Added tasks 3627-3630

Author: javadev

src/main/java/g3601_3700/s3627_maximum_median_sum_of_subsequences_of_size_3/Solution.java

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+package g3601_3700.s3627_maximum_median_sum_of_subsequences_of_size_3;
+
+// #Medium #Weekly_Contest_460 #2025_07_27_Time_22_ms_(100.00%)_Space_129.50_MB_(86.95%)
+
+import java.util.Arrays;
+
+public class Solution {
+    public long maximumMedianSum(int[] nums) {
+        int n = nums.length;
+        Arrays.sort(nums);
+        int m = n / 3;
+        long sum = 0;
+        for (int i = n - 2; i >= n - 2 * m; i = i - 2) {
+            sum = sum + nums[i];
+        }
+        return sum;
+    }
+}

src/main/java/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/java/g3601_3700/s3628_maximum_number_of_subsequences_after_one_inserting/Solution.java

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+package g3601_3700.s3628_maximum_number_of_subsequences_after_one_inserting;
+
+// #Medium #Weekly_Contest_460 #2025_07_27_Time_12_ms_(100.00%)_Space_45.76_MB_(72.28%)
+
+public class Solution {
+    public long numOfSubsequences(String s) {
+        long tc = 0;
+        char[] chs = s.toCharArray();
+        for (char c : chs) {
+            tc += (c == 'T') ? 1 : 0;
+        }
+        long ls = 0;
+        long cs = 0;
+        long lcf = 0;
+        long ctf = 0;
+        long lct = 0;
+        long ocg = 0;
+        long tp = 0;
+        for (char curr : chs) {
+            long rt = tc - tp;
+            long cg = ls * rt;
+            ocg = (cg > ocg) ? cg : ocg;
+            if (curr == 'L') {
+                ls++;
+            } else {
+                if (curr == 'C') {
+                    cs++;
+                    lcf += ls;
+                } else {
+                    if (curr == 'T') {
+                        lct += lcf;
+                        ctf += cs;
+                        tp++;
+                    }
+                }
+            }
+        }
+        long fcg = ls * (tc - tp);
+        ocg = fcg > ocg ? fcg : ocg;
+        long maxi = 0;
+        long[] bo = {lcf, ctf, ocg};
+        for (long op : bo) {
+            maxi = op > maxi ? op : maxi;
+        }
+        return lct + maxi;
+    }
+}

src/main/java/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/java/g3601_3700/s3629_minimum_jumps_to_reach_end_via_prime_teleportation/Solution.java

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+package g3601_3700.s3629_minimum_jumps_to_reach_end_via_prime_teleportation;
+
+// #Medium #Weekly_Contest_460 #2025_07_27_Time_116_ms_(99.81%)_Space_76.00_MB_(67.96%)
+
+import java.util.ArrayDeque;
+import java.util.ArrayList;
+import java.util.Arrays;
+
+public class Solution {
+    public int minJumps(int[] nums) {
+        int n = nums.length;
+        if (n == 1) {
+            return 0;
+        }
+        int maxVal = 0;
+        for (int v : nums) {
+            maxVal = Math.max(maxVal, v);
+        }
+        boolean[] isPrime = sieve(maxVal);
+        @SuppressWarnings("unchecked")
+        ArrayList<Integer>[] posOfValue = new ArrayList[maxVal + 1];
+        for (int i = 0; i < n; i++) {
+            int v = nums[i];
+            if (posOfValue[v] == null) {
+                posOfValue[v] = new ArrayList<>();
+            }
+            posOfValue[v].add(i);
+        }
+        boolean[] primeProcessed = new boolean[maxVal + 1];
+        int[] dist = new int[n];
+        Arrays.fill(dist, -1);
+        ArrayDeque<Integer> q = new ArrayDeque<>();
+        q.add(0);
+        dist[0] = 0;
+        while (!q.isEmpty()) {
+            int i = q.poll();
+            int 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);
+            }
+            int v = nums[i];
+            if (v <= maxVal && isPrime[v] && !primeProcessed[v]) {
+                for (int mult = v; mult <= maxVal; mult += v) {
+                    ArrayList<Integer> list = posOfValue[mult];
+                    if (list != null) {
+                        for (int idx : list) {
+                            if (dist[idx] == -1) {
+                                dist[idx] = d + 1;
+                                q.add(idx);
+                            }
+                        }
+                    }
+                }
+                primeProcessed[v] = true;
+            }
+        }
+        return -1;
+    }
+
+    private boolean[] sieve(int n) {
+        boolean[] prime = new boolean[n + 1];
+        if (n >= 2) {
+            Arrays.fill(prime, true);
+        }
+        if (n >= 0) {
+            prime[0] = false;
+        }
+        if (n >= 1) {
+            prime[1] = false;
+        }
+        for (int i = 2; (long) i * i <= n; i++) {
+            if (prime[i]) {
+                for (int j = i * i; j <= n; j += i) {
+                    prime[j] = false;
+                }
+            }
+        }
+        return prime;
+    }
+}

src/main/java/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/java/g3601_3700/s3630_partition_array_for_maximum_xor_and_and/Solution.java

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+package g3601_3700.s3630_partition_array_for_maximum_xor_and_and;
+
+// #Hard #Weekly_Contest_460 #2025_07_27_Time_77_ms_(100.00%)_Space_51.20_MB_(30.86%)
+
+public class Solution {
+    public long maximizeXorAndXor(int[] nums) {
+        int n = nums.length;
+        int full = 1 << n;
+        int[] xorMask = new int[full];
+        int[] andMask = new int[full];
+        int[] orMask = new int[full];
+        for (int mask = 1; mask < full; mask++) {
+            int lb = mask & -mask;
+            int i = Integer.numberOfTrailingZeros(lb);
+            int prev = mask ^ lb;
+            xorMask[mask] = xorMask[prev] ^ nums[i];
+            andMask[mask] = prev == 0 ? nums[i] : andMask[prev] & nums[i];
+            orMask[mask] = orMask[prev] | nums[i];
+        }
+        long best = 0;
+        int all = full - 1;
+        for (int b = 0; b < full; b++) {
+            long andB = andMask[b];
+            int rest = all ^ b;
+            if (andB + 2L * orMask[rest] <= best) {
+                continue;
+            }
+            for (int a = rest; ; a = (a - 1) & rest) {
+                int c = rest ^ a;
+                long sum = xorMask[a] + andB + xorMask[c];
+                if (sum > best) {
+                    best = sum;
+                }
+                if (a == 0) {
+                    break;
+                }
+            }
+        }
+        return best;
+    }
+}