Added tasks 3572-3579

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 #Array #Hash_Table #Sorting #Greedy #Heap_Priority_Queue
+// #2025_06_10_Time_2_ms_(100.00%)_Space_64.25_MB_(40.62%)
+
+public class Solution {
+    public int maxSumDistinctTriplet(int[] x, int[] y) {
+        int index = -1;
+        int max = -1;
+        int sum = 0;
+        for (int i = 0; i < y.length; i++) {
+            if (y[i] > max) {
+                max = y[i];
+                index = i;
+            }
+        }
+        sum += max;
+        if (max == -1) {
+            return -1;
+        }
+        int index2 = -1;
+        max = -1;
+        for (int i = 0; i < y.length; i++) {
+            if (y[i] > max && x[i] != x[index]) {
+                max = y[i];
+                index2 = i;
+            }
+        }
+        sum += max;
+        if (max == -1) {
+            return -1;
+        }
+        max = -1;
+        for (int i = 0; i < y.length; i++) {
+            if (y[i] > max && x[i] != x[index] && x[i] != x[index2]) {
+                max = y[i];
+            }
+        }
+        if (max == -1) {
+            return -1;
+        }
+        sum += max;
+        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 #Array #Dynamic_Programming #2025_06_10_Time_10_ms_(99.46%)_Space_44.46_MB_(97.36%)
+
+public class Solution {
+    public long maximumProfit(int[] prices, int k) {
+        int n = prices.length;
+        long[] prev = new long[n];
+        long[] curr = new long[n];
+        for (int t = 1; t <= k; t++) {
+            long bestLong = -prices[0];
+            long bestShort = prices[0];
+            curr[0] = 0;
+            for (int i = 1; i < n; i++) {
+                long res = curr[i - 1];
+                res = Math.max(res, prices[i] + bestLong);
+                res = Math.max(res, -prices[i] + bestShort);
+                curr[i] = res;
+                bestLong = Math.max(bestLong, prev[i - 1] - prices[i]);
+                bestShort = Math.max(bestShort, prev[i - 1] + prices[i]);
+            }
+            long[] tmp = prev;
+            prev = curr;
+            curr = tmp;
+        }
+        return prev[n - 1];
+    }
+}

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 #Array #Math #Enumeration #Number_Theory
+// #2025_06_10_Time_13_ms_(100.00%)_Space_45.07_MB_(78.08%)
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
+
+@SuppressWarnings("unchecked")
+public class Solution {
+    public long maxGCDScore(int[] nums, int k) {
+        int mx = 0;
+        for (int x : nums) {
+            mx = Math.max(mx, x);
+        }
+        int width = 32 - Integer.numberOfLeadingZeros(mx);
+        List<Integer>[] lowbitPos = new List[width];
+        Arrays.setAll(lowbitPos, i -> new ArrayList<>());
+        int[][] intervals = new int[width + 1][3];
+        int size = 0;
+        long ans = 0;
+        for (int i = 0; i < nums.length; i++) {
+            int x = nums[i];
+            int tz = Integer.numberOfTrailingZeros(x);
+            lowbitPos[tz].add(i);
+            for (int j = 0; j < size; j++) {
+                intervals[j][0] = gcd(intervals[j][0], x);
+            }
+            intervals[size][0] = x;
+            intervals[size][1] = i - 1;
+            intervals[size][2] = i;
+            size++;
+            int idx = 1;
+            for (int j = 1; j < size; j++) {
+                if (intervals[j][0] != intervals[j - 1][0]) {
+                    intervals[idx][0] = intervals[j][0];
+                    intervals[idx][1] = intervals[j][1];
+                    intervals[idx][2] = intervals[j][2];
+                    idx++;
+                } else {
+                    intervals[idx - 1][2] = intervals[j][2];
+                }
+            }
+            size = idx;
+            for (int j = 0; j < size; j++) {
+                int g = intervals[j][0];
+                int l = intervals[j][1];
+                int r = intervals[j][2];
+                ans = Math.max(ans, (long) g * (i - l));
+                List<Integer> pos = lowbitPos[Integer.numberOfTrailingZeros(g)];
+                int minL = pos.size() > k ? Math.max(l, pos.get(pos.size() - k - 1)) : l;
+                if (minL < r) {
+                    ans = Math.max(ans, (long) g * 2 * (i - minL));
+                }
+            }
+        }
+        return ans;
+    }
+
+    private int gcd(int a, int b) {
+        while (a != 0) {
+            int tmp = a;
+            a = b % a;
+            b = tmp;
+        }
+        return b;
+    }
+}

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 #Array #Dynamic_Programming #Depth_First_Search #Tree #Bit_Manipulation #Bitmask
+// #2025_06_10_Time_92_ms_(98.73%)_Space_55.23_MB_(11.71%)
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
+
+@SuppressWarnings("unchecked")
+public class Solution {
+    private int digits = 10;
+    private int full = 1 << digits;
+    private long neg = Long.MIN_VALUE / 4;
+    private long mod = (long) 1e9 + 7;
+    private List<Integer>[] tree;
+    private int[] val;
+    private int[] mask;
+    private boolean[] isOk;
+    private long res = 0;
+
+    public int goodSubtreeSum(int[] vals, int[] par) {
+        int n = vals.length;
+        val = vals;
+        mask = new int[n];
+        isOk = new boolean[n];
+        for (int i = 0; i < n; i++) {
+            int m = 0;
+            int v = vals[i];
+            boolean valid = true;
+            while (v > 0) {
+                int d = v % 10;
+                if (((m >> d) & 1) == 1) {
+                    valid = false;
+                    break;
+                }
+                m |= 1 << d;
+                v /= 10;
+            }
+            mask[i] = m;
+            isOk[i] = valid;
+        }
+        tree = new ArrayList[n];
+        Arrays.setAll(tree, ArrayList::new);
+        int root = 0;
+        for (int i = 1; i < n; i++) {
+            tree[par[i]].add(i);
+        }
+        dfs(root);
+        return (int) (res % mod);
+    }
+
+    private long[] dfs(int u) {
+        long[] dp = new long[full];
+        Arrays.fill(dp, neg);
+        dp[0] = 0;
+        if (isOk[u]) {
+            dp[mask[u]] = val[u];
+        }
+        for (int v : tree[u]) {
+            long[] child = dfs(v);
+            long[] newDp = Arrays.copyOf(dp, full);
+            for (int m1 = 0; m1 < full; m1++) {
+                if (dp[m1] < 0) {
+                    continue;
+                }
+                int remain = full - 1 - m1;
+                for (int m2 = remain; m2 > 0; m2 = (m2 - 1) & remain) {
+                    if (child[m2] < 0) {
+                        continue;
+                    }
+                    int newM = m1 | m2;
+                    newDp[newM] = Math.max(newDp[newM], dp[m1] + child[m2]);
+                }
+            }
+            dp = newDp;
+        }
+        long best = 0;
+        for (long v : dp) {
+            best = Math.max(best, v);
+        }
+        res = (res + best) % mod;
+        return dp;
+    }
+}