Fast Exponentiation Power in Log N in DSA Go - Time & Space Complexity

Time Complexity: Fast Exponentiation Power in Log N

O(log n)

Understanding Time Complexity

We want to understand how fast the power calculation grows as the exponent gets bigger.

How does the number of steps change when we calculate x to the power n?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func fastPower(x int, n int) int {
    if n == 0 {
        return 1
    }
    half := fastPower(x, n / 2)
    if n % 2 == 0 {
        return half * half
    } else {
        return half * half * x
    }
}

This code calculates x raised to the power n using a fast method by splitting the problem in half each time.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls that halve the exponent n each time.
• How many times: About log base 2 of n times, because n is divided by 2 each step.

How Execution Grows With Input

Each time we call the function, the exponent gets cut in half, so the number of steps grows slowly.

Input Size (n)	Approx. Operations
10	About 4 recursive calls
100	About 7 recursive calls
1000	About 10 recursive calls

Pattern observation: Doubling n adds only one more step, so growth is very slow.

Final Time Complexity

Time Complexity: O(log n)

This means the steps needed grow slowly as the exponent gets bigger, making it very efficient.

Common Mistake

[X] Wrong: "Calculating power always takes n steps because we multiply x n times."

[OK] Correct: This method uses splitting to reduce steps to about log n, not n, so it is much faster.

Interview Connect

Knowing how to speed up power calculations shows you understand how to reduce repeated work, a key skill in many problems.

Self-Check

"What if we changed the code to use a loop instead of recursion? How would the time complexity change?"

Practice

1. What is the main advantage of using fast exponentiation (power in log n) over the simple repeated multiplication method?

easy

A. It reduces the number of multiplications from n to log n

B. It uses less memory by storing all intermediate powers

C. It works only for even powers

D. It avoids multiplication completely

Solution

1. Step 1: Understand simple exponentiation

   Simple method multiplies the base by itself n times, so it takes n multiplications.

2. Step 2: Understand fast exponentiation

   Fast exponentiation uses repeated squaring, reducing multiplications to about log n times.

3. Final Answer:

   It reduces the number of multiplications from n to log n -> Option A

4. Quick Check:

   Fast exponentiation = fewer multiplications [OK]

Hint: Fast exponentiation cuts steps from n to log n [OK]

Common Mistakes:

• Thinking it only works for even powers
• Confusing memory use with speed
• Believing it avoids multiplication

2. Which of the following Go function signatures correctly represents a fast exponentiation function that calculates base^exp (integer power)?

easy

A. func fastPower(base float64, exp int) float64

B. func fastPower(base int, exp float64) int

C. func fastPower(base int, exp int) int

D. func fastPower(base int, exp int) float64

Solution

1. Step 1: Identify parameter types for integer power

   Base and exponent should be integers for integer power calculation.

2. Step 2: Check return type

   Return type should be int to match integer power result.

3. Final Answer:

   func fastPower(base int, exp int) int -> Option C

4. Quick Check:

   Integer base and exponent with int return [OK]

Hint: Use int for base, exponent, and return for integer powers [OK]

Common Mistakes:

• Using float64 for exponent when integer is needed
• Returning float64 instead of int
• Mixing base and exponent types

3. What is the output of the following Go code snippet?

func fastPower(base int, exp int) int {
    if exp == 0 {
        return 1
    }
    half := fastPower(base, exp/2)
    if exp%2 == 0 {
        return half * half
    } else {
        return half * half * base
    }
}

fmt.Println(fastPower(2, 5))

medium

A. 10

B. 32

C. 25

D. 16

Solution

1. Step 1: Trace recursive calls for fastPower(2, 5)

   fastPower(2,5) calls fastPower(2,2), which calls fastPower(2,1), which calls fastPower(2,0)=1.

2. Step 2: Calculate results bottom-up

   fastPower(2,1) = 1*1*2=2; fastPower(2,2) = 2*2=4; fastPower(2,5) = 4*4*2=32.

3. Final Answer:

   32 -> Option B

4. Quick Check:

   2^5 = 32 [OK]

Hint: Remember 2^5 = 32; fastPower uses recursion [OK]

Common Mistakes:

• Confusing multiplication order
• Forgetting to multiply by base for odd exponent
• Miscomputing base cases

4. The following Go code for fast exponentiation has a bug. What is the bug?

func fastPower(base int, exp int) int {
    if exp == 0 {
        return 0
    }
    half := fastPower(base, exp/2)
    if exp%2 == 0 {
        return half * half
    } else {
        return half * half * base
    }
}

medium

A. Function does not handle negative exponents

B. Missing multiplication by base for even exponents

C. Incorrect integer division for exp/2

D. Base case returns 0 instead of 1

Solution

1. Step 1: Check base case for exponent 0

   Exponent 0 means power is 1, but code returns 0, which is incorrect.

2. Step 2: Verify other parts

   Multiplication and division logic is correct; negative exponent handling is not required here.

3. Final Answer:

   Base case returns 0 instead of 1 -> Option D

4. Quick Check:

   Power of 0 exponent = 1 [OK]

Hint: Base case for exp=0 must return 1 [OK]

Common Mistakes:

• Returning 0 for exp=0 base case
• Ignoring negative exponent cases
• Misplacing multiplication logic

5. You want to compute (3^13) mod 7 efficiently using fast exponentiation in Go. Which modification is needed in the fastPower function to correctly compute this without overflow?

hard

A. Add modulo operation after each multiplication to keep numbers small

B. Use float64 type for base and exponent to handle large numbers

C. Multiply only once at the end after all recursion completes

D. Increase the base to base*mod to avoid overflow

Solution

1. Step 1: Understand modular exponentiation

   To avoid overflow and keep numbers small, apply modulo after each multiplication.

2. Step 2: Modify fastPower to include modulo

   Insert modulo operation after multiplying halves and base to keep intermediate results within mod.

3. Final Answer:

   Add modulo operation after each multiplication to keep numbers small -> Option A

4. Quick Check:

   Modulo after multiplication prevents overflow [OK]

Hint: Apply modulo after every multiplication step [OK]

Common Mistakes:

• Delaying modulo until end causes overflow
• Using float64 instead of int causes precision loss
• Multiplying base by mod increases value incorrectly