Added tasks 3572-3579

Author: javadev

src/main/java/g3501_3600/s3572_maximize_ysum_by_picking_a_triplet_of_distinct_xvalues/Solution.java

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+package g3501_3600.s3572_maximize_ysum_by_picking_a_triplet_of_distinct_xvalues;
+
+// #Medium #2025_06_08_Time_51_ms_(100.00%)_Space_57.50_MB_(100.00%)
+
+import java.util.Collections;
+import java.util.HashMap;
+import java.util.Map;
+import java.util.PriorityQueue;
+
+public class Solution {
+    public int maxSumDistinctTriplet(int[] x, int[] y) {
+        Map<Integer, Integer> map = new HashMap<>();
+        for (int i = 0; i < x.length; i++) {
+            map.put(x[i], Math.max(map.getOrDefault(x[i], 0), y[i]));
+        }
+        PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());
+        for (int val : map.values()) {
+            maxHeap.add(val);
+        }
+        if (maxHeap.size() < 3) {
+            return -1;
+        }
+        int sum = 0;
+        for (int i = 0; i < 3; i++) {
+            sum += maxHeap.poll();
+        }
+        return sum;
+    }
+}

src/main/java/g3501_3600/s3572_maximize_ysum_by_picking_a_triplet_of_distinct_xvalues/readme.md

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+3572\. Maximize Y‑Sum by Picking a Triplet of Distinct X‑Values
+
+Medium
+
+You are given two integer arrays `x` and `y`, each of length `n`. You must choose three **distinct** indices `i`, `j`, and `k` such that:
+
+*   `x[i] != x[j]`
+*   `x[j] != x[k]`
+*   `x[k] != x[i]`
+
+Your goal is to **maximize** the value of `y[i] + y[j] + y[k]` under these conditions. Return the **maximum** possible sum that can be obtained by choosing such a triplet of indices.
+
+If no such triplet exists, return -1.
+
+**Example 1:**
+
+**Input:** x = [1,2,1,3,2], y = [5,3,4,6,2]
+
+**Output:** 14
+
+**Explanation:**
+
+*   Choose `i = 0` (`x[i] = 1`, `y[i] = 5`), `j = 1` (`x[j] = 2`, `y[j] = 3`), `k = 3` (`x[k] = 3`, `y[k] = 6`).
+*   All three values chosen from `x` are distinct. `5 + 3 + 6 = 14` is the maximum we can obtain. Hence, the output is 14.
+
+**Example 2:**
+
+**Input:** x = [1,2,1,2], y = [4,5,6,7]
+
+**Output:** \-1
+
+**Explanation:**
+
+*   There are only two distinct values in `x`. Hence, the output is -1.
+
+**Constraints:**
+
+*   `n == x.length == y.length`
+*   <code>3 <= n <= 10<sup>5</sup></code>
+*   <code>1 <= x[i], y[i] <= 10<sup>6</sup></code>

src/main/java/g3501_3600/s3573_best_time_to_buy_and_sell_stock_v/Solution.java

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+package g3501_3600.s3573_best_time_to_buy_and_sell_stock_v;
+
+// #Medium #2025_06_08_Time_259_ms_(100.00%)_Space_87.72_MB_(100.00%)
+
+import java.util.Arrays;
+
+public class Solution {
+    private static final long MN = (long) -1e14;
+    private long[][][] dp;
+    private int[] prices;
+
+    private long f(int i, int k, int state) {
+        if (i == prices.length) {
+            return (state == 0) ? 0 : MN;
+        }
+        if (dp[i][k][state] != MN) {
+            return dp[i][k][state];
+        }
+        long p = prices[i];
+        long profit = MN;
+        profit = Math.max(profit, f(i + 1, k, state));
+        if (state == 0) {
+            profit = Math.max(profit, f(i + 1, k, 1) - p);
+            profit = Math.max(profit, f(i + 1, k, 2) + p);
+        } else if (k > 0) {
+            if (state == 1) {
+                profit = Math.max(profit, f(i + 1, k - 1, 0) + p);
+            } else {
+                profit = Math.max(profit, f(i + 1, k - 1, 0) - p);
+            }
+        }
+        return dp[i][k][state] = profit;
+    }
+
+    public long maximumProfit(int[] prices, int k) {
+        this.prices = prices;
+        int n = prices.length;
+        dp = new long[n + 1][k + 1][3];
+        for (long[][] twoD : dp) {
+            for (long[] oneD : twoD) {
+                Arrays.fill(oneD, MN);
+            }
+        }
+        return f(0, k, 0);
+    }
+}

