Added tasks 3678-3686

src/main/kotlin/g3601_3700/s3678_smallest_absent_positive_greater_than_average/Solution.kt

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+package g3601_3700.s3678_smallest_absent_positive_greater_than_average
+
+// #Easy #Biweekly_Contest_165 #2025_09_20_Time_3_ms_(100.00%)_Space_47.84_MB_(100.00%)
+
+class Solution {
+    fun smallestAbsent(nums: IntArray): Int {
+        var sum = 0
+        for (j in nums) {
+            sum += j
+        }
+        val avg = sum.toDouble() / nums.size
+        var num: Int
+        if (avg < 0) {
+            num = 1
+        } else {
+            num = avg.toInt() + 1
+        }
+        while (true) {
+            var flag = false
+            for (j in nums) {
+                if (num == j) {
+                    flag = true
+                    break
+                }
+            }
+            if (!flag && num > avg) {
+                return num
+            }
+            num++
+        }
+    }
+}

src/main/kotlin/g3601_3700/s3678_smallest_absent_positive_greater_than_average/readme.md

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+3678\. Smallest Absent Positive Greater Than Average
+
+Easy
+
+You are given an integer array `nums`.
+
+Return the **smallest absent positive** integer in `nums` such that it is **strictly greater** than the **average** of all elements in `nums`.
+
+The **average** of an array is defined as the sum of all its elements divided by the number of elements.
+
+**Example 1:**
+
+**Input:** nums = [3,5]
+
+**Output:** 6
+
+**Explanation:**
+
+*   The average of `nums` is `(3 + 5) / 2 = 8 / 2 = 4`.
+*   The smallest absent positive integer greater than 4 is 6.
+
+**Example 2:**
+
+**Input:** nums = [-1,1,2]
+
+**Output:** 3
+
+**Explanation:**
+
+*   The average of `nums` is `(-1 + 1 + 2) / 3 = 2 / 3 = 0.667`.
+*   The smallest absent positive integer greater than 0.667 is 3.
+
+**Example 3:**
+
+**Input:** nums = [4,-1]
+
+**Output:** 2
+
+**Explanation:**
+
+*   The average of `nums` is `(4 + (-1)) / 2 = 3 / 2 = 1.50`.
+*   The smallest absent positive integer greater than 1.50 is 2.
+
+**Constraints:**
+
+*   `1 <= nums.length <= 100`
+*   `-100 <= nums[i] <= 100`

src/main/kotlin/g3601_3700/s3679_minimum_discards_to_balance_inventory/Solution.kt

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+package g3601_3700.s3679_minimum_discards_to_balance_inventory
+
+// #Medium #Biweekly_Contest_165 #2025_09_20_Time_6_ms_(100.00%)_Space_68.59_MB_(76.92%)
+
+import kotlin.math.max
+
+class Solution {
+    fun minArrivalsToDiscard(arrivals: IntArray, w: Int, m: Int): Int {
+        val n = arrivals.size
+        var dis = 0
+        val removed = BooleanArray(n)
+        var maxVal = 0
+        for (v in arrivals) {
+            maxVal = max(maxVal, v)
+        }
+        val freq = IntArray(maxVal + 1)
+        for (i in 0..<n) {
+            val outIdx = i - w
+            if (outIdx >= 0 && !removed[outIdx]) {
+                val oldVal = arrivals[outIdx]
+                freq[oldVal]--
+            }
+            val `val` = arrivals[i]
+            if (freq[`val`] >= m) {
+                dis++
+                removed[i] = true
+            } else {
+                freq[`val`]++
+            }
+        }
+        return dis
+    }
+}

