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Commitf433c16

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Added tasks 2677-2681
1 parent35c1405 commitf433c16

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2677\. Chunk Array
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Easy
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Given an array`arr` and a chunk size`size`, return a**chunked** array. A**chunked** array contains the original elements in`arr`, but consists of subarrays each of length`size`. The length of the last subarray may be less than`size` if`arr.length` is not evenly divisible by`size`.
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You may assume the array is the output of`JSON.parse`. In other words, it is valid JSON.
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Please solve it without using lodash's`_.chunk` function.
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**Example 1:**
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**Input:** arr =[1,2,3,4,5], size = 1
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**Output:**[[1],[2],[3],[4],[5]]
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**Explanation:** The arr has been split into subarrays each with 1 element.
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**Example 2:**
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**Input:** arr =[1,9,6,3,2], size = 3
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**Output:**[[1,9,6],[3,2]]
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**Explanation:** The arr has been split into subarrays with 3 elements. However, only two elements are left for the 2nd subarray.
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**Example 3:**
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**Input:** arr =[8,5,3,2,6], size = 6
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**Output:**[[8,5,3,2,6]]
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**Explanation:** Size is greater than arr.length thus all elements are in the first subarray.
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**Example 4:**
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**Input:** arr =[], size = 1
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**Output:**[]
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**Explanation:** There are no elements to be chunked so an empty array is returned.
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**Constraints:**
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*`arr is a valid JSON array`
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* <code>2 <= JSON.stringify(arr).length <= 10<sup>5</sup></code>
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*`1 <= size <= arr.length + 1`
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// #Easy #2023_09_11_Time_49_ms_(96.15%)_Space_44.4_MB_(96.63%)
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functionchunk(arr:any[],size:number):any[][]{
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if(arr.length===0)return[]
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if(size>=arr.length)return[arr]
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leti:number=0
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letres:Array<Array<number>>=[]
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while(i<arr.length){
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res.push(arr.slice(i,i+size))
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i+=size
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}
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returnres
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}
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export{chunk}
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packageg2601_2700.s2678_number_of_senior_citizens;
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// #Easy #Array #String #2023_09_11_Time_0_ms_(100.00%)_Space_40.7_MB_(97.65%)
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publicclassSolution {
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publicintcountSeniors(String[]details) {
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intcount =0;
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for (Stringdetail :details) {
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if (((detail.charAt(11) -'0' ==6) && (detail.charAt(12) -'0' >0))
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|| (detail.charAt(11) -'0' >6)) {
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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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2678\. Number of Senior Citizens
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Easy
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You are given a**0-indexed** array of strings`details`. Each element of`details` provides information about a given passenger compressed into a string of length`15`. The system is such that:
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* The first ten characters consist of the phone number of passengers.
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* The next character denotes the gender of the person.
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* The following two characters are used to indicate the age of the person.
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* The last two characters determine the seat allotted to that person.
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Return_the number of passengers who are**strictly****more than 60 years old**._
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**Example 1:**
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**Input:** details =["7868190130M7522","5303914400F9211","9273338290F4010"]
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**Output:** 2
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**Explanation:** The passengers at indices 0, 1, and 2 have ages 75, 92, and 40. Thus, there are 2 people who are over 60 years old.
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**Example 2:**
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**Input:** details =["1313579440F2036","2921522980M5644"]
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**Output:** 0
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**Explanation:** None of the passengers are older than 60.
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**Constraints:**
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*`1 <= details.length <= 100`
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*`details[i].length == 15`
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*`details[i] consists of digits from '0' to '9'.`
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*`details[i][10] is either 'M' or 'F' or 'O'.`
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* The phone numbers and seat numbers of the passengers are distinct.
