Given a number ‘n’, we need to find all of its prime factors. The Brute Force Approach: for(int i = 1; i <= n; i++) { if(n % i == 0) { if(prime(i)) v.push_back(i); } } The...

Given a number ‘n’, we need to print all the prime numbers up to ‘n’. for(int i = 2; i * i <= n; i++) { if(prime(i)) v.push_back(i); } This is the normal way to find primes up to ‘n’, but...

📝 Note If the number of iteration is based on division, then the time complexity will be in logarithmic. The numbers those has exactly two factors / divisors, 1 and the number it...

I am currently following Striver’s DSA playlist, a highly recommended resource for anyone aiming to build a strong foundation in Data Structures and Algorithms (DSA). The playlist provides a struct...