Added tasks 2144, 2145, 2146, 2147, 2148.

Author: ThanhNIT

src/main/java/g2101_2200/s2144_minimum_cost_of_buying_candies_with_discount/Solution.java

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+package g2101_2200.s2144_minimum_cost_of_buying_candies_with_discount;
+
+// #Easy #Array #Sorting #Greedy #2022_06_07_Time_2_ms_(97.50%)_Space_41.4_MB_(99.04%)
+
+import java.util.Arrays;
+
+public class Solution {
+    public int minimumCost(int[] cost) {
+        Arrays.sort(cost);
+        int size = 0;
+        int sum = 0;
+        for (int i = cost.length - 1; i >= 0; i--) {
+            size++;
+            if (size % 3 != 0) {
+                sum += cost[i];
+            }
+        }
+
+        return sum;
+    }
+}

src/main/java/g2101_2200/s2144_minimum_cost_of_buying_candies_with_discount/readme.md

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+2144\. Minimum Cost of Buying Candies With Discount
+
+Easy
+
+A shop is selling candies at a discount. For **every two** candies sold, the shop gives a **third** candy for **free**.
+
+The customer can choose **any** candy to take away for free as long as the cost of the chosen candy is less than or equal to the **minimum** cost of the two candies bought.
+
+*   For example, if there are `4` candies with costs `1`, `2`, `3`, and `4`, and the customer buys candies with costs `2` and `3`, they can take the candy with cost `1` for free, but not the candy with cost `4`.
+
+Given a **0-indexed** integer array `cost`, where `cost[i]` denotes the cost of the <code>i<sup>th</sup></code> candy, return _the **minimum cost** of buying **all** the candies_.
+
+**Example 1:**
+
+**Input:** cost = [1,2,3]
+
+**Output:** 5
+
+**Explanation:** We buy the candies with costs 2 and 3, and take the candy with cost 1 for free. 
+
+The total cost of buying all candies is 2 + 3 = 5. This is the **only** way we can buy the candies. 
+
+Note that we cannot buy candies with costs 1 and 3, and then take the candy with cost 2 for free. 
+
+The cost of the free candy has to be less than or equal to the minimum cost of the purchased candies.
+
+**Example 2:**
+
+**Input:** cost = [6,5,7,9,2,2]
+
+**Output:** 23
+
+**Explanation:** The way in which we can get the minimum cost is described below: 
+
+- Buy candies with costs 9 and 7 
+
+- Take the candy with cost 6 for free 
+
+- We buy candies with costs 5 and 2 
+
+- Take the last remaining candy with cost 2 for free 
+  
+Hence, the minimum cost to buy all candies is 9 + 7 + 5 + 2 = 23.
+
+**Example 3:**
+
+**Input:** cost = [5,5]
+
+**Output:** 10
+
+**Explanation:** Since there are only 2 candies, we buy both of them. There is not a third candy we can take for free. 
+
+Hence, the minimum cost to buy all candies is 5 + 5 = 10.
+
+**Constraints:**
+
+*   `1 <= cost.length <= 100`
+*   `1 <= cost[i] <= 100`

src/main/java/g2101_2200/s2145_count_the_hidden_sequences/Solution.java

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+package g2101_2200.s2145_count_the_hidden_sequences;
+
+// #Medium #Array #Prefix_Sum #2022_06_07_Time_7_ms_(36.03%)_Space_111.3_MB_(13.23%)
+
+public class Solution {
+    public int numberOfArrays(int[] diff, int lower, int upper) {
+        int n = diff.length;
+        if (lower == upper) {
+            for (int j : diff) {
+                if (j != 0) {
+                    return 0;
+                }
+            }
+        }
+        int max = -(int) 1e9;
+        int min = (int) 1e9;
+        int[] hidden = new int[n + 1];
+        hidden[0] = 0;
+        for (int i = 1; i <= n; i++) {
+            hidden[i] = hidden[i - 1] + diff[i - 1];
+        }
+        for (int i = 0; i <= n; i++) {
+            if (hidden[i] > max) {
+                max = hidden[i];
+            }
+            if (hidden[i] < min) {
+                min = hidden[i];
+            }
+        }
+        int low = lower - min;
+        int high = upper - max;
+        if (low > high) {
+            return 0;
+        }
+        return (high - low) + 1;
+    }
+}

