LeetCode 650: 2 Keys Keyboard Solution in Python – A Step-by-Step Guide

What Is LeetCode 650: 2 Keys Keyboard?

In LeetCode 650: 2 Keys Keyboard, you start with a notepad containing one 'A' and can perform two operations:

• Copy All: Copy the current string to a clipboard (e.g., "A" → clipboard = "A").
• Paste: Append the clipboard contents to the notepad (e.g., "A" + "A" = "AA").

Your task is to find the minimum number of steps to produce exactly n 'A's, returning this count as an integer. For example, with n = 3, you can copy "A" (1 step), paste to "AA" (2 steps), paste again to "AAA" (3 steps)—3 steps total. The problem tests your ability to optimize a sequence of operations to hit a precise target.

Problem Statement

• Input:
• n: An integer (target number of 'A's).
• Output: An integer—minimum steps to get n 'A's.
• Rules:
• Start with one 'A'.
• Use "Copy All" and "Paste" only.
• Reach exactly n 'A's, no more or less.

Constraints

• 1 ≤ n ≤ 1000.

Examples

• Input: n = 3
• Output: 3 (Copy "A", Paste "AA", Paste "AAA")
• Input: n = 1
• Output: 0 (Already have 1 'A')
• Input: n = 4
• Output: 4 (Copy "A", Paste "AA", Copy "AA", Paste "AAAA")

These examples set the notepad—let’s solve it!

Understanding the Problem: Minimizing Steps

To solve LeetCode 650: 2 Keys Keyboard in Python, we need to determine the fewest steps to produce n 'A's using "Copy All" and "Paste" from a single 'A'. A brute-force approach—trying all possible sequences—would be exponential, impractical for n ≤ 1000. Since each copy sets the paste size, and we build n through multiplications and additions, we can optimize with greedy or DP techniques. We’ll use:

• Best Solution (Greedy with Prime Factorization): O(√n) time, O(1) space—fast and elegant.
• Alternative Solution (Dynamic Programming): O(n²) time, O(n) space—systematic but slower.

Let’s start with the greedy solution, breaking it down for beginners.

Best Solution: Greedy with Prime Factorization

Why This Is the Best Solution

The greedy with prime factorization approach is the top choice for LeetCode 650 because it’s efficient—O(√n) time with O(1) space—and leverages a key insight: the minimum steps to reach n is the sum of its prime factors, as each factor represents a multiplication (copy-paste sequence) to build n. It fits constraints (n ≤ 1000) perfectly and is surprisingly simple once you see the pattern. It’s like breaking n into its building blocks to count the steps!

How It Works

Think of this as a factor counter:

• Step 1: Handle Base Case:
• If n = 1, return 0 (no steps needed).
• Step 2: Prime Factorization:
• Find smallest prime factor d of n.
• Each factor d means: Copy current string (length m), paste d-1 times to reach d * m.
• Steps += d, recurse on n // d.
• Step 3: Sum Factors:
• Total steps = sum of prime factors.
• Step 4: Return Result:
• Return step count.

It’s like a multiplication stepper—factor and add!

Step-by-Step Example

Example: n = 6

• Step 1: n > 1, proceed.
• Step 2: Factorize:
• Smallest factor = 2 (6 % 2 = 0).
• Steps = 2 (Copy "A" → "A", Paste → "AA").
• n = 6 // 2 = 3.
• Smallest factor = 3 (3 % 3 = 0).
• Steps += 3 (Copy "AA" → "AA", Paste → "AAAA", Paste → "AAAAAA").
• n = 3 // 3 = 1, stop.
• Step 3: Total = 2 + 3 = 5.
• Output: 5 (Verified: Copy A, Paste AA, Copy AA, Paste AAAA, Paste AAAAAA).

