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Commit9182a52

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packageg3201_3300.s3206_alternating_groups_i;
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// #Easy #Array #Sliding_Window #2024_07_09_Time_1_ms_(97.24%)_Space_42.8_MB_(90.31%)
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publicclassSolution {
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publicintnumberOfAlternatingGroups(int[]colors) {
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intn =colors.length;
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intcount =0;
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if (colors[n -1] !=colors[0] &&colors[0] !=colors[1]) {
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count++;
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}
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if (colors[n -1] !=colors[0] &&colors[n -1] !=colors[n -2]) {
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count++;
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}
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for (inti =1;i <n -1;i++) {
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if (colors[i] !=colors[i -1] &&colors[i] !=colors[i +1]) {
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count++;
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}
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}
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returncount;
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}
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}
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3206\. Alternating Groups I
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Easy
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There is a circle of red and blue tiles. You are given an array of integers`colors`. The color of tile`i` is represented by`colors[i]`:
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*`colors[i] == 0` means that tile`i` is**red**.
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*`colors[i] == 1` means that tile`i` is**blue**.
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Every 3 contiguous tiles in the circle with**alternating** colors (the middle tile has a different color from its**left** and**right** tiles) is called an**alternating** group.
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Return the number of**alternating** groups.
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**Note** that since`colors` represents a**circle**, the**first** and the**last** tiles are considered to be next to each other.
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**Example 1:**
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**Input:** colors =[1,1,1]
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**Output:** 0
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**Explanation:**
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![](https://assets.leetcode.com/uploads/2024/05/16/image_2024-05-16_23-53-171.png)
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**Example 2:**
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**Input:** colors =[0,1,0,0,1]
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**Output:** 3
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**Explanation:**
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![](https://assets.leetcode.com/uploads/2024/05/16/image_2024-05-16_23-47-491.png)
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Alternating groups:
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**![](https://assets.leetcode.com/uploads/2024/05/16/image_2024-05-16_23-50-441.png)**![](https://assets.leetcode.com/uploads/2024/05/16/image_2024-05-16_23-48-211.png)**![](https://assets.leetcode.com/uploads/2024/05/16/image_2024-05-16_23-49-351.png)**
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**Constraints:**
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*`3 <= colors.length <= 100`
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*`0 <= colors[i] <= 1`
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packageg3201_3300.s3207_maximum_points_after_enemy_battles;
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// #Medium #Array #Greedy #2024_07_09_Time_1_ms_(100.00%)_Space_55.5_MB_(99.34%)
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publicclassSolution {
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publiclongmaximumPoints(int[]enemyEnergies,intcurrentEnergy) {
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intn =enemyEnergies.length;
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intmin =enemyEnergies[0];
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for (inti =1;i <n;i++) {
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min =Math.min(min,enemyEnergies[i]);
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}
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if (currentEnergy ==0 ||currentEnergy <min) {
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return0;
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}
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longsum =currentEnergy;
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for (inti =n -1;i >=0;i--) {
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sum +=enemyEnergies[i];
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}
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sum -=min;
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returnsum /min;
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}
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}
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3207\. Maximum Points After Enemy Battles
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Medium
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You are given an integer array`enemyEnergies` denoting the energy values of various enemies.
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You are also given an integer`currentEnergy` denoting the amount of energy you have initially.
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You start with 0 points, and all the enemies are unmarked initially.
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You can perform**either** of the following operations**zero** or multiple times to gain points:
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* Choose an**unmarked** enemy,`i`, such that`currentEnergy >= enemyEnergies[i]`. By choosing this option:
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* You gain 1 point.
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* Your energy is reduced by the enemy's energy, i.e.`currentEnergy = currentEnergy - enemyEnergies[i]`.
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* If you have**at least** 1 point, you can choose an**unmarked** enemy,`i`. By choosing this option:
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* Your energy increases by the enemy's energy, i.e.`currentEnergy = currentEnergy + enemyEnergies[i]`.
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* The enemy`i` is**marked**.
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Return an integer denoting the**maximum** points you can get in the end by optimally performing operations.