src/main/java/g2101_2200/s2145_count_the_hidden_sequences/readme.md

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+2145\. Count the Hidden Sequences
+
+Medium
+
+You are given a **0-indexed** array of `n` integers `differences`, which describes the **differences** between each pair of **consecutive** integers of a **hidden** sequence of length `(n + 1)`. More formally, call the hidden sequence `hidden`, then we have that `differences[i] = hidden[i + 1] - hidden[i]`.
+
+You are further given two integers `lower` and `upper` that describe the **inclusive** range of values `[lower, upper]` that the hidden sequence can contain.
+
+*   For example, given `differences = [1, -3, 4]`, `lower = 1`, `upper = 6`, the hidden sequence is a sequence of length `4` whose elements are in between `1` and `6` (**inclusive**).
+    *   `[3, 4, 1, 5]` and `[4, 5, 2, 6]` are possible hidden sequences.
+    *   `[5, 6, 3, 7]` is not possible since it contains an element greater than `6`.
+    *   `[1, 2, 3, 4]` is not possible since the differences are not correct.
+
+Return _the number of **possible** hidden sequences there are._ If there are no possible sequences, return `0`.
+
+**Example 1:**
+
+**Input:** differences = [1,-3,4], lower = 1, upper = 6
+
+**Output:** 2
+
+**Explanation:** The possible hidden sequences are: 
+
+- [3, 4, 1, 5] 
+
+- [4, 5, 2, 6] 
+  
+Thus, we return 2.
+
+**Example 2:**
+
+**Input:** differences = [3,-4,5,1,-2], lower = -4, upper = 5
+
+**Output:** 4
+
+**Explanation:** The possible hidden sequences are: 
+
+- [-3, 0, -4, 1, 2, 0] 
+
+- [-2, 1, -3, 2, 3, 1] 
+
+- [-1, 2, -2, 3, 4, 2] 
+
+- [0, 3, -1, 4, 5, 3] 
+  
+Thus, we return 4.
+
+**Example 3:**
+
+**Input:** differences = [4,-7,2], lower = 3, upper = 6
+
+**Output:** 0
+
+**Explanation:** There are no possible hidden sequences. Thus, we return 0.
+
+**Constraints:**
+
+*   `n == differences.length`
+*   <code>1 <= n <= 10<sup>5</sup></code>
+*   <code>-10<sup>5</sup> <= differences[i] <= 10<sup>5</sup></code>
+*   <code>-10<sup>5</sup> <= lower <= upper <= 10<sup>5</sup></code>

src/main/java/g2101_2200/s2146_k_highest_ranked_items_within_a_price_range/Solution.java

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+package g2101_2200.s2146_k_highest_ranked_items_within_a_price_range;
+
+// #Medium #Array #Sorting #Breadth_First_Search #Matrix #Heap_Priority_Queue
+// #2022_06_07_Time_81_ms_(88.84%)_Space_65.7_MB_(99.30%)
+
+import java.util.Arrays;
+import java.util.LinkedList;
+import java.util.List;
+import java.util.Queue;
+
+public class Solution {
+    static class Item {
+        int row;
+        int col;
+        int dist;
+        int price;
+
+        public Item(int row, int col, int dist, int price) {
+            this.row = row;
+            this.col = col;
+            this.dist = dist;
+            this.price = price;
+        }
+    }
+
+    public List<List<Integer>> highestRankedKItems(
+            int[][] grid, int[] pricing, int[] start, int k) {
+        int n = grid.length;
+        int m = grid[0].length;
+        Queue<int[]> bfs = new LinkedList<>();
+        LinkedList<Item> items = new LinkedList<>();
+
+        bfs.add(start);
+        if (grid[start[0]][start[1]] >= pricing[0] && grid[start[0]][start[1]] <= pricing[1]) {
+            items.add(new Item(start[0], start[1], 0, grid[start[0]][start[1]]));
+        }
+        grid[start[0]][start[1]] = -1;
+
+        int distance = 0;
+        while (!bfs.isEmpty()) {
+            int size = bfs.size();
+            distance++;
+            while (size-- > 0) {
+                int[] loc = bfs.poll();
+                int[] dirX = {0, 1, -1, 0};
+                int[] dirY = {-1, 0, 0, 1};
+                for (int i = 0; i < 4; i++) {
+                    int newX = loc[0] + dirX[i];
+                    int newY = loc[1] + dirY[i];
+                    if (newX < 0
+                            || newX >= n
+                            || newY < 0
+                            || newY >= m
+                            || grid[newX][newY] == -1
+                            || grid[newX][newY] == 0) {
+                        continue;
+                    }
+                    if (grid[newX][newY] >= pricing[0] && grid[newX][newY] <= pricing[1]) {
+                        items.add(new Item(newX, newY, distance, grid[newX][newY]));
+                    }
+                    grid[newX][newY] = -1;
+                    bfs.add(new int[] {newX, newY});
+                }
+            }
+        }
+        items.sort(
+                (a, b) -> {
+                    int distDiff = a.dist - b.dist;
+                    if (distDiff == 0) {
+                        int priceDiff = a.price - b.price;
+                        if (priceDiff == 0) {
+                            int rowDiff = a.row - b.row;
+                            if (rowDiff == 0) {
+                                return a.col - b.col;
+                            }
+                            return rowDiff;
+                        }
+                        return priceDiff;
+                    }
+                    return distDiff;
+                });
+        List<List<Integer>> ans = new LinkedList<>();
+        while (k-- > 0 && !items.isEmpty()) {
+            Item item = items.poll();
+            ans.add(Arrays.asList(item.row, item.col));
+        }
+        return ans;
+    }
+}

src/main/java/g2101_2200/s2146_k_highest_ranked_items_within_a_price_range/readme.md