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained clearly:

Copy codeclass Solution:
    def minSteps(self, n: int) -> int:
        # Step 1: Handle base case
        if n == 1:
            return 0

        # Step 2: Greedy factorization
        steps = 0
        d = 2  # Start with smallest prime
        while n > 1:
            while n % d == 0:
                steps += d  # Add factor to steps
                n //= d     # Reduce n
            d += 1  # Try next divisor
            # Optimization: If d exceeds sqrt(n), n is prime
            if d * d > n:
                if n > 1:
                    steps += n  # n is prime, add it
                break

        # Step 3: Return total steps
        return steps

• Lines 4-5: Base case: n = 1 needs 0 steps.
• Lines 8-19: Factorize:
• Start with d = 2, add d each time n is divisible, reduce n.
• Optimize: If d² > n and n > 1, n is prime, add n.
• Line 22: Return sum of factors.
• Time Complexity: O(√n)—prime factorization up to √n.
• Space Complexity: O(1)—few variables.

This is like a factor adder—greedy and fast!

Alternative Solution: Dynamic Programming

Why an Alternative Approach?

The dynamic programming (DP) approach computes minimum steps for all numbers up to n—O(n²) time, O(n) space. It’s more systematic, building solutions from smaller subproblems, but slower and memory-heavy. It’s like planning every step count up to your target!

How It Works

Picture this as a step planner:

• Step 1: Init DP array:
• dp[i]: Min steps to reach i 'A's.
• dp[1] = 0.
• Step 2: Fill DP:
• For i from 2 to n:
• Try all j < i where i % j = 0 (j is clipboard size).
• dp[i] = min(dp[i], dp[j] + i//j).
• Step 3: Return dp[n].

It’s like a step builder—plan and compute!

Step-by-Step Example

Example: n = 4

• Step 1: Init: dp = [inf, 0, inf, inf, inf].
• Step 2: Fill:
• i=2: dp[2] = dp[1] + 2//1 = 0 + 2 = 2.
• i=3: dp[3] = dp[1] + 3//1 = 0 + 3 = 3.
• i=4:
• j=1: dp[1] + 4//1 = 0 + 4 = 4.
• j=2: dp[2] + 4//2 = 2 + 2 = 4.
• dp[4] = min(4, 4) = 4.
• Step 3: Return dp[4] = 4.
• Output: 4.

Code for DP Approach

Copy codeclass Solution:
    def minSteps(self, n: int) -> int:
        # Step 1: Initialize DP array
        dp = [float('inf')] * (n + 1)
        dp[1] = 0

        # Step 2: Fill DP
        for i in range(2, n + 1):
            for j in range(1, i):
                if i % j == 0:
                    dp[i] = min(dp[i], dp[j] + i // j)

        # Step 3: Return minimum steps for n
        return dp[n]

• Lines 4-6: Init DP with infinity, dp[1] = 0.
• Lines 9-11: Fill:
• For each i, try divisors j, update min steps.
• Line 14: Return dp[n].
• Time Complexity: O(n²)—n states, O(n) divisors each.
• Space Complexity: O(n)—DP array.

It’s a step calculator—systematic but heavy!

Comparing the Two Solutions

• Greedy (Best):
• Pros: O(√n) time, O(1) space, fast and minimal.
• Cons: Factorization logic less obvious.
• DP (Alternative):
• Pros: O(n²) time, O(n) space, systematic.
• Cons: Slower, more memory.

Greedy wins for efficiency.

Additional Examples and Edge Cases

• Input: n = 1
• Output: 0.
• Input: n = 9
• Output: 6 (3 * 3 = 9, steps = 3 + 3).

Both handle these well.

Complexity Breakdown

• Greedy: Time O(√n), Space O(1).
• DP: Time O(n²), Space O(n).

Greedy is optimal.

Key Takeaways

• Greedy: Prime factors—smart!
• DP: Step-by-step—clear!
• Steps: Building is fun.
• Python Tip: Math rocks—see Python Basics.

Final Thoughts: Type Those 'A's

LeetCode 650: 2 Keys Keyboard in Python is a neat optimization challenge. Greedy with prime factorization offers speed and elegance, while DP provides a thorough alternative. Want more? Try LeetCode 279: Perfect Squares or LeetCode 322: Coin Change. Ready to paste? Head to Solve LeetCode 650 on LeetCode and minimize those steps today!