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Commit d5ed891

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Added tasks 1331, 1332, 1333, 1334, 1335.
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package g1301_1400.s1331_rank_transform_of_an_array;
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// #Easy #Array #Hash_Table #Sorting #2022_03_19_Time_22_ms_(98.50%)_Space_58.4_MB_(93.72%)
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import java.util.Arrays;
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import java.util.HashMap;
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@SuppressWarnings({})
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public class Solution {
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public int[] arrayRankTransform(int[] arr) {
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int[] tmp = Arrays.copyOf(arr, arr.length);
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Arrays.sort(tmp);
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HashMap<Integer, Integer> mp = new HashMap<>();
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int i = 1;
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for (Integer x : tmp) {
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if (!mp.containsKey(x)) {
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mp.put(x, i++);
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}
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}
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i = 0;
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for (Integer x : arr) {
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arr[i++] = mp.get(x);
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}
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return arr;
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}
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}
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1331\. Rank Transform of an Array
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Easy
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Given an array of integers `arr`, replace each element with its rank.
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The rank represents how large the element is. The rank has the following rules:
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* Rank is an integer starting from 1.
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* The larger the element, the larger the rank. If two elements are equal, their rank must be the same.
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* Rank should be as small as possible.
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**Example 1:**
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**Input:** arr = [40,10,20,30]
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**Output:** [4,1,2,3]
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**Explanation:**: 40 is the largest element. 10 is the smallest. 20 is the second smallest. 30 is the third smallest.
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**Example 2:**
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**Input:** arr = [100,100,100]
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**Output:** [1,1,1]
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**Explanation:**: Same elements share the same rank.
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**Example 3:**
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**Input:** arr = [37,12,28,9,100,56,80,5,12]
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**Output:** [5,3,4,2,8,6,7,1,3]
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**Constraints:**
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* <code>0 <= arr.length <= 10<sup>5</sup></code>
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* <code>-10<sup>9</sup> <= arr[i] <= 10<sup>9</sup></code>
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package g1301_1400.s1332_remove_palindromic_subsequences;
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// #Easy #String #Two_Pointers #2022_03_19_Time_0_ms_(100.00%)_Space_39.9_MB_(78.40%)
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public class Solution {
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public int removePalindromeSub(String s) {
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if (s.isEmpty()) {
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return 0;
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}
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if (s.equals((new StringBuilder(s)).reverse().toString())) {
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return 1;
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}
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return 2;
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}
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}
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1332\. Remove Palindromic Subsequences
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Easy
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You are given a string `s` consisting **only** of letters `'a'` and `'b'`. In a single step you can remove one **palindromic subsequence** from `s`.
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Return _the **minimum** number of steps to make the given string empty_.
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A string is a **subsequence** of a given string if it is generated by deleting some characters of a given string without changing its order. Note that a subsequence does **not** necessarily need to be contiguous.
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A string is called **palindrome** if is one that reads the same backward as well as forward.
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**Example 1:**
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**Input:** s = "ababa"
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**Output:** 1
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**Explanation:** s is already a palindrome, so its entirety can be removed in a single step.
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**Example 2:**
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**Input:** s = "abb"
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**Output:** 2
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**Explanation:** "abb" -> "bb" -> "". Remove palindromic subsequence "a" then "bb".
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**Example 3:**
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**Input:** s = "baabb"
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**Output:** 2
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**Explanation:** "baabb" -> "b" -> "". Remove palindromic subsequence "baab" then "b".
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**Constraints:**
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* `1 <= s.length <= 1000`
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* `s[i]` is either `'a'` or `'b'`.
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package g1301_1400.s1333_filter_restaurants_by_vegan_friendly_price_and_distance;
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// #Medium #Array #Sorting #2022_03_19_Time_10_ms_(55.43%)_Space_58_MB_(52.00%)
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import java.util.ArrayList;
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import java.util.List;
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import java.util.stream.Collectors;
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public class Solution {
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public List<Integer> filterRestaurants(
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int[][] restaurants, int veganFriendly, int maxPrice, int maxDistance) {
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List<int[]> list = new ArrayList<>();
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for (int[] restaurant : restaurants) {
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if (((veganFriendly == 1 && restaurant[2] == 1) || veganFriendly == 0)
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&& restaurant[3] <= maxPrice
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&& restaurant[4] <= maxDistance) {
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list.add(restaurant);
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}
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}
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list.sort((a, b) -> b[1] - a[1] == 0 ? b[0] - a[0] : b[1] - a[1]);
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return list.stream().map(restaurant -> restaurant[0]).collect(Collectors.toList());
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}
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}
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1333\. Filter Restaurants by Vegan-Friendly, Price and Distance
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Medium
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Given the array `restaurants` where <code>restaurants[i] = [id<sub>i</sub>, rating<sub>i</sub>, veganFriendly<sub>i</sub>, price<sub>i</sub>, distance<sub>i</sub>]</code>. You have to filter the restaurants using three filters.
