coderEdge

Recommended GitHub Repository for Node.js Servers If you are planning to build a backend application, here is a good GitHub repository to refer to before you make a new server: 🔗 callicoder / node-easy-notes-app This repository serves as an excellent starting point and reference guide for building RESTful APIs using Node.js, Express, and MongoDB.

  #!/bin/python3 import math import os import random import re import sys # # Complete the 'getOneBits' function below. # # The function is expected to return an INTEGER_ARRAY. # The function accepts INTEGER n as parameter. # def getOneBits ( n ):     # Write your code here     if n == 0 :         return [ 0 ]             binary = bin ( n )[ 2 :]     d = []     pos = 1     temp = n     for idx , bit in enumerate ( binary , 1 ) :         if bit == '1' :                         d . append ( idx )             return [ len ( d )] + d     if __name__ == '__main__' :     fptr = open ( os . environ [ 'OUTPUT_PATH' ], 'w' )     n = int ( input () . strip ())     result = getOneBits ( n )     fptr ...

  #!/bin/python3 import math import os import random import re import sys # # Complete the 'romanizer' function below. # # The function is expected to return a STRING_ARRAY. # The function accepts INTEGER_ARRAY numbers as parameter. # def romanizer ( numbers ):     # Write your code here     def intToRoman ( num ):         val = [             1000 , 900 , 500 , 400 , 100 , 90 , 50 , 40 , 10 , 9 , 5 , 4 , 1         ]         syms = [             "M" , "CM" , "D" , "CD" ,             "C" , "XC" , "L" , "XL" ,             "X" , "IX" , "V" , "IV" ,             "I"         ]         roman = ""         i = 0         while num > 0 :           ...

Guide: Binary Search is used to find target integer in a sorted array quickly. Binary search has left , right , and mid variables, but the target variable is what binary search is used for. Assume num2 is a sorted array in which we are searching for target. The Match: If nums2[mid] is exactly equal to your target, you've found a common number! You can immediately return it. Go Right: If your target is greater than nums2[mid], that means your target has to be in the right half of nums2. Move your left pointer to mid + 1. Go Left: If your target is smaller than nums2[mid], your target must be in the left half. Move your right pointer to mid - 1. //java int left = 0; int right = nums2.length - 1; int mid = 0; int target = 0; while(left <= right) { mid = left + (right - left) / 2; if(nums1[i] == nums2[mid]) { return target; } else { if(target < nums2[mid]) { right = mid - 1; } else { ...

hi foks ,If you are too trying to get started with nodejs then these tricks will Help you out : Getting started with Nodejs npm init -y Conclusion: Hope these tricks,you have found helpful alot.

  Union-Find (or Disjoint Set Union, DSU) is a data structure that tracks elements partitioned into disjoint subsets, supporting rapid merging () and representative finding () operations. Optimized with path compression and union by rank/size, it achieves near-constant amortized time complexity, $O(\alpha(n))$ , making it efficient for Kruskal’s algorithm and dynamic connectivity. [ 1 , 2 , 3 , 4 ] Key Concepts and Operations Find(x): Determines the representative (root) of the set containing element . Union(x, y): Merges the sets containing elements and . MakeSet(x): Initializes a new set containing only element . Structure: Typically implemented as a forest of trees, where each node points to its parent. The root of a tree is its own parent and acts as the representative. [ 4 , 5 , 6 , 7 , 8 ] Optimizations Path Compression: During a operation, makes every node on the path point directly to the root, flattening the tree. Union by Rank/Size: Always attaches the smaller tr...

  Balancing brackets  = checking whether every opening bracket has a matching closing bracket in the correct order. Types usually considered: () [] {} Core idea (stack method) Use a  stack : Read characters left → right. If opening bracket → push onto stack. If closing bracket → stack empty → invalid top doesn’t match → invalid else pop. End → stack must be empty. Time:  O(n) Space:  O(n) Example Input: {[()]} Process: { push [ push ( push ) pop ] pop } pop Stack empty → balanced. Python (clean version) def is_balanced(s): stack = [] pairs = {')': '(', ']': '[', '}': '{'} for ch in s: if ch in "([{": stack.append(ch) elif ch in ")]}": if not stack or stack[-1] != pairs[ch]: return False stack.pop() return len(stack) == 0 Edge cases people miss "(]"  → order mismatch "((("  → leftover opens "))...