src/main/java/g3501_3600/s3573_best_time_to_buy_and_sell_stock_v/readme.md

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+3573\. Best Time to Buy and Sell Stock V
+
+Medium
+
+You are given an integer array `prices` where `prices[i]` is the price of a stock in dollars on the <code>i<sup>th</sup></code> day, and an integer `k`.
+
+You are allowed to make at most `k` transactions, where each transaction can be either of the following:
+
+*   **Normal transaction**: Buy on day `i`, then sell on a later day `j` where `i < j`. You profit `prices[j] - prices[i]`.
+    
+*   **Short selling transaction**: Sell on day `i`, then buy back on a later day `j` where `i < j`. You profit `prices[i] - prices[j]`.
+    
+
+**Note** that you must complete each transaction before starting another. Additionally, you can't buy or sell on the same day you are selling or buying back as part of a previous transaction.
+
+Return the **maximum** total profit you can earn by making **at most** `k` transactions.
+
+**Example 1:**
+
+**Input:** prices = [1,7,9,8,2], k = 2
+
+**Output:** 14
+
+**Explanation:**
+
+We can make $14 of profit through 2 transactions:
+
+*   A normal transaction: buy the stock on day 0 for $1 then sell it on day 2 for $9.
+*   A short selling transaction: sell the stock on day 3 for $8 then buy back on day 4 for $2.
+
+**Example 2:**
+
+**Input:** prices = [12,16,19,19,8,1,19,13,9], k = 3
+
+**Output:** 36
+
+**Explanation:**
+
+We can make $36 of profit through 3 transactions:
+
+*   A normal transaction: buy the stock on day 0 for $12 then sell it on day 2 for $19.
+*   A short selling transaction: sell the stock on day 3 for $19 then buy back on day 4 for $8.
+*   A normal transaction: buy the stock on day 5 for $1 then sell it on day 6 for $19.
+
+**Constraints:**
+
+*   <code>2 <= prices.length <= 10<sup>3</sup></code>
+*   <code>1 <= prices[i] <= 10<sup>9</sup></code>
+*   `1 <= k <= prices.length / 2`

src/main/java/g3501_3600/s3574_maximize_subarray_gcd_score/Solution.java

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+package g3501_3600.s3574_maximize_subarray_gcd_score;
+
+// #Hard #2025_06_08_Time_364_ms_(100.00%)_Space_44.70_MB_(100.00%)
+
+public class Solution {
+    public long maxGCDScore(int[] nums, int k) {
+        long ans = 0;
+        int n = nums.length;
+        for (int i = 0; i < n; i++) {
+            long countGCD = 0;
+            long oddCount = 0;
+            long ongoingGCD = 0;
+            for (int j = i; j < n; j++) {
+                long currentGCD = gcd(ongoingGCD, nums[j]);
+                if (currentGCD != ongoingGCD) {
+                    ongoingGCD = currentGCD;
+                    countGCD = 1;
+                } else if (nums[j] == ongoingGCD) {
+                    countGCD++;
+                }
+                if (nums[j] % 2 != 0) {
+                    oddCount++;
+                }
+                int len = j - i + 1;
+                long res = ongoingGCD * len;
+                if (ongoingGCD % 2 != 0) {
+                    if (k >= oddCount) {
+                        res *= 2L;
+                    }
+                } else if (k >= countGCD) {
+                    res *= 2L;
+                }
+                ans = Math.max(ans, res);
+            }
+        }
+        return ans;
+    }
+
+    private long gcd(long a, long b) {
+        if (a == 0) {
+            return b;
+        }
+        return gcd(b % a, a);
+    }
+}

src/main/java/g3501_3600/s3574_maximize_subarray_gcd_score/readme.md

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+3574\. Maximize Subarray GCD Score
+
+Hard
+
+You are given an array of positive integers `nums` and an integer `k`.
+
+You may perform at most `k` operations. In each operation, you can choose one element in the array and **double** its value. Each element can be doubled **at most** once.
+
+The **score** of a contiguous **subarray** is defined as the **product** of its length and the _greatest common divisor (GCD)_ of all its elements.
+
+Your task is to return the **maximum** **score** that can be achieved by selecting a contiguous subarray from the modified array.
+
+**Note:**
+
+*   The **greatest common divisor (GCD)** of an array is the largest integer that evenly divides all the array elements.
+
+**Example 1:**
+
+**Input:** nums = [2,4], k = 1
+
+**Output:** 8
+
+**Explanation:**
+
+*   Double `nums[0]` to 4 using one operation. The modified array becomes `[4, 4]`.
+*   The GCD of the subarray `[4, 4]` is 4, and the length is 2.
+*   Thus, the maximum possible score is `2 × 4 = 8`.
+
+**Example 2:**
+
+**Input:** nums = [3,5,7], k = 2
+
+**Output:** 14
+
+**Explanation:**
+
+*   Double `nums[2]` to 14 using one operation. The modified array becomes `[3, 5, 14]`.
+*   The GCD of the subarray `[14]` is 14, and the length is 1.
+*   Thus, the maximum possible score is `1 × 14 = 14`.
+
+**Example 3:**
+
+**Input:** nums = [5,5,5], k = 1
+
+**Output:** 15
+
+**Explanation:**
+
+*   The subarray `[5, 5, 5]` has a GCD of 5, and its length is 3.
+*   Since doubling any element doesn't improve the score, the maximum score is `3 × 5 = 15`.
+
+**Constraints:**
+
+*   `1 <= n == nums.length <= 1500`
+*   <code>1 <= nums[i] <= 10<sup>9</sup></code>
+*   `1 <= k <= n`

