstaticorg.hamcrest.MatcherAssert.assertThat;
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importorg.junit.jupiter.api.Test;
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classSolutionTest {
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@Test
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voidminSizeSubarray() {
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assertThat(newSolution().minSizeSubarray(newint[] {1,2,3},5),equalTo(2));
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}
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@Test
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voidminSizeSubarray2() {
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assertThat(newSolution().minSizeSubarray(newint[] {1,1,1,2,3},4),equalTo(2));
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}
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@Test
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voidminSizeSubarray3() {
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assertThat(newSolution().minSizeSubarray(newint[] {2,4,6,8},3),equalTo(-1));
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}
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@Test
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voidminSizeSubarray4() {
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assertThat(newSolution().minSizeSubarray(newint[] {0},1),equalTo(-1));
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}
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}
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packageg2801_2900.s2876_count_visited_nodes_in_a_directed_graph;
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importstaticorg.hamcrest.CoreMatchers.equalTo;
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importstaticorg.hamcrest.MatcherAssert.assertThat;
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importjava.util.Arrays;
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importorg.junit.jupiter.api.Test;
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classSolutionTest {
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@Test
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voidcountVisitedNodes() {
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assertThat(
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newSolution().countVisitedNodes(Arrays.asList(1,2,0,0)),
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equalTo(newint[] {3,3,3,4}));
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}
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@Test
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voidcountVisitedNodes2() {
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assertThat(
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newSolution().countVisitedNodes(Arrays.asList(1,2,3,4,0)),
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equalTo(newint[] {5,5,5,5,5}));
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}
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}

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