src/main/kotlin/g3601_3700/s3679_minimum_discards_to_balance_inventory/readme.md

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+3679\. Minimum Discards to Balance Inventory
+
+Medium
+
+You are given two integers `w` and `m`, and an integer array `arrivals`, where `arrivals[i]` is the type of item arriving on day `i` (days are **1-indexed**).
+
+Items are managed according to the following rules:
+
+*   Each arrival may be **kept** or **discarded**; an item may only be discarded on its arrival day.
+*   For each day `i`, consider the window of days `[max(1, i - w + 1), i]` (the `w` most recent days up to day `i`):
+    *   For **any** such window, each item type may appear **at most** `m` times among kept arrivals whose arrival day lies in that window.
+    *   If keeping the arrival on day `i` would cause its type to appear **more than** `m` times in the window, that arrival **must** be discarded.
+
+Return the **minimum** number of arrivals to be discarded so that every `w`\-day window contains at most `m` occurrences of each type.
+
+**Example 1:**
+
+**Input:** arrivals = [1,2,1,3,1], w = 4, m = 2
+
+**Output:** 0
+
+**Explanation:**
+
+*   On day 1, Item 1 arrives; the window contains no more than `m` occurrences of this type, so we keep it.
+*   On day 2, Item 2 arrives; the window of days 1 - 2 is fine.
+*   On day 3, Item 1 arrives, window `[1, 2, 1]` has item 1 twice, within limit.
+*   On day 4, Item 3 arrives, window `[1, 2, 1, 3]` has item 1 twice, allowed.
+*   On day 5, Item 1 arrives, window `[2, 1, 3, 1]` has item 1 twice, still valid.
+
+There are no discarded items, so return 0.
+
+**Example 2:**
+
+**Input:** arrivals = [1,2,3,3,3,4], w = 3, m = 2
+
+**Output:** 1
+
+**Explanation:**
+
+*   On day 1, Item 1 arrives. We keep it.
+*   On day 2, Item 2 arrives, window `[1, 2]` is fine.
+*   On day 3, Item 3 arrives, window `[1, 2, 3]` has item 3 once.
+*   On day 4, Item 3 arrives, window `[2, 3, 3]` has item 3 twice, allowed.
+*   On day 5, Item 3 arrives, window `[3, 3, 3]` has item 3 three times, exceeds limit, so the arrival must be discarded.
+*   On day 6, Item 4 arrives, window `[3, 4]` is fine.
+
+Item 3 on day 5 is discarded, and this is the minimum number of arrivals to discard, so return 1.
+
+**Constraints:**
+
+*   <code>1 <= arrivals.length <= 10<sup>5</sup></code>
+*   <code>1 <= arrivals[i] <= 10<sup>5</sup></code>
+*   `1 <= w <= arrivals.length`
+*   `1 <= m <= w`

src/main/kotlin/g3601_3700/s3680_generate_schedule/Solution.kt

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+package g3601_3700.s3680_generate_schedule
+
+// #Medium #Biweekly_Contest_165 #2025_09_20_Time_3_ms_(100.00%)_Space_48.94_MB_(100.00%)
+
+class Solution {
+    fun generateSchedule(n: Int): Array<IntArray> {
+        if (n < 5) {
+            return Array<IntArray>(0) { IntArray(0) }
+        }
+        val res = Array<IntArray>(n * (n - 1)) { IntArray(2) }
+        var idx = 0
+        for (i in 2..<n - 1) {
+            for (j in 0..<n) {
+                res[idx++] = intArrayOf(j, (j + i) % n)
+            }
+        }
+        for (i in 0..<n) {
+            res[idx++] = intArrayOf(i, (i + 1) % n)
+            res[idx++] = intArrayOf((i + 4) % n, (i + 3) % n)
+        }
+        return res
+    }
+}

src/main/kotlin/g3601_3700/s3680_generate_schedule/readme.md

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+3680\. Generate Schedule
+
+Medium
+
+You are given an integer `n` representing `n` teams. You are asked to generate a schedule such that:
+
+*   Each team plays every other team **exactly twice**: once at home and once away.
+*   There is **exactly one** match per day; the schedule is a list of **consecutive** days and `schedule[i]` is the match on day `i`.
+*   No team plays on **consecutive** days.
+
+Return a 2D integer array `schedule`, where `schedule[i][0]` represents the home team and `schedule[i][1]` represents the away team. If multiple schedules meet the conditions, return **any** one of them.
+
+If no schedule exists that meets the conditions, return an empty array.
+
+**Example 1:**
+
+**Input:** n = 3
+
+**Output:** []
+
+**Explanation:**
+
+Since each team plays every other team exactly twice, a total of 6 matches need to be played: `[0,1],[0,2],[1,2],[1,0],[2,0],[2,1]`.
+
+It's not possible to create a schedule without at least one team playing consecutive days.
+
+**Example 2:**
+
+**Input:** n = 5
+
+**Output:** [[0,1],[2,3],[0,4],[1,2],[3,4],[0,2],[1,3],[2,4],[0,3],[1,4],[2,0],[3,1],[4,0],[2,1],[4,3],[1,0],[3,2],[4,1],[3,0],[4,2]]
+
+**Explanation:**
+
+Since each team plays every other team exactly twice, a total of 20 matches need to be played.
+
+The output shows one of the schedules that meet the conditions. No team plays on consecutive days.
+
+**Constraints:**
+
+*   `2 <= n <= 50`

