Added tasks 3627-3630

src/main/kotlin/g3601_3700/s3627_maximum_median_sum_of_subsequences_of_size_3/Solution.kt

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+package g3601_3700.s3627_maximum_median_sum_of_subsequences_of_size_3
+
+// #Medium #Weekly_Contest_460 #2025_07_27_Time_46_ms_(91.67%)_Space_133.87_MB_(16.67%)
+
+class Solution {
+    fun maximumMedianSum(nums: IntArray): Long {
+        nums.sort()
+        var i = 0
+        var j = nums.size
+        var sum = 0L
+        while (i < j) {
+            sum += nums[j - 2]
+            j = j - 2
+            i++
+        }
+        return sum
+    }
+}

src/main/kotlin/g3601_3700/s3627_maximum_median_sum_of_subsequences_of_size_3/readme.md

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+3627\. Maximum Median Sum of Subsequences of Size 3
+
+Medium
+
+You are given an integer array `nums` with a length divisible by 3.
+
+You want to make the array empty in steps. In each step, you can select any three elements from the array, compute their **median**, and remove the selected elements from the array.
+
+The **median** of an odd-length sequence is defined as the middle element of the sequence when it is sorted in non-decreasing order.
+
+Return the **maximum** possible sum of the medians computed from the selected elements.
+
+**Example 1:**
+
+**Input:** nums = [2,1,3,2,1,3]
+
+**Output:** 5
+
+**Explanation:**
+
+*   In the first step, select elements at indices 2, 4, and 5, which have a median 3. After removing these elements, `nums` becomes `[2, 1, 2]`.
+*   In the second step, select elements at indices 0, 1, and 2, which have a median 2. After removing these elements, `nums` becomes empty.
+
+Hence, the sum of the medians is `3 + 2 = 5`.
+
+**Example 2:**
+
+**Input:** nums = [1,1,10,10,10,10]
+
+**Output:** 20
+
+**Explanation:**
+
+*   In the first step, select elements at indices 0, 2, and 3, which have a median 10. After removing these elements, `nums` becomes `[1, 10, 10]`.
+*   In the second step, select elements at indices 0, 1, and 2, which have a median 10. After removing these elements, `nums` becomes empty.
+
+Hence, the sum of the medians is `10 + 10 = 20`.
+
+**Constraints:**
+
+*   <code>1 <= nums.length <= 5 * 10<sup>5</sup></code>
+*   `nums.length % 3 == 0`
+*   <code>1 <= nums[i] <= 10<sup>9</sup></code>

src/main/kotlin/g3601_3700/s3628_maximum_number_of_subsequences_after_one_inserting/Solution.kt

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+package g3601_3700.s3628_maximum_number_of_subsequences_after_one_inserting
+
+// #Medium #Weekly_Contest_460 #2025_07_27_Time_13_ms_(100.00%)_Space_48.00_MB_(75.00%)
+
+class Solution {
+    fun numOfSubsequences(s: String): Long {
+        var tc: Long = 0
+        val chs = s.toCharArray()
+        for (c in chs) {
+            tc += (if (c == 'T') 1 else 0).toLong()
+        }
+        var ls: Long = 0
+        var cs: Long = 0
+        var lcf: Long = 0
+        var ctf: Long = 0
+        var lct: Long = 0
+        var ocg: Long = 0
+        var tp: Long = 0
+        for (curr in chs) {
+            val rt = tc - tp
+            val cg = ls * rt
+            ocg = if (cg > ocg) cg else ocg
+            if (curr == 'L') {
+                ls++
+            } else {
+                if (curr == 'C') {
+                    cs++
+                    lcf += ls
+                } else {
+                    if (curr == 'T') {
+                        lct += lcf
+                        ctf += cs
+                        tp++
+                    }
+                }
+            }
+        }
+        val fcg = ls * (tc - tp)
+        ocg = if (fcg > ocg) fcg else ocg
+        var maxi: Long = 0
+        val bo = longArrayOf(lcf, ctf, ocg)
+        for (op in bo) {
+            maxi = if (op > maxi) op else maxi
+        }
+        return lct + maxi
+    }
+}

src/main/kotlin/g3601_3700/s3628_maximum_number_of_subsequences_after_one_inserting/readme.md