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packageg2601_2700.s2679_sum_in_a_matrix;
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// #Medium #Array #Sorting #Matrix #Heap_Priority_Queue #Simulation
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// #2023_09_11_Time_13_ms_(99.66%)_Space_56.9_MB_(31.71%)
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importjava.util.Arrays;
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publicclassSolution {
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publicintmatrixSum(int[][]nums) {
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intresult =0;
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for (int[]row :nums) {
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Arrays.sort(row);
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reverseArray(row);
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}
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for (inti =0;i <nums[0].length;i++) {
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intmax =0;
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for (int[]num :nums) {
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max =Math.max(max,num[i]);
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}
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result +=max;
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}
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returnresult;
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}
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privatevoidreverseArray(int[]arr) {
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intleft =0;
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intright =arr.length -1;
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while (left <right) {
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inttemp =arr[left];
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arr[left] =arr[right];
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arr[right] =temp;
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left++;
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right--;
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}
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}
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}
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2679\. Sum in a Matrix
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Medium
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You are given a**0-indexed** 2D integer array`nums`. Initially, your score is`0`. Perform the following operations until the matrix becomes empty:
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1. From each row in the matrix, select the largest number and remove it. In the case of a tie, it does not matter which number is chosen.
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2. Identify the highest number amongst all those removed in step 1. Add that number to your**score**.
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Return_the final**score**._
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**Example 1:**
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**Input:** nums =[[7,2,1],[6,4,2],[6,5,3],[3,2,1]]
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**Output:** 15
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**Explanation:** In the first operation, we remove 7, 6, 6, and 3. We then add 7 to our score. Next, we remove 2, 4, 5, and 2. We add 5 to our score. Lastly, we remove 1, 2, 3, and 1. We add 3 to our score. Thus, our final score is 7 + 5 + 3 = 15.
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**Example 2:**
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**Input:** nums =[[1]]
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**Output:** 1
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**Explanation:** We remove 1 and add it to the answer. We return 1.
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**Constraints:**
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*`1 <= nums.length <= 300`
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*`1 <= nums[i].length <= 500`
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* <code>0 <= nums[i][j] <= 10<sup>3</sup></code>
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packageg2601_2700.s2680_maximum_or;
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// #Medium #Array #Greedy #Bit_Manipulation #Prefix_Sum
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// #2023_09_11_Time_2_ms_(100.00%)_Space_55.2_MB_(56.19%)
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publicclassSolution {
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publiclongmaximumOr(int[]nums,intk) {
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int[]suffix =newint[nums.length];
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for (inti =nums.length -2;i >=0;i--) {
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suffix[i] =suffix[i +1] |nums[i +1];
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}
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longprefix =0L;
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longmax =0L;
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for (inti =0;i <=nums.length -1;i++) {
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longnum =nums[i];
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max =Math.max(max,prefix | (num <<k) |suffix[i]);
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prefix |=num;
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}
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returnmax;
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}
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}
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2680\. Maximum OR
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Medium
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You are given a**0-indexed** integer array`nums` of length`n` and an integer`k`. In an operation, you can choose an element and multiply it by`2`.
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Return_the maximum possible value of_`nums[0] | nums[1] | ... | nums[n - 1]`_that can be obtained after applying the operation on nums at most_`k`_times_.
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Note that`a | b` denotes the**bitwise or** between two integers`a` and`b`.
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**Example 1:**
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**Input:** nums =[12,9], k = 1
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**Output:** 30
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**Explanation:** If we apply the operation to index 1, our new array nums will be equal to[12,18]. Thus, we return the bitwise or of 12 and 18, which is 30.
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**Example 2:**
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**Input:** nums =[8,1,2], k = 2
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**Output:** 35
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**Explanation:** If we apply the operation twice on index 0, we yield a new array of[32,1,2]. Thus, we return 32|1|2 = 35.
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**Constraints:**
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* <code>1 <= nums.length <= 10<sup>5</sup></code>
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* <code>1 <= nums[i] <= 10<sup>9</sup></code>
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*`1 <= k <= 15`
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packageg2601_2700.s2681_power_of_heroes;
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// #Hard #Array #Math #Sorting #Prefix_Sum #2023_09_11_Time_13_ms_(88.24%)_Space_54.4_MB_(55.29%)
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importjava.util.Arrays;
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publicclassSolution {
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publicintsumOfPower(int[]nums) {
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Arrays.sort(nums);
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longsumOfMin =0L;
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longres =0L;
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for (intnum :nums) {
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longmod =1_000_000_007L;
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longmax = (long)num *num %mod;
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longmySumOfMin = (sumOfMin +num) %mod;
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res = (res +max *mySumOfMin) %mod;
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sumOfMin = (sumOfMin +mySumOfMin) %mod;
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}
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return (int)res;
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}
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}
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2681\. Power of Heroes
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Hard
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You are given a**0-indexed** integer array`nums` representing the strength of some heroes. The**power** of a group of heroes is defined as follows:
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* Let <code>i<sub>0</sub></code>, <code>i<sub>1</sub></code>, ... ,<code>i<sub>k</sub></code> be the indices of the heroes in a group. Then, the power of this group is <code>max(nums[i<sub>0</sub>], nums[i<sub>1</sub>], ... ,nums[i<sub>k</sub>])<sup>2</sup> * min(nums[i<sub>0</sub>], nums[i<sub>1</sub>], ... ,nums[i<sub>k</sub>])</code>.
