Added tasks 3572-3579

Author: javadev

src/main/kotlin/g3501_3600/s3572_maximize_ysum_by_picking_a_triplet_of_distinct_xvalues/Solution.kt

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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_5_ms_(100.00%)_Space_82.11_MB_(56.00%)
+
+class Solution {
+    fun maxSumDistinctTriplet(x: IntArray, y: IntArray): Int {
+        var index = -1
+        var max = -1
+        var sum = 0
+        for (i in y.indices) {
+            if (y[i] > max) {
+                max = y[i]
+                index = i
+            }
+        }
+        sum += max
+        if (max == -1) {
+            return -1
+        }
+        var index2 = -1
+        max = -1
+        for (i in y.indices) {
+            if (y[i] > max && x[i] != x[index]) {
+                max = y[i]
+                index2 = i
+            }
+        }
+        sum += max
+        if (max == -1) {
+            return -1
+        }
+        max = -1
+        for (i in y.indices) {
+            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/kotlin/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/kotlin/g3501_3600/s3573_best_time_to_buy_and_sell_stock_v/Solution.kt

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+package g3501_3600.s3573_best_time_to_buy_and_sell_stock_v
+
+// #Medium #Array #Dynamic_Programming #2025_06_10_Time_27_ms_(100.00%)_Space_48.69_MB_(80.00%)
+
+import kotlin.math.max
+
+class Solution {
+    fun maximumProfit(prices: IntArray, k: Int): Long {
+        val n = prices.size
+        var prev = LongArray(n)
+        var curr = LongArray(n)
+        for (t in 1..k) {
+            var bestLong = -prices[0].toLong()
+            var bestShort = prices[0].toLong()
+            curr[0] = 0
+            for (i in 1..<n) {
+                var res = curr[i - 1]
+                res = max(res, prices[i] + bestLong)
+                res = max(res, -prices[i] + bestShort)
+                curr[i] = res
+                bestLong = max(bestLong, prev[i - 1] - prices[i])
+                bestShort = max(bestShort, prev[i - 1] + prices[i])
+            }
+            val tmp = prev
+            prev = curr
+            curr = tmp
+        }
+        return prev[n - 1]
+    }
+}

src/main/kotlin/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/kotlin/g3501_3600/s3574_maximize_subarray_gcd_score/Solution.kt

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+package g3501_3600.s3574_maximize_subarray_gcd_score
+
+// #Hard #Array #Math #Enumeration #Number_Theory
+// #2025_06_10_Time_19_ms_(100.00%)_Space_50.12_MB_(100.00%)
+
+import kotlin.math.max
+
+class Solution {
+    fun maxGCDScore(nums: IntArray, k: Int): Long {
+        var mx = 0
+        for (x in nums) {
+            mx = max(mx, x)
+        }
+        val width = 32 - Integer.numberOfLeadingZeros(mx)
+        val lowBitPos: Array<MutableList<Int>> = Array<MutableList<Int>>(width) { _ -> ArrayList<Int>() }
+        val intervals = Array<IntArray>(width + 1) { IntArray(3) }
+        var size = 0
+        var ans: Long = 0
+        for (i in nums.indices) {
+            val x = nums[i]
+            val tz = Integer.numberOfTrailingZeros(x)
+            lowBitPos[tz].add(i)
+            for (j in 0..<size) {
+                intervals[j][0] = gcd(intervals[j][0], x)
+            }
+            intervals[size][0] = x
+            intervals[size][1] = i - 1
+            intervals[size][2] = i
+            size++
+            var idx = 1
+            for (j in 1..<size) {
+                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 (j in 0..<size) {
+                val g = intervals[j][0]
+                val l = intervals[j][1]
+                val r = intervals[j][2]
+                ans = max(ans, g.toLong() * (i - l))
+                val pos = lowBitPos[Integer.numberOfTrailingZeros(g)]
+                val minL = if (pos.size > k) max(l, pos[pos.size - k - 1]) else l
+                if (minL < r) {
+                    ans = max(ans, g.toLong() * 2 * (i - minL))
+                }
+            }
+        }
+        return ans
+    }
+
+    private fun gcd(a: Int, b: Int): Int {
+        var a = a
+        var b = b
+        while (a != 0) {
+            val tmp = a
+            a = b % a
+            b = tmp
+        }
+        return b
+    }
+}

src/main/kotlin/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/kotlin/g3501_3600/s3575_maximum_good_subtree_score/Solution.kt

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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_71_ms_(100.00%)_Space_78.07_MB_(0.00%)
+
+import kotlin.math.max
+
+class Solution {
+    private val digits = 10
+    private val full = 1 shl digits
+    private val neg = Long.Companion.MIN_VALUE / 4
+    private val mod = 1e9.toLong() + 7
+    private lateinit var tree: Array<ArrayList<Int>>
+    private lateinit var `val`: IntArray
+    private lateinit var mask: IntArray
+    private lateinit var isOk: BooleanArray
+    private var res: Long = 0
+
+    fun goodSubtreeSum(vals: IntArray, par: IntArray): Int {
+        val n = vals.size
+        `val` = vals
+        mask = IntArray(n)
+        isOk = BooleanArray(n)
+        for (i in 0..<n) {
+            var m = 0
+            var v = vals[i]
+            var valid = true
+            while (v > 0) {
+                val d = v % 10
+                if (((m shr d) and 1) == 1) {
+                    valid = false
+                    break
+                }
+                m = m or (1 shl d)
+                v /= 10
+            }
+            mask[i] = m
+            isOk[i] = valid
+        }
+        tree = Array(n) { initialCapacity: Int -> ArrayList(initialCapacity) }
+        val root = 0
+        for (i in 1..<n) {
+            tree[par[i]].add(i)
+        }
+        dfs(root)
+        return (res % mod).toInt()
+    }
+
+    private fun dfs(u: Int): LongArray {
+        var dp = LongArray(full)
+        dp.fill(neg)
+        dp[0] = 0
+        if (isOk[u]) {
+            dp[mask[u]] = `val`[u].toLong()
+        }
+        for (v in tree[u]) {
+            val child = dfs(v)
+            val newDp = dp.copyOf(full)
+            for (m1 in 0..<full) {
+                if (dp[m1] < 0) {
+                    continue
+                }
+                val remain = full - 1 - m1
+                var m2 = remain
+                while (m2 > 0) {
+                    if (child[m2] < 0) {
+                        m2 = (m2 - 1) and remain
+                        continue
+                    }
+                    val newM = m1 or m2
+                    newDp[newM] = max(newDp[newM], dp[m1] + child[m2])
+                    m2 = (m2 - 1) and remain
+                }
+            }
+            dp = newDp
+        }
+        var best: Long = 0
+        for (v in dp) {
+            best = max(best, v)
+        }
+        res = (res + best) % mod
+        return dp
+    }
+}