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Standard Priority Queue IX: Vowels Frequency

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Tally the frequency of the vowels. Push that into a pQueue, the caveat is when the frequencies are the same, you need to keep track of the first occurrence of each vowel and do some math to insert into the priority queue with the right order. After that, it is just dequeuing and inserting into the proper places. Use StringBuilder to avoid unnecessary string allocations. Code is down below, cheers, ACC. Sort Vowels by Frequency - LeetCode You are given a string  s  consisting of lowercase English characters. Rearrange only the  vowels  in the string so that they appear in  non-increasing  order of their frequency. If multiple vowels have the same  frequency , order them by the position of their  first occurrence  in  s . Return the modified string. Vowels are  'a' ,  'e' ,  'i' ,  'o' , and  'u' . The  frequency  of a letter is the number of times it occurs in the string.   Example 1: Input:   ...
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Mapping a Tree to a HashTable data structure

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I think a HashTable (of a HashSet) is a very convenient way to represent a tree. It gives you instant access to any element of the tree if there is a unique id representing each node. The problem (and code) below exemplifies this point. Once the representation is there, performing a Depth-First Search becomes very trivial. Code is down below, cheers, ACC. Kill Process - LeetCode You have  n  processes forming a rooted tree structure. You are given two integer arrays  pid  and  ppid , where  pid[i]  is the ID of the  i th  process and  ppid[i]  is the ID of the  i th  process's parent process. Each process has only  one parent process  but may have multiple children processes. Only one process has  ppid[i] = 0 , which means this process has  no parent process  (the root of the tree). When a process is  killed , all of its children processes will also be killed. Given an integer  kill ...
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Binary Search to Find Next Greater Element V

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You can use the .BinarySearch method of List<T>. It works well, there is a bit of a trick that you need to do when the return index is negative (just assign it to its complement, ~index). No need to perform the Binary Search "by hand" anymore. Achieves a good NLogN, fast solution. LC suggests a Tree to be used but this approach (two lists, one for even, one for odd) works well. Code is down below, cheers, ACC. Count Smaller Elements With Opposite Parity - LeetCode You are given an integer array  nums  of length  n . The  score  of an index  i  is defined as the number of indices  j  such that: i < j < n nums[j] < nums[i] nums[i]  and  nums[j]  have different parity (one is even and the other is odd). Return an integer array  answer  of length  n , where  answer[i]  is the score of index  i .   Example 1: Input:   nums = [5,2,4,1,3] Output:   [2,1,2,0,0] Explanation: Fo...
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