LeetCode 651: 4 Keys Keyboard Solution in Python – A Step-by-Step Guide

What Is LeetCode 651: 4 Keys Keyboard?

In LeetCode 651: 4 Keys Keyboard, you’re given an integer n representing the number of keystrokes allowed on a notepad initially empty. You can use four operations:

• Type 'A': Add one 'A' (1 keystroke).
• Ctrl+A: Select all current text (1 keystroke).
• Ctrl+C: Copy selected text to clipboard (1 keystroke).
• Ctrl+V: Paste clipboard contents (1 keystroke).

Your task is to find the maximum number of 'A's you can produce with exactly n keystrokes, returning this count as an integer. For example, with n = 3, you can type "AAA" (3 'A's) in 3 steps, but with n = 7, smarter sequences yield 9 'A's. This problem tests your ability to optimize a sequence of operations under constraints.

Problem Statement

• Input:
• n: An integer (number of keystrokes).
• Output: An integer—maximum number of 'A's with n keystrokes.
• Rules:
• Start with empty notepad.
• Use Type 'A', Ctrl+A, Ctrl+C, Ctrl+V.
• Exactly n keystrokes, maximize 'A's.

Constraints

• 1 ≤ n ≤ 50.

Examples

• Input: n = 3
• Output: 3 (Type "A" thrice: "AAA")
• Input: n = 7
• Output: 9 (Type "AA", Ctrl+A, Ctrl+C, Ctrl+V thrice: "AAAAAAAAA")
• Input: n = 1
• Output: 1 (Type "A")

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

Understanding the Problem: Maximizing 'A's

To solve LeetCode 651: 4 Keys Keyboard in Python, we need to maximize the number of 'A's produced with exactly n keystrokes using four operations. A brute-force approach—trying all possible sequences—would be exponential (O(4^n)), impractical for n ≤ 50. Since operations build on copying and pasting previous states, we can optimize with DP or greedy strategies. We’ll use:

• Best Solution (Dynamic Programming with Optimization): O(n²) time, O(n) space—fast and precise.
• Alternative Solution (Greedy with Iteration): O(n) time, O(1) space—simpler but heuristic-based.

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

Best Solution: Dynamic Programming with Optimization

Why This Is the Best Solution

The dynamic programming (DP) with optimization approach is the top choice for LeetCode 651 because it’s efficient—O(n²) time with O(n) space—and systematically computes the maximum 'A's for each keystroke count by considering all prior states, ensuring optimality. It fits constraints (n ≤ 50) well and balances complexity with accuracy. It’s like planning every keystroke to get the most out of your moves!

How It Works

Think of this as a keystroke planner:

• Step 1: Initialize DP Array:
• dp[i]: Max 'A's with i keystrokes.
• Base: dp[0] = 0, dp[1] = 1, etc., up to dp[i] = i (typing 'A' i times).
• Step 2: Fill DP:
• For i from 1 to n:
• Option 1: Type 'A' (dp[i-1] + 1).
• Option 2: Copy-paste sequence:
• For j < i (last copy at j):
• Steps = j + 2 (to copy) + k (pastes).
• If steps ≤ i, 'A's = dp[j] * (k + 1).
• Take max of all options.
• Step 3: Return Result:
• Return dp[n].

It’s like a step maximizer—plan and grow!

Step-by-Step Example

Example: n = 7

• Step 1: Init: dp = [0, 1, 2, 3, 4, 5, 6, 7].
• Step 2: Fill:
• i=1: dp[1] = 1 (type A).
• i=2: dp[2] = 2 (type AA).
• i=3: dp[3] = 3 (type AAA).
• i=4:
• Type: dp[3] + 1 = 4.
• j=1: 1+2+1=4, dp[1]*2 = 2.
• dp[4] = max(4, 2) = 4.
• i=5:
• Type: dp[4] + 1 = 5.
• j=2: 2+2+1=5, dp[2]*2 = 4.
• dp[5] = 5.
• i=6:
• Type: dp[5] + 1 = 6.
• j=3: 3+2+1=6, dp[3]*2 = 6.
• j=2: 2+2+2=6, dp[2]*3 = 6.
• dp[6] = 6.
• i=7:
• Type: dp[6] + 1 = 7.
• j=4: 4+2+1=7, dp[4]*2 = 8.
• j=3: 3+2+2=7, dp[3]*3 = 9.
• dp[7] = 9.
• Step 3: Return dp[7] = 9.
• Output: 9.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained clearly:

Copy codeclass Solution:
    def maxA(self, n: int) -> int:
        # Step 1: Initialize DP array
        dp = [0] * (n + 1)
        for i in range(1, min(n + 1, 7)):  # Base case: type A up to 6
            dp[i] = i

        # Step 2: Fill DP with optimization
        for i in range(7, n + 1):
            dp[i] = dp[i - 1] + 1  # Type A
            # Try copy-paste sequences
            for j in range(i - 3, 0, -1):  # Last copy at j
                steps_needed