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+2146\. K Highest Ranked Items Within a Price Range
+
+Medium
+
+You are given a **0-indexed** 2D integer array `grid` of size `m x n` that represents a map of the items in a shop. The integers in the grid represent the following:
+
+*   `0` represents a wall that you cannot pass through.
+*   `1` represents an empty cell that you can freely move to and from.
+*   All other positive integers represent the price of an item in that cell. You may also freely move to and from these item cells.
+
+It takes `1` step to travel between adjacent grid cells.
+
+You are also given integer arrays `pricing` and `start` where `pricing = [low, high]` and `start = [row, col]` indicates that you start at the position `(row, col)` and are interested only in items with a price in the range of `[low, high]` (**inclusive**). You are further given an integer `k`.
+
+You are interested in the **positions** of the `k` **highest-ranked** items whose prices are **within** the given price range. The rank is determined by the **first** of these criteria that is different:
+
+1.  Distance, defined as the length of the shortest path from the `start` (**shorter** distance has a higher rank).
+2.  Price (**lower** price has a higher rank, but it must be **in the price range**).
+3.  The row number (**smaller** row number has a higher rank).
+4.  The column number (**smaller** column number has a higher rank).
+
+Return _the_ `k` _highest-ranked items within the price range **sorted** by their rank (highest to lowest)_. If there are fewer than `k` reachable items within the price range, return _**all** of them_.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2021/12/16/example1drawio.png)
+
+**Input:** grid = [[1,2,0,1],[1,3,0,1],[0,2,5,1]], pricing = [2,5], start = [0,0], k = 3
+
+**Output:** [[0,1],[1,1],[2,1]]
+
+**Explanation:** You start at (0,0). 
+
+With a price range of [2,5], we can take items from (0,1), (1,1), (2,1) and (2,2). 
+
+The ranks of these items are: 
+
+- (0,1) with distance 1 
+
+- (1,1) with distance 2 
+
+- (2,1) with distance 3 
+
+- (2,2) with distance 4 
+  
+Thus, the 3 highest ranked items in the price range are (0,1), (1,1), and (2,1).
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2021/12/16/example2drawio1.png)
+
+**Input:** grid = [[1,2,0,1],[1,3,3,1],[0,2,5,1]], pricing = [2,3], start = [2,3], k = 2
+
+**Output:** [[2,1],[1,2]]
+
+**Explanation:** You start at (2,3). 
+
+With a price range of [2,3], we can take items from (0,1), (1,1), (1,2) and (2,1). 
+
+The ranks of these items are: 
+
+- (2,1) with distance 2, price 2 
+
+- (1,2) with distance 2, price 3 
+
+- (1,1) with distance 3 
+
+- (0,1) with distance 4 
+  
+Thus, the 2 highest ranked items in the price range are (2,1) and (1,2).
+
+**Example 3:**
+
+![](https://assets.leetcode.com/uploads/2021/12/30/example3.png)
+
+**Input:** grid = [[1,1,1],[0,0,1],[2,3,4]], pricing = [2,3], start = [0,0], k = 3
+
+**Output:** [[2,1],[2,0]]
+
+**Explanation:** You start at (0,0). 
+
+With a price range of [2,3], we can take items from (2,0) and (2,1). 
+
+The ranks of these items are: 
+
+- (2,1) with distance 5 
+
+- (2,0) with distance 6 
+  
+Thus, the 2 highest ranked items in the price range are (2,1) and (2,0). 
+
+    Note that k = 3 but there are only 2 reachable items within the price range.
+
+**Constraints:**
+
+*   `m == grid.length`
+*   `n == grid[i].length`
+*   <code>1 <= m, n <= 10<sup>5</sup></code>
+*   <code>1 <= m * n <= 10<sup>5</sup></code>
+*   <code>0 <= grid[i][j] <= 10<sup>5</sup></code>
+*   `pricing.length == 2`
+*   <code>2 <= low <= high <= 10<sup>5</sup></code>
+*   `start.length == 2`
+*   `0 <= row <= m - 1`
+*   `0 <= col <= n - 1`
+*   `grid[row][col] > 0`
+*   `1 <= k <= m * n`