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Codesignal

Aracde > Intro > Edge of the Ocean

Challenge: shapeArea

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Logic: The approach to this question differs to that of the classic Two Sum problem in that we have some direction with how we want to search for our target.

Since the array is sorted, we can make some general observations:

Smaller sums would come from the left half of the array Larger sums would come from the right half of the array Therefore, using two pointers starting at the end points of the array, we can choose to increase or decrease our current sum however we like. Pay attention to the example below:

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// SPDX-License-Identifier: MIT
pragma solidity >= 0.6.0 <0.9.0;

contract SimpleStorage {
    // this will get initialized as 0
    uint256 favoriteNumber;

    struct People {
        uint256 favoriteNumber;
        string name;

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The official Leetcode page talks about four following approaches:

1. Approach 1: Brute Force
2. Approach 2: Using Extra Array
3. Approach 3: Using Cyclic Replacements
4. Approach 4: Using Reverse

The second approach can be done like this: (but only works in irb and not on leetcode somehow)

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The Optimized implementation of Bubble Sort from GeeksforGeeks failed the below test case:

Code:

# @param {Integer[]} nums
# @return {Integer[]}
def sorted_squares(nums)
  nums = nums.map { |n| n*n } # square the numbers
  
  # now lets sort with bubble sort

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=begin
Codility Challange: PrefixSet
https://app.codility.com/programmers/task/prefix_set/

Solution: https://codility.com/media/train/solution-prefix-set.pdf
=end

def prefixSetGolden(arr)
  n = arr.size
  occur = {}

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/**
 * These methods only work for an array of numbers
 */


function findMinimum(array) {
  let length = array.length;
  let min = Infinity;
  while(length--) {
    if(array[length] < min) {

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module CounterIncrementor
  def next(counter_name)
    counter = self.find_by(name: counter_name)
    counter.increment
  end
end

class Counter
  include Mongoid::Document
  include Mongoid::Timestamps