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The `veganFriendly` filter will be either _true_ (meaning you should only include restaurants with <code>veganFriendly<sub>i</sub></code> set to true) or _false_ (meaning you can include any restaurant). In addition, you have the filters `maxPrice` and `maxDistance` which are the maximum value for price and distance of restaurants you should consider respectively.
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Return the array of restaurant _**IDs**_ after filtering, ordered by **rating** from highest to lowest. For restaurants with the same rating, order them by _**id**_ from highest to lowest. For simplicity <code>veganFriendly<sub>i</sub></code> and `veganFriendly` take value _1_ when it is _true_, and _0_ when it is _false_.
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**Example 1:**
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**Input:** restaurants = [[1,4,1,40,10],[2,8,0,50,5],[3,8,1,30,4],[4,10,0,10,3],[5,1,1,15,1]], veganFriendly = 1, maxPrice = 50, maxDistance = 10
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**Output:** [3,1,5]
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**Explanation:** The restaurants are:
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Restaurant 1 [id=1, rating=4, veganFriendly=1, price=40, distance=10]
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Restaurant 2 [id=2, rating=8, veganFriendly=0, price=50, distance=5]
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Restaurant 3 [id=3, rating=8, veganFriendly=1, price=30, distance=4]
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Restaurant 4 [id=4, rating=10, veganFriendly=0, price=10, distance=3]
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Restaurant 5 [id=5, rating=1, veganFriendly=1, price=15, distance=1]
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After filter restaurants with veganFriendly = 1, maxPrice = 50 and maxDistance = 10 we have restaurant 3, restaurant 1 and restaurant 5 (ordered by rating from highest to lowest).
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**Example 2:**
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**Input:** restaurants = [[1,4,1,40,10],[2,8,0,50,5],[3,8,1,30,4],[4,10,0,10,3],[5,1,1,15,1]], veganFriendly = 0, maxPrice = 50, maxDistance = 10
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**Output:** [4,3,2,1,5]
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**Explanation:** The restaurants are the same as in example 1, but in this case the filter veganFriendly = 0, therefore all restaurants are considered.
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**Example 3:**
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**Input:** restaurants = [[1,4,1,40,10],[2,8,0,50,5],[3,8,1,30,4],[4,10,0,10,3],[5,1,1,15,1]], veganFriendly = 0, maxPrice = 30, maxDistance = 3
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**Output:** [4,5]
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**Constraints:**
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* `1 <= restaurants.length <= 10^4`
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* `restaurants[i].length == 5`
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* <code>1 <= id<sub>i</sub>, rating<sub>i</sub>, price<sub>i</sub>, distance<sub>i</sub> <= 10^5</code>
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* `1 <= maxPrice, maxDistance <= 10^5`
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* <code>veganFriendly<sub>i</sub></code> and `veganFriendly` are 0 or 1.
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* All <code>id<sub>i</sub></code> are distinct.
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package g1301_1400.s1334_find_the_city_with_the_smallest_number_of_neighbors_at_a_threshold_distance;
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// #Medium #Dynamic_Programming #Graph #Shortest_Path
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// #2022_03_19_Time_21_ms_(49.75%)_Space_47_MB_(36.59%)
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import java.util.ArrayList;
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import java.util.HashMap;
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import java.util.List;
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public class Solution {
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public int findTheCity(int n, int[][] edges, int maxDist) {
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int[][] graph = new int[n][n];
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for (int[] edge : edges) {
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graph[edge[0]][edge[1]] = edge[2];
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graph[edge[1]][edge[0]] = edge[2];
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}
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return fllowdWarshall(graph, n, maxDist);
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}
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private int fllowdWarshall(int[][] graph, int n, int maxDist) {
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int inf = 10001;
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int[][] dist = new int[n][n];
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < n; j++) {
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if (i != j && graph[i][j] == 0) {
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dist[i][j] = inf;
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} else {
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dist[i][j] = graph[i][j];
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}
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}
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}
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for (int k = 0; k < n; k++) {
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < n; j++) {
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if (dist[i][k] + dist[k][j] < dist[i][j]) {
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dist[i][j] = dist[i][k] + dist[k][j];
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}
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}
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}
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}
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return getList(dist, n, maxDist);
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}
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private int getList(int[][] dist, int n, int maxDist) {
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HashMap<Integer, List<Integer>> map = new HashMap<>();
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < n; j++) {
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if (!map.containsKey(i)) {
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map.put(i, new ArrayList<>());
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if (dist[i][j] <= maxDist && i != j) {
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map.get(i).add(j);
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}
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} else if (map.containsKey(i) && dist[i][j] <= maxDist && i != j) {
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map.get(i).add(j);
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}
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}
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}
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int numOfEle = Integer.MAX_VALUE;
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int ans = 0;
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for (int i = 0; i < n; i++) {
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if (numOfEle >= map.get(i).size()) {
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numOfEle = Math.min(numOfEle, map.get(i).size());
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ans = i;
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}
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}
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return ans;
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}
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}
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1334\. Find the City With the Smallest Number of Neighbors at a Threshold Distance
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Medium
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There are `n` cities numbered from `0` to `n-1`. Given the array `edges` where <code>edges[i] = [from<sub>i</sub>, to<sub>i</sub>, weight<sub>i</sub>]</code> represents a bidirectional and weighted edge between cities <code>from<sub>i</sub></code> and <code>to<sub>i</sub></code>, and given the integer `distanceThreshold`.
