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Commit5d76ba7

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packageg3601_3700.s3678_smallest_absent_positive_greater_than_average;
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// #Easy #Biweekly_Contest_165 #2025_09_20_Time_2_ms_(100.00%)_Space_45.28_MB_(53.71%)
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publicclassSolution {
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publicintsmallestAbsent(int[]nums) {
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intsum =0;
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for (intj :nums) {
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sum +=j;
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}
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doubleavg = (double)sum /nums.length;
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intnum;
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if (avg <0) {
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num =1;
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}else {
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num = (int)avg +1;
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}
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while (true) {
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booleanflag =false;
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for (intj :nums) {
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if (num ==j) {
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flag =true;
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break;
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}
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}
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if (!flag &&num >avg) {
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returnnum;
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}
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num++;
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}
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}
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}
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3678\. Smallest Absent Positive Greater Than Average
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Easy
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You are given an integer array`nums`.
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Return the**smallest absent positive** integer in`nums` such that it is**strictly greater** than the**average** of all elements in`nums`.
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The**average** of an array is defined as the sum of all its elements divided by the number of elements.
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**Example 1:**
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**Input:** nums =[3,5]
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**Output:** 6
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**Explanation:**
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* The average of`nums` is`(3 + 5) / 2 = 8 / 2 = 4`.
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* The smallest absent positive integer greater than 4 is 6.
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**Example 2:**
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**Input:** nums =[-1,1,2]
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**Output:** 3
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**Explanation:**
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* The average of`nums` is`(-1 + 1 + 2) / 3 = 2 / 3 = 0.667`.
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* The smallest absent positive integer greater than 0.667 is 3.
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**Example 3:**
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**Input:** nums =[4,-1]
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**Output:** 2
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**Explanation:**
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* The average of`nums` is`(4 + (-1)) / 2 = 3 / 2 = 1.50`.
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* The smallest absent positive integer greater than 1.50 is 2.
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**Constraints:**
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*`1 <= nums.length <= 100`
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*`-100 <= nums[i] <= 100`
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packageg3601_3700.s3679_minimum_discards_to_balance_inventory;
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// #Medium #Biweekly_Contest_165 #2025_09_20_Time_2_ms_(100.00%)_Space_60.72_MB_(75.43%)
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publicclassSolution {
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publicintminArrivalsToDiscard(int[]arrivals,intw,intm) {
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intn =arrivals.length;
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intdis =0;
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boolean[]removed =newboolean[n];
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intmaxVal =0;
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for (intv :arrivals) {
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maxVal =Math.max(maxVal,v);
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}
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int[]freq =newint[maxVal +1];
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for (inti =0;i <n;i++) {
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intoutIdx =i -w;
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if (outIdx >=0 && !removed[outIdx]) {
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intoldVal =arrivals[outIdx];
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freq[oldVal]--;
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}
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intval =arrivals[i];
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if (freq[val] >=m) {
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dis++;
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removed[i] =true;
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}else {
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freq[val]++;
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}
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}
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returndis;
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}
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}
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3679\. Minimum Discards to Balance Inventory
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Medium
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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**).
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Items are managed according to the following rules:
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* Each arrival may be**kept** or**discarded**; an item may only be discarded on its arrival day.
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* For each day`i`, consider the window of days`[max(1, i - w + 1), i]` (the`w` most recent days up to day`i`):
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* For**any** such window, each item type may appear**at most**`m` times among kept arrivals whose arrival day lies in that window.
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* 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.
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Return the**minimum** number of arrivals to be discarded so that every`w`\-day window contains at most`m` occurrences of each type.
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**Example 1:**
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**Input:** arrivals =[1,2,1,3,1], w = 4, m = 2
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**Output:** 0
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**Explanation:**
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* On day 1, Item 1 arrives; the window contains no more than`m` occurrences of this type, so we keep it.
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* On day 2, Item 2 arrives; the window of days 1 - 2 is fine.
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* On day 3, Item 1 arrives, window`[1, 2, 1]` has item 1 twice, within limit.
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* On day 4, Item 3 arrives, window`[1, 2, 1, 3]` has item 1 twice, allowed.
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* On day 5, Item 1 arrives, window`[2, 1, 3, 1]` has item 1 twice, still valid.
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There are no discarded items, so return 0.
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**Example 2:**
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**Input:** arrivals =[1,2,3,3,3,4], w = 3, m = 2
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**Output:** 1
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**Explanation:**
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* On day 1, Item 1 arrives. We keep it.
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* On day 2, Item 2 arrives, window`[1, 2]` is fine.
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* On day 3, Item 3 arrives, window`[1, 2, 3]` has item 3 once.
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* On day 4, Item 3 arrives, window`[2, 3, 3]` has item 3 twice, allowed.
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* On day 5, Item 3 arrives, window`[3, 3, 3]` has item 3 three times, exceeds limit, so the arrival must be discarded.
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* On day 6, Item 4 arrives, window`[3, 4]` is fine.
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Item 3 on day 5 is discarded, and this is the minimum number of arrivals to discard, so return 1.