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+3628\. Maximum Number of Subsequences After One Inserting
+
+Medium
+
+You are given a string `s` consisting of uppercase English letters.
+
+You are allowed to insert **at most one** uppercase English letter at **any** position (including the beginning or end) of the string.
+
+Return the **maximum** number of `"LCT"` subsequences that can be formed in the resulting string after **at most one insertion**.
+
+**Example 1:**
+
+**Input:** s = "LMCT"
+
+**Output:** 2
+
+**Explanation:**
+
+We can insert a `"L"` at the beginning of the string s to make `"LLMCT"`, which has 2 subsequences, at indices [0, 3, 4] and [1, 3, 4].
+
+**Example 2:**
+
+**Input:** s = "LCCT"
+
+**Output:** 4
+
+**Explanation:**
+
+We can insert a `"L"` at the beginning of the string s to make `"LLCCT"`, which has 4 subsequences, at indices [0, 2, 4], [0, 3, 4], [1, 2, 4] and [1, 3, 4].
+
+**Example 3:**
+
+**Input:** s = "L"
+
+**Output:** 0
+
+**Explanation:**
+
+Since it is not possible to obtain the subsequence `"LCT"` by inserting a single letter, the result is 0.
+
+**Constraints:**
+
+*   <code>1 <= s.length <= 10<sup>5</sup></code>
+*   `s` consists of uppercase English letters.

src/main/kotlin/g3601_3700/s3629_minimum_jumps_to_reach_end_via_prime_teleportation/Solution.kt

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+package g3601_3700.s3629_minimum_jumps_to_reach_end_via_prime_teleportation
+
+// #Medium #Weekly_Contest_460 #2025_07_27_Time_406_ms_(100.00%)_Space_153.64_MB_(100.00%)
+
+import java.util.ArrayDeque
+import kotlin.math.max
+
+class Solution {
+    fun minJumps(nums: IntArray): Int {
+        val n = nums.size
+        if (n == 1) {
+            return 0
+        }
+        var maxVal = 0
+        for (v in nums) {
+            maxVal = max(maxVal, v)
+        }
+        val isPrime = sieve(maxVal)
+        val posOfValue: Array<ArrayList<Int>> = Array<ArrayList<Int>>(maxVal + 1) { ArrayList<Int>() }
+        for (i in 0..<n) {
+            val v = nums[i]
+            posOfValue[v].add(i)
+        }
+        val primeProcessed = BooleanArray(maxVal + 1)
+        val dist = IntArray(n)
+        dist.fill(-1)
+        val q = ArrayDeque<Int>()
+        q.add(0)
+        dist[0] = 0
+        while (q.isNotEmpty()) {
+            val i: Int = q.poll()!!
+            val d = dist[i]
+            if (i == n - 1) {
+                return d
+            }
+            if (i + 1 < n && dist[i + 1] == -1) {
+                dist[i + 1] = d + 1
+                q.add(i + 1)
+            }
+            if (i - 1 >= 0 && dist[i - 1] == -1) {
+                dist[i - 1] = d + 1
+                q.add(i - 1)
+            }
+            val v = nums[i]
+            if (v <= maxVal && isPrime[v] && !primeProcessed[v]) {
+                var mult = v
+                while (mult <= maxVal) {
+                    val list = posOfValue[mult]
+                    for (idx in list) {
+                        if (dist[idx] == -1) {
+                            dist[idx] = d + 1
+                            q.add(idx)
+                        }
+                    }
+                    mult += v
+                }
+                primeProcessed[v] = true
+            }
+        }
+        return -1
+    }
+
+    private fun sieve(n: Int): BooleanArray {
+        val prime = BooleanArray(n + 1)
+        if (n >= 2) {
+            prime.fill(true)
+        }
+        if (n >= 0) {
+            prime[0] = false
+        }
+        if (n >= 1) {
+            prime[1] = false
+        }
+        var i = 2
+        while (i.toLong() * i <= n) {
+            if (prime[i]) {
+                var j = i * i
+                while (j <= n) {
+                    prime[j] = false
+                    j += i
+                }
+            }
+            i++
+        }
+        return prime
+    }
+}

src/main/kotlin/g3601_3700/s3629_minimum_jumps_to_reach_end_via_prime_teleportation/readme.md