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Return_the sum of the**power** of all**non-empty** groups of heroes possible._ Since the sum could be very large, return it**modulo** <code>10<sup>9</sup> + 7</code>.
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**Example 1:**
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**Input:** nums =[2,1,4]
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**Output:** 141
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**Explanation:**
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1<sup>st</sup> group:[2] has power = 2<sup>2</sup>\* 2 = 8.
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2<sup>nd</sup> group:[1] has power = 1<sup>2</sup>\* 1 = 1.
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3<sup>rd</sup> group:[4] has power = 4<sup>2</sup>\* 4 = 64.
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4<sup>th</sup> group:[2,1] has power = 2<sup>2</sup>\* 1 = 4.
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5<sup>th</sup> group:[2,4] has power = 4<sup>2</sup>\* 2 = 32.
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6<sup>th</sup> group:[1,4] has power = 4<sup>2</sup>\* 1 = 16.
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7<sup>th</sup> group:[2,1,4] has power = 4<sup>2</sup>\* 1 = 16.
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The sum of powers of all groups is 8 + 1 + 64 + 4 + 32 + 16 + 16 = 141.
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**Example 2:**
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**Input:** nums =[1,1,1]
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**Output:** 7
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**Explanation:** A total of 7 groups are possible, and the power of each group will be 1. Therefore, the sum of the powers of all groups is 7.
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**Constraints:**
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* <code>1 <= nums.length <= 10<sup>5</sup></code>
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* <code>1 <= nums[i] <= 10<sup>9</sup></code>
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// tslint:disable:no-magic-numbers
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import{chunk}from'src/main/java/g2601_2700/s2677_chunk_array/solution'
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import{expect,test}from'vitest'
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test('chunk',()=>{
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expect(chunk([1,2,3,4,5],1)).toEqual([[1],[2],[3],[4],[5]])
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})
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test('chunk2',()=>{
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expect(chunk([1,9,6,3,2],3)).toEqual([
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[1,9,6],
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[3,2],
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])
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})
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test('chunk3',()=>{
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expect(chunk([8,5,3,2,6],6)).toEqual([[8,5,3,2,6]])
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})
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test('chunk4',()=>{
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expect(chunk([],1)).toEqual([])
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})
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packageg2601_2700.s2678_number_of_senior_citizens;
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importstaticorg.hamcrest.CoreMatchers.equalTo;
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importstaticorg.hamcrest.MatcherAssert.assertThat;
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importorg.junit.jupiter.api.Test;
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classSolutionTest {
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@Test
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voidcountSeniors() {
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assertThat(
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newSolution()
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.countSeniors(
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newString[] {
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"7868190130M7522","5303914400F9211","9273338290F4010"
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}),
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equalTo(2));
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}
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@Test
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voidcountSeniors2() {
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assertThat(
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newSolution().countSeniors(newString[] {"1313579440F2036","2921522980M5644"}),
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equalTo(0));
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}
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}
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packageg2601_2700.s2679_sum_in_a_matrix;
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importstaticorg.hamcrest.CoreMatchers.equalTo;
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importstaticorg.hamcrest.MatcherAssert.assertThat;
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importcom_github_leetcode.CommonUtils;
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importorg.junit.jupiter.api.Test;
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classSolutionTest {
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@Test
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voidmatrixSum() {
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assertThat(
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newSolution()
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.matrixSum(
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CommonUtils
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.convertLeetCodeIrregularLengths2DArrayInputIntoJavaArray(
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"[7,2,1],[6,4,2],[6,5,3],[3,2,1]")),
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equalTo(15));
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}
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@Test
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voidmatrixSum2() {
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assertThat(newSolution().matrixSum(newint[][] {{1}}),equalTo(1));
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}
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}

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