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**Constraints:**
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* <code>1 <= arrivals.length <= 10<sup>5</sup></code>
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* <code>1 <= arrivals[i] <= 10<sup>5</sup></code>
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*`1 <= w <= arrivals.length`
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*`1 <= m <= w`
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packageg3601_3700.s3680_generate_schedule;
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// #Medium #Biweekly_Contest_165 #2025_09_20_Time_2_ms_(100.00%)_Space_45.33_MB_(59.29%)
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publicclassSolution {
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publicint[][]generateSchedule(intn) {
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if (n <5) {
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returnnewint[0][];
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}
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int[][]res =newint[n * (n -1)][];
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intidx =0;
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for (inti =2;i <n -1;i++) {
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for (intj =0;j <n;j++) {
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res[idx++] =newint[] {j, (j +i) %n};
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}
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}
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for (inti =0;i <n;i++) {
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res[idx++] =newint[] {i, (i +1) %n};
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res[idx++] =newint[] {(i +4) %n, (i +3) %n};
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}
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returnres;
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}
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}
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3680\. Generate Schedule
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Medium
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You are given an integer`n` representing`n` teams. You are asked to generate a schedule such that:
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* Each team plays every other team**exactly twice**: once at home and once away.
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* There is**exactly one** match per day; the schedule is a list of**consecutive** days and`schedule[i]` is the match on day`i`.
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* No team plays on**consecutive** days.
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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.
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If no schedule exists that meets the conditions, return an empty array.
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**Example 1:**
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**Input:** n = 3
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**Output:**[]
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**Explanation:**
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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]`.
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It's not possible to create a schedule without at least one team playing consecutive days.
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**Example 2:**
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**Input:** n = 5
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**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]]
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**Explanation:**
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Since each team plays every other team exactly twice, a total of 20 matches need to be played.
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The output shows one of the schedules that meet the conditions. No team plays on consecutive days.
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**Constraints:**
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*`2 <= n <= 50`
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packageg3601_3700.s3681_maximum_xor_of_subsequences;
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// #Hard #Biweekly_Contest_165 #2025_09_20_Time_27_ms_(99.73%)_Space_58.35_MB_(30.31%)
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publicclassSolution {
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publicintmaxXorSubsequences(int[]nums) {
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intn =nums.length;
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if (n ==0) {
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return0;
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}
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intx =0;
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while (true) {
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inty =0;
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for (intv :nums) {
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if (v >y) {
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y =v;
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}
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}
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if (y ==0) {
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returnx;
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}
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x =Math.max(x,x ^y);
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for (inti =0;i <n;i++) {
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intv =nums[i];
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nums[i] =Math.min(v,v ^y);
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}
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}
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}
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}
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3681\. Maximum XOR of Subsequences
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Hard
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You are given an integer array`nums` of length`n` where each element is a non-negative integer.
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Select**two****subsequences** of`nums` (they may be empty and are**allowed** to**overlap**), each preserving the original order of elements, and let:
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*`X` be the bitwise XOR of all elements in the first subsequence.
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*`Y` be the bitwise XOR of all elements in the second subsequence.
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Return the**maximum** possible value of`X XOR Y`.
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**Note:** The XOR of an**empty** subsequence is 0.
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**Example 1:**
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**Input:** nums =[1,2,3]
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**Output:** 3
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**Explanation:**
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Choose subsequences:
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* First subsequence`[2]`, whose XOR is 2.
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* Second subsequence`[2,3]`, whose XOR is 1.
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Then, XOR of both subsequences =`2 XOR 1 = 3`.
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This is the maximum XOR value achievable from any two subsequences.
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**Example 2:**
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**Input:** nums =[5,2]
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**Output:** 7
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**Explanation:**
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Choose subsequences:
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* First subsequence`[5]`, whose XOR is 5.
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* Second subsequence`[2]`, whose XOR is 2.
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Then, XOR of both subsequences =`5 XOR 2 = 7`.
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This is the maximum XOR value achievable from any two subsequences.
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**Constraints:**
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* <code>2 <= nums.length <= 10<sup>5</sup></code>
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* <code>0 <= nums[i] <= 10<sup>9</sup></code>
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packageg3601_3700.s3683_earliest_time_to_finish_one_task;
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// #Easy #Weekly_Contest_467 #2025_09_20_Time_0_ms_(100.00%)_Space_45.29_MB_(39.62%)
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publicclassSolution {
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publicintearliestTime(int[][]tasks) {
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intans =1000;
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for (inti =0;i <tasks.length;i++) {
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intst =tasks[i][0];
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inttm =tasks[i][1];
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ans =Math.min(ans,st +tm);
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}
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returnans;
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}
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}
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3683\. Earliest Time to Finish One Task
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Easy
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You are given a 2D integer array`tasks` where <code>tasks[i] =[s<sub>i</sub>, t<sub>i</sub>]</code>.
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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.
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Return the earliest time at which at least one task is finished.
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**Example 1:**
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**Input:** tasks =[[1,6],[2,3]]
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**Output:** 5
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**Explanation:**
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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.
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**Example 2:**
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**Input:** tasks =[[100,100],[100,100],[100,100]]
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**Output:** 200
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**Explanation:**
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All three tasks finish at time`100 + 100 = 200`.
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**Constraints:**
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*`1 <= tasks.length <= 100`
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* <code>tasks[i] =[s<sub>i</sub>, t<sub>i</sub>]</code>
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* <code>1 <= s<sub>i</sub>, t<sub>i</sub> <= 100</code>

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