src/main/java/g3501_3600/s3575_maximum_good_subtree_score/Solution.java

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+package g3501_3600.s3575_maximum_good_subtree_score;
+
+// #Hard #2025_06_08_Time_564_ms_(100.00%)_Space_45.22_MB_(100.00%)
+
+import java.util.ArrayList;
+import java.util.HashMap;
+import java.util.HashSet;
+import java.util.List;
+import java.util.Map;
+import java.util.Set;
+
+@SuppressWarnings("unchecked")
+public class Solution {
+    private long ans;
+    private static final int MOD = 1_000_000_007;
+
+    public int goodSubtreeSum(int[] vals, int[] par) {
+        int n = vals.length;
+        List<Integer>[] adj = new ArrayList[n];
+        for (int i = 0; i < n; i++) {
+            adj[i] = new ArrayList<>();
+        }
+        for (int i = 1; i < n; i++) {
+            adj[par[i]].add(i);
+        }
+        this.ans = 0;
+        dfs(0, vals, adj);
+        return (int) ((this.ans % MOD + MOD) % MOD);
+    }
+
+    private Map<Integer, Integer> dfs(int u, int[] vals, List<Integer>[] adj) {
+        // du: The DP map for the subtree at node u.
+        // Key: bitmask of digits. Value: max sum for that combination of digits.
+        Map<Integer, Integer> du = new HashMap<>();
+        // Base case: A sum of 0 is possible with an empty set of digits (mask 0).
+        du.put(0, 0);
+        // Process the current node's value.
+        String s = String.valueOf(Math.abs(vals[u]));
+        if (hasUniqueDigits(s)) {
+            int mask = 0;
+            for (char c : s.toCharArray()) {
+                mask |= (1 << (c - '0'));
+            }
+            du.put(mask, vals[u]);
+        }
+        for (int v : adj[u]) {
+            Map<Integer, Integer> dv = dfs(v, vals, adj);
+            Map<Integer, Integer> duSnapshot = new HashMap<>(du);
+            for (Map.Entry<Integer, Integer> entryV : dv.entrySet()) {
+                int mv = entryV.getKey();
+                int sv = entryV.getValue();
+                for (Map.Entry<Integer, Integer> entryU : duSnapshot.entrySet()) {
+                    int mu = entryU.getKey();
+                    int su = entryU.getValue();
+                    // If the digit sets are disjoint (no common bits in masks), we can combine
+                    // them.
+                    if ((mu & mv) == 0) {
+                        int newMask = mu | mv;
+                        int newSum = su + sv;
+                        // Update `du` with the best possible sum for the new combined mask.
+                        du.put(
+                                newMask,
+                                Math.max(du.getOrDefault(newMask, Integer.MIN_VALUE), newSum));
+                    }
+                }
+            }
+        }
+        // After processing all children, the max value in `du` is the "good" sum for the subtree at
+        // u.
+        // Initialize with a very small number to correctly find the maximum, even if sums are
+        // negative.
+        int maxSubtreeSum = Integer.MIN_VALUE;
+        for (int sum : du.values()) {
+            maxSubtreeSum = Math.max(maxSubtreeSum, sum);
+        }
+        // Add this subtree's best sum to the total answer.
+        // If du is empty (should not happen due to {0:0}), we add 0.
+        this.ans += (maxSubtreeSum == Integer.MIN_VALUE ? 0 : maxSubtreeSum);
+        return du;
+    }
+
+    private boolean hasUniqueDigits(String s) {
+        Set<Character> digits = new HashSet<>();
+        for (char c : s.toCharArray()) {
+            if (!digits.add(c)) {
+                return false;
+            }
+        }
+        return true;
+    }
+}