GCD and LCM Euclidean Algorithm in DSA C# - Time & Space Complexity

Time Complexity: GCD and LCM Euclidean Algorithm

O(log min(a, b))

Understanding Time Complexity

We want to understand how fast the Euclidean algorithm finds the greatest common divisor (GCD) of two numbers.

How does the number of steps grow as the input numbers get bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int GCD(int a, int b) {
    while (b != 0) {
        int temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}

int LCM(int a, int b) {
    return (a / GCD(a, b)) * b;
}
    

This code finds the GCD using the Euclidean algorithm and then calculates the LCM using the GCD.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop repeatedly calculates the remainder (a % b).
• How many times: The loop runs until b becomes zero, which depends on how quickly the numbers reduce.

How Execution Grows With Input

The number of steps depends on how many times we can replace the pair (a, b) with smaller numbers using the remainder.

Input Size (n)	Approx. Operations
10	Up to 5 steps
100	Up to 10 steps
1000	Up to 15 steps

Pattern observation: The steps grow slowly, roughly proportional to the number of digits, not the value itself.

Final Time Complexity

Time Complexity: O(log min(a, b))

This means the time to find the GCD grows slowly, roughly with the number of digits of the smaller input.

Common Mistake

[X] Wrong: "The Euclidean algorithm takes time proportional to the input values themselves (like O(n))."

[OK] Correct: The algorithm reduces the numbers quickly by using remainders, so the steps grow with the number of digits (logarithmically), not the full value.

Interview Connect

Understanding this algorithm's time complexity shows you can analyze efficient number-based algorithms, a useful skill for many coding problems.

Self-Check

"What if we used subtraction instead of remainder in the GCD calculation? How would the time complexity change?"

Practice

1. What does the Euclidean algorithm primarily calculate between two numbers a and b?

easy

A. The difference between a and b

B. The Greatest Common Divisor (GCD)

C. The sum of a and b

D. The Least Common Multiple (LCM)

Solution

1. Step 1: Understand the Euclidean algorithm purpose

   The Euclidean algorithm is designed to find the greatest common divisor (GCD) of two numbers by repeatedly applying the remainder operation.

2. Step 2: Differentiate GCD from LCM

   While GCD finds the largest number dividing both, LCM finds the smallest number divisible by both. The Euclidean algorithm directly calculates GCD.

3. Final Answer:

   The Greatest Common Divisor (GCD) -> Option B

4. Quick Check:

   Euclidean algorithm = GCD [OK]

Hint: Euclidean algorithm always finds GCD, not LCM [OK]

Common Mistakes:

• Confusing GCD with LCM
• Thinking it calculates sum or difference
• Assuming it finds prime factors directly

2. Which of the following is the correct C# syntax to calculate the GCD of two integers a and b using the built-in method?

easy

A. int gcd = Math.GreatestCommonDivisor(a, b);

B. int gcd = Math.GCD(a, b);

C. int gcd = Math.gcd(a, b);

D. int gcd = Math.Gcd(a, b);

Solution

1. Step 1: Identify the correct method name in C#

   The .NET 6+ Math class provides the method Math.Gcd(int, int) to calculate the greatest common divisor.

2. Step 2: Check method name casing and spelling

   Method names in C# are PascalCase, so Math.Gcd is correct, while others are incorrect or do not exist.

3. Final Answer:

   int gcd = Math.Gcd(a, b); -> Option D

4. Quick Check:

   Correct method is Math.Gcd [OK]

Hint: Use Math.Gcd with capital G and C# PascalCase [OK]

Common Mistakes:

• Using incorrect method names like Math.gcd or Math.GCD
• Assuming method is GreatestCommonDivisor
• Wrong casing causing compile errors

3. What is the output of the following C# code?

int a = 48;
int b = 18;
int gcd = Math.Gcd(a, b);
int lcm = (a * b) / gcd;
Console.WriteLine($"GCD: {gcd}, LCM: {lcm}");

medium

A. GCD: 18, LCM: 48

B. GCD: 12, LCM: 72

C. GCD: 6, LCM: 144

D. GCD: 3, LCM: 288

Solution

1. Step 1: Calculate GCD of 48 and 18

   Using Euclidean algorithm: 48 % 18 = 12, 18 % 12 = 6, 12 % 6 = 0, so GCD = 6.

2. Step 2: Calculate LCM using formula

   LCM = (48 * 18) / 6 = 864 / 6 = 144.

3. Final Answer:

   GCD: 6, LCM: 144 -> Option C

4. Quick Check:

   GCD=6, LCM=144 [OK]

Hint: LCM = (a * b) / GCD always [OK]

Common Mistakes:

• Calculating GCD incorrectly
• Swapping GCD and LCM values
• Forgetting to divide product by GCD for LCM

4. Identify the error in this C# code snippet that tries to compute the GCD using Euclidean algorithm:

int Gcd(int a, int b) {
    while (b != 0) {
        int temp = a % b;
        a = b;
        b = temp;
    }
    return b;
}

medium

A. Returns b instead of a after loop ends

B. Uses wrong operator for remainder

C. Infinite loop because b never changes

D. Missing base case for zero inputs

Solution

1. Step 1: Trace the Euclidean algorithm loop

   The loop updates a and b until b becomes zero. At loop end, a holds the GCD.

2. Step 2: Check return statement correctness

   The code returns b, which is zero at loop end. It should return a which holds the GCD.

3. Final Answer:

   Returns b instead of a after loop ends -> Option A

4. Quick Check:

   Return a after loop [OK]

Hint: Return 'a' after loop, not 'b' [OK]

Common Mistakes:

• Returning the wrong variable after loop
• Confusing remainder operator with division
• Not updating variables inside loop

5. Given two arrays A = [12, 15, 21] and B = [18, 35, 14], which C# code snippet correctly computes an array of LCMs for each pair (A[i], B[i]) using the Euclidean algorithm?

hard

A. int[] lcms = new int[A.Length];
for (int i = 0; i < A.Length; i++) {
int gcd = Math.Gcd(A[i], B[i]);
lcms[i] = (A[i] * B[i]) / gcd;
}

B. int[] lcms = new int[A.Length];
for (int i = 0; i < A.Length; i++) {
int gcd = Math.Gcd(A[i], B[i]);
lcms[i] = gcd * A[i] * B[i];
}

C. int[] lcms = new int[A.Length];
for (int i = 0; i < A.Length; i++) {
lcms[i] = Math.Gcd(A[i], B[i]);
}

D. int[] lcms = new int[A.Length];
for (int i = 0; i < A.Length; i++) {
lcms[i] = (A[i] + B[i]) / Math.Gcd(A[i], B[i]);
}

Solution

1. Step 1: Calculate GCD for each pair

   Use Math.Gcd(A[i], B[i]) to find the greatest common divisor for each pair.

2. Step 2: Compute LCM using formula

   LCM = (A[i] * B[i]) / GCD. Store this in the lcms array at index i.

3. Final Answer:

   Option A code snippet correctly computes LCM array -> Option A

4. Quick Check:

   LCM = (a*b)/GCD for each pair [OK]

Hint: LCM = (a*b)/GCD for each pair in loop [OK]

Common Mistakes:

• Multiplying GCD with product instead of dividing
• Assigning GCD instead of LCM to result array
• Using sum instead of product in LCM formula