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**Example 1:**
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**Input:** enemyEnergies =[3,2,2], currentEnergy = 2
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**Output:** 3
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**Explanation:**
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The following operations can be performed to get 3 points, which is the maximum:
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* First operation on enemy 1:`points` increases by 1, and`currentEnergy` decreases by 2. So,`points = 1`, and`currentEnergy = 0`.
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* Second operation on enemy 0:`currentEnergy` increases by 3, and enemy 0 is marked. So,`points = 1`,`currentEnergy = 3`, and marked enemies =`[0]`.
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* First operation on enemy 2:`points` increases by 1, and`currentEnergy` decreases by 2. So,`points = 2`,`currentEnergy = 1`, and marked enemies =`[0]`.
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* Second operation on enemy 2:`currentEnergy` increases by 2, and enemy 2 is marked. So,`points = 2`,`currentEnergy = 3`, and marked enemies =`[0, 2]`.
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* First operation on enemy 1:`points` increases by 1, and`currentEnergy` decreases by 2. So,`points = 3`,`currentEnergy = 1`, and marked enemies =`[0, 2]`.
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**Example 2:**
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**Input:** enemyEnergies =[2], currentEnergy = 10
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**Output:** 5
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**Explanation:**
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Performing the first operation 5 times on enemy 0 results in the maximum number of points.
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**Constraints:**
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* <code>1 <= enemyEnergies.length <= 10<sup>5</sup></code>
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* <code>1 <= enemyEnergies[i] <= 10<sup>9</sup></code>
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* <code>0 <= currentEnergy <= 10<sup>9</sup></code>
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packageg3201_3300.s3208_alternating_groups_ii;
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// #Medium #Array #Sliding_Window #2024_07_09_Time_2_ms_(99.02%)_Space_63.3_MB_(22.96%)
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publicclassSolution {
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publicintnumberOfAlternatingGroups(int[]colors,intk) {
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inti =0;
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intlen =0;
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inttotal =0;
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while (i <colors.length -1) {
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intj =i +1;
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if (colors[j] !=colors[i]) {
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len =2;
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j++;
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while (j <colors.length &&colors[j] !=colors[j -1]) {
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j++;
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len++;
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}
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if (j ==colors.length) {
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break;
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}
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total +=Math.max(0, (len -k +1));
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}
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i =j;
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len =0;
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}
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if (colors[0] !=colors[colors.length -1]) {
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// if(len == colors.length) {
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// return Math.max(0, colors.length);
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// }
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len =len ==0 ?2 :len +1;
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intj =1;
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while (j <colors.length &&colors[j] !=colors[j -1]) {
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j++;
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len++;
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}
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if (j >=k) {
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len -= (j -k +1);
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}
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}
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total +=Math.max(0, (len -k +1));
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returntotal;
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}
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}
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3208\. Alternating Groups II
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Medium
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There is a circle of red and blue tiles. You are given an array of integers`colors` and an integer`k`. The color of tile`i` is represented by`colors[i]`:
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*`colors[i] == 0` means that tile`i` is**red**.
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*`colors[i] == 1` means that tile`i` is**blue**.
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An**alternating** group is every`k` contiguous tiles in the circle with**alternating** colors (each tile in the group except the first and last one has a different color from its**left** and**right** tiles).
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Return the number of**alternating** groups.
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**Note** that since`colors` represents a**circle**, the**first** and the**last** tiles are considered to be next to each other.