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Return the city with the smallest number of cities that are reachable through some path and whose distance is **at most** `distanceThreshold`, If there are multiple such cities, return the city with the greatest number.
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Notice that the distance of a path connecting cities _**i**_ and _**j**_ is equal to the sum of the edges' weights along that path.
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**Example 1:**
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![](https://assets.leetcode.com/uploads/2020/01/16/find_the_city_01.png)
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**Input:** n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4
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**Output:** 3
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**Explanation:** The figure above describes the graph.
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The neighboring cities at a distanceThreshold = 4 for each city are:
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City 0 -> [City 1, City 2]
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City 1 -> [City 0, City 2, City 3]
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City 2 -> [City 0, City 1, City 3]
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City 3 -> [City 1, City 2]
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Cities 0 and 3 have 2 neighboring cities at a distanceThreshold = 4, but we have to return city 3 since it has the greatest number.
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**Example 2:**
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![](https://assets.leetcode.com/uploads/2020/01/16/find_the_city_02.png)
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**Input:** n = 5, edges = [[0,1,2],[0,4,8],[1,2,3],[1,4,2],[2,3,1],[3,4,1]], distanceThreshold = 2
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**Output:** 0
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**Explanation:** The figure above describes the graph.
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The neighboring cities at a distanceThreshold = 2 for each city are:
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City 0 -> [City 1]
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City 1 -> [City 0, City 4]
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City 2 -> [City 3, City 4]
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City 3 -> [City 2, City 4]
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City 4 -> [City 1, City 2, City 3]
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The city 0 has 1 neighboring city at a distanceThreshold = 2.
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**Constraints:**
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* `2 <= n <= 100`
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* `1 <= edges.length <= n * (n - 1) / 2`
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* `edges[i].length == 3`
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* <code>0 <= from<sub>i</sub> < to<sub>i</sub> < n</code>
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* <code>1 <= weight<sub>i</sub>, distanceThreshold <= 10^4</code>
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* All pairs <code>(from<sub>i</sub>, to<sub>i</sub>)</code> are distinct.
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package g1301_1400.s1335_minimum_difficulty_of_a_job_schedule;
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// #Hard #Array #Dynamic_Programming #2022_03_19_Time_11_ms_(79.28%)_Space_41.6_MB_(59.99%)
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public class Solution {
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public int minDifficulty(int[] jobDifficulty, int d) {
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int totalJobs = jobDifficulty.length;
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if (totalJobs < d) {
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return -1;
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}
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int maxJobsOneDay = totalJobs - d + 1;
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int[] map = new int[totalJobs];
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int maxDiff = Integer.MIN_VALUE;
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for (int k = totalJobs - 1; k > totalJobs - 1 - maxJobsOneDay; k--) {
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maxDiff = Math.max(maxDiff, jobDifficulty[k]);
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map[k] = maxDiff;
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}
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for (int day = d - 1; day > 0; day--) {
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int maxEndIndex = (totalJobs - 1) - (d - day);
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int maxStartIndex = maxEndIndex - maxJobsOneDay + 1;
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for (int startIndex = maxStartIndex; startIndex <= maxEndIndex; startIndex++) {
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map[startIndex] = Integer.MAX_VALUE;
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int maxDiffOfTheDay = Integer.MIN_VALUE;
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for (int endIndex = startIndex; endIndex <= maxEndIndex; endIndex++) {
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maxDiffOfTheDay = Math.max(maxDiffOfTheDay, jobDifficulty[endIndex]);
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int totalDiff = maxDiffOfTheDay + map[endIndex + 1];
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map[startIndex] = Math.min(map[startIndex], totalDiff);
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}
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}
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}
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return map[0];
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}
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}

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