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+3629\. Minimum Jumps to Reach End via Prime Teleportation
+
+Medium
+
+You are given an integer array `nums` of length `n`.
+
+You start at index 0, and your goal is to reach index `n - 1`.
+
+From any index `i`, you may perform one of the following operations:
+
+*   **Adjacent Step**: Jump to index `i + 1` or `i - 1`, if the index is within bounds.
+*   **Prime Teleportation**: If `nums[i]` is a prime number `p`, you may instantly jump to any index `j != i` such that `nums[j] % p == 0`.
+
+Return the **minimum** number of jumps required to reach index `n - 1`.
+
+**Example 1:**
+
+**Input:** nums = [1,2,4,6]
+
+**Output:** 2
+
+**Explanation:**
+
+One optimal sequence of jumps is:
+
+*   Start at index `i = 0`. Take an adjacent step to index 1.
+*   At index `i = 1`, `nums[1] = 2` is a prime number. Therefore, we teleport to index `i = 3` as `nums[3] = 6` is divisible by 2.
+
+Thus, the answer is 2.
+
+**Example 2:**
+
+**Input:** nums = [2,3,4,7,9]
+
+**Output:** 2
+
+**Explanation:**
+
+One optimal sequence of jumps is:
+
+*   Start at index `i = 0`. Take an adjacent step to index `i = 1`.
+*   At index `i = 1`, `nums[1] = 3` is a prime number. Therefore, we teleport to index `i = 4` since `nums[4] = 9` is divisible by 3.
+
+Thus, the answer is 2.
+
+**Example 3:**
+
+**Input:** nums = [4,6,5,8]
+
+**Output:** 3
+
+**Explanation:**
+
+*   Since no teleportation is possible, we move through `0 → 1 → 2 → 3`. Thus, the answer is 3.
+
+**Constraints:**
+
+*   <code>1 <= n == nums.length <= 10<sup>5</sup></code>
+*   <code>1 <= nums[i] <= 10<sup>6</sup></code>

src/main/kotlin/g3601_3700/s3630_partition_array_for_maximum_xor_and_and/Solution.kt

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+package g3601_3700.s3630_partition_array_for_maximum_xor_and_and
+
+// #Hard #Array #Math #Greedy #Enumeration #Weekly_Contest_460
+// #2025_08_03_Time_57_ms_(100.00%)_Space_51.81_MB_(100.00%)
+
+class Solution {
+    fun maximizeXorAndXor(nums: IntArray): Long {
+        val n = nums.size
+        val full = 1 shl n
+        val xorMask = IntArray(full)
+        val andMask = IntArray(full)
+        val orMask = IntArray(full)
+        for (mask in 1..<full) {
+            val lb = mask and -mask
+            val i = Integer.numberOfTrailingZeros(lb)
+            val prev = mask xor lb
+            xorMask[mask] = xorMask[prev] xor nums[i]
+            andMask[mask] = if (prev == 0) nums[i] else andMask[prev] and nums[i]
+            orMask[mask] = orMask[prev] or nums[i]
+        }
+        var best: Long = 0
+        val all = full - 1
+        for (b in 0..<full) {
+            val andB = andMask[b].toLong()
+            val rest = all xor b
+            if (andB + 2L * orMask[rest] <= best) {
+                continue
+            }
+            var a = rest
+            while (true) {
+                val c = rest xor a
+                val sum = xorMask[a] + andB + xorMask[c]
+                if (sum > best) {
+                    best = sum
+                }
+                if (a == 0) {
+                    break
+                }
+                a = (a - 1) and rest
+            }
+        }
+        return best
+    }
+}