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**Example 1:**
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**Input:** colors =[0,1,0,1,0], k = 3
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**Output:** 3
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**Explanation:**
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**![](https://assets.leetcode.com/uploads/2024/06/19/screenshot-2024-05-28-183519.png)**
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Alternating groups:
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![](https://assets.leetcode.com/uploads/2024/05/28/screenshot-2024-05-28-182448.png)![](https://assets.leetcode.com/uploads/2024/05/28/screenshot-2024-05-28-182844.png)![](https://assets.leetcode.com/uploads/2024/05/28/screenshot-2024-05-28-183057.png)
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**Example 2:**
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**Input:** colors =[0,1,0,0,1,0,1], k = 6
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**Output:** 2
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**Explanation:**
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**![](https://assets.leetcode.com/uploads/2024/06/19/screenshot-2024-05-28-183907.png)**
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Alternating groups:
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![](https://assets.leetcode.com/uploads/2024/06/19/screenshot-2024-05-28-184128.png)![](https://assets.leetcode.com/uploads/2024/06/19/screenshot-2024-05-28-184240.png)
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**Example 3:**
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**Input:** colors =[1,1,0,1], k = 4
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**Output:** 0
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**Explanation:**
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![](https://assets.leetcode.com/uploads/2024/06/19/screenshot-2024-05-28-184516.png)
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**Constraints:**
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* <code>3 <= colors.length <= 10<sup>5</sup></code>
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*`0 <= colors[i] <= 1`
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*`3 <= k <= colors.length`
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packageg3201_3300.s3209_number_of_subarrays_with_and_value_of_k;
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// #Hard #Array #Binary_Search #Bit_Manipulation #Segment_Tree
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// #2024_07_09_Time_7_ms_(100.00%)_Space_62.9_MB_(11.74%)
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publicclassSolution {
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publiclongcountSubarrays(int[]nums,intk) {
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longans =0;
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intleft =0;
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intright =0;
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for (inti =0;i <nums.length;i++) {
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intx =nums[i];
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for (intj =i -1;j >=0 && (nums[j] &x) !=nums[j];j--) {
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nums[j] &=x;
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}
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while (left <=i &&nums[left] <k) {
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left++;
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}
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while (right <=i &&nums[right] <=k) {
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right++;
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}
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ans +=right -left;
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}
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returnans;
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}
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}
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3209\. Number of Subarrays With AND Value of K
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Hard
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Given an array of integers`nums` and an integer`k`, return the number of subarrays of`nums` where the bitwise`AND` of the elements of the subarray equals`k`.
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**Example 1:**
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**Input:** nums =[1,1,1], k = 1
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**Output:** 6
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**Explanation:**
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All subarrays contain only 1's.
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**Example 2:**
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**Input:** nums =[1,1,2], k = 1
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**Output:** 3
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**Explanation:**
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Subarrays having an`AND` value of 1 are: <code>[<ins>**1**</ins>,1,2]</code>, <code>[1,<ins>**1**</ins>,2]</code>, <code>[<ins>**1,1**</ins>,2]</code>.
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**Example 3:**
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**Input:** nums =[1,2,3], k = 2
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**Output:** 2
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**Explanation:**
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Subarrays having an`AND` value of 2 are: <code>[1,**<ins>2</ins>**,3]</code>, <code>[1,<ins>**2,3**</ins>]</code>.
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**Constraints:**
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* <code>1 <= nums.length <= 10<sup>5</sup></code>
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* <code>0 <= nums[i], k <= 10<sup>9</sup></code>
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packageg3201_3300.s3210_find_the_encrypted_string;
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// #Easy #String #2024_07_09_Time_1_ms_(100.00%)_Space_42.8_MB_(34.96%)
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publicclassSolution {
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publicStringgetEncryptedString(Strings,intk) {
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intn =s.length();
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k =k %n;
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StringBuilderstr =newStringBuilder(s.substring(k,n));
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str.append(s.substring(0,k));
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returnstr.toString();
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}
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}
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3210\. Find the Encrypted String
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Easy
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You are given a string`s` and an integer`k`. Encrypt the string using the following algorithm:
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* For each character`c` in`s`, replace`c` with the <code>k<sup>th</sup></code> character after`c` in the string (in a cyclic manner).
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Return the_encrypted string_.
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**Example 1:**
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**Input:** s = "dart", k = 3
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**Output:** "tdar"
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**Explanation:**
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* For`i = 0`, the 3<sup>rd</sup> character after`'d'` is`'t'`.
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* For`i = 1`, the 3<sup>rd</sup> character after`'a'` is`'d'`.
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* For`i = 2`, the 3<sup>rd</sup> character after`'r'` is`'a'`.
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* For`i = 3`, the 3<sup>rd</sup> character after`'t'` is`'r'`.
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**Example 2:**
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**Input:** s = "aaa", k = 1
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**Output:** "aaa"
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**Explanation:**
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As all the characters are the same, the encrypted string will also be the same.
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**Constraints:**
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*`1 <= s.length <= 100`
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* <code>1 <= k <= 10<sup>4</sup></code>
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*`s` consists only of lowercase English letters.

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