This course is designed to help learners master the core concepts of Data Structures and Algorithms (DSA) using the C programming language. Starting from the basics and advancing to complex topics like graphs, dynamic programming, and memory optimization, this course is ideal for students, job seekers, and aspiring developers. You’ll learn how to structure and manipulate data efficiently, solve real-world coding problems, and prepare for technical interviews at top companies. The content is structured step-by-step, combining theory with hands-on coding examples and practice problems to reinforce understanding. Whether you're preparing for university exams, campus placements, or competitive programming, this course provides a strong foundation in logic building, code efficiency, and problem-solving using C. Key Highlights: Covers all major DSA topics from beginner to advanced level 100+ coding examples with explanations Focus on time and space complexity optimization Designed for coding interviews, competitive exams, and CS fundamentals
Mathematics is the backbone of efficient algorithms in Data Structures. In this chapter, we cover essential concepts such as GCD, LCM, and Modulo Arithmetic that are frequently used in number theory, combinatorics, and optimization problems.
GCD of two numbers is the largest number that divides both numbers without leaving a remainder.
Example: GCD of 12 and 18 is 6.
Efficient method to compute GCD:
int gcd(int a, int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
LCM of two numbers is the smallest number divisible by both numbers.
Formula: LCM(a, b) = (a × b) / GCD(a, b)
int lcm(int a, int b) {
return (a * b) / gcd(a, b);
}
Modulo (%) returns the remainder of a division. Used heavily in competitive programming to avoid overflow and in cyclic behavior problems (like clocks, circular arrays).
// Examples 7 % 3 = 1 10 % 5 = 0 -5 % 3 = -2 (in C)
(a + b) % m = ((a % m) + (b % m)) % m(a * b) % m = ((a % m) * (b % m)) % m(a - b + m) % m handles negative results
For solving equations like (a / b) % m, we use the Modular Inverse of b under m:
Use Fermat's Theorem when m is prime: b⁻¹ ≡ b^(m-2) mod m
// Fast Power to compute (base^exp) % mod
int power(int base, int exp, int mod) {
int res = 1;
base %= mod;
while (exp > 0) {
if (exp % 2) res = (1LL * res * base) % mod;
base = (1LL * base * base) % mod;
exp /= 2;
}
return res;
}
In the next chapter, we’ll cover Arrays: Basics and Operations — the foundational structure for storing collections of data in DSA.