src/main/kotlin/g3601_3700/s3681_maximum_xor_of_subsequences/Solution.kt

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+package g3601_3700.s3681_maximum_xor_of_subsequences
+
+// #Hard #Biweekly_Contest_165 #2025_09_20_Time_26_ms_(100.00%)_Space_76.00_MB_(77.78%)
+
+import kotlin.math.max
+import kotlin.math.min
+
+class Solution {
+    fun maxXorSubsequences(nums: IntArray): Int {
+        val n = nums.size
+        if (n == 0) {
+            return 0
+        }
+        var x = 0
+        while (true) {
+            var y = 0
+            for (v in nums) {
+                if (v > y) {
+                    y = v
+                }
+            }
+            if (y == 0) {
+                return x
+            }
+            x = max(x, x xor y)
+            for (i in 0..<n) {
+                val v = nums[i]
+                nums[i] = min(v, v xor y)
+            }
+        }
+    }
+}

src/main/kotlin/g3601_3700/s3681_maximum_xor_of_subsequences/readme.md

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+3681\. Maximum XOR of Subsequences
+
+Hard
+
+You are given an integer array `nums` of length `n` where each element is a non-negative integer.
+
+Select **two** **subsequences** of `nums` (they may be empty and are **allowed** to **overlap**), each preserving the original order of elements, and let:
+
+*   `X` be the bitwise XOR of all elements in the first subsequence.
+*   `Y` be the bitwise XOR of all elements in the second subsequence.
+
+Return the **maximum** possible value of `X XOR Y`.
+
+**Note:** The XOR of an **empty** subsequence is 0.
+
+**Example 1:**
+
+**Input:** nums = [1,2,3]
+
+**Output:** 3
+
+**Explanation:**
+
+Choose subsequences:
+
+*   First subsequence `[2]`, whose XOR is 2.
+*   Second subsequence `[2,3]`, whose XOR is 1.
+
+Then, XOR of both subsequences = `2 XOR 1 = 3`.
+
+This is the maximum XOR value achievable from any two subsequences.
+
+**Example 2:**
+
+**Input:** nums = [5,2]
+
+**Output:** 7
+
+**Explanation:**
+
+Choose subsequences:
+
+*   First subsequence `[5]`, whose XOR is 5.
+*   Second subsequence `[2]`, whose XOR is 2.
+
+Then, XOR of both subsequences = `5 XOR 2 = 7`.
+
+This is the maximum XOR value achievable from any two subsequences.
+
+**Constraints:**
+
+*   <code>2 <= nums.length <= 10<sup>5</sup></code>
+*   <code>0 <= nums[i] <= 10<sup>9</sup></code>

src/main/kotlin/g3601_3700/s3683_earliest_time_to_finish_one_task/Solution.kt

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+package g3601_3700.s3683_earliest_time_to_finish_one_task
+
+// #Easy #Weekly_Contest_467 #2025_09_20_Time_1_ms_(100.00%)_Space_52.19_MB_(60.38%)
+
+import kotlin.math.min
+
+class Solution {
+    fun earliestTime(tasks: Array<IntArray>): Int {
+        var ans = 1000
+        for (i in tasks.indices) {
+            val st = tasks[i][0]
+            val tm = tasks[i][1]
+            ans = min(ans, st + tm)
+        }
+        return ans
+    }
+}

src/main/kotlin/g3601_3700/s3683_earliest_time_to_finish_one_task/readme.md

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+3683\. Earliest Time to Finish One Task
+
+Easy
+
+You are given a 2D integer array `tasks` where <code>tasks[i] = [s<sub>i</sub>, t<sub>i</sub>]</code>.
+
+Each <code>[s<sub>i</sub>, t<sub>i</sub>]</code> in `tasks` represents a task with start time <code>s<sub>i</sub></code> that takes <code>t<sub>i</sub></code> units of time to finish.
+
+Return the earliest time at which at least one task is finished.
+
+**Example 1:**
+
+**Input:** tasks = [[1,6],[2,3]]
+
+**Output:** 5
+
+**Explanation:**
+
+The first task starts at time `t = 1` and finishes at time `1 + 6 = 7`. The second task finishes at time `2 + 3 = 5`. You can finish one task at time 5.
+
+**Example 2:**
+
+**Input:** tasks = [[100,100],[100,100],[100,100]]
+
+**Output:** 200
+
+**Explanation:**
+
+All three tasks finish at time `100 + 100 = 200`.
+
+**Constraints:**
+
+*   `1 <= tasks.length <= 100`
+*   <code>tasks[i] = [s<sub>i</sub>, t<sub>i</sub>]</code>
+*   <code>1 <= s<sub>i</sub>, t<sub>i</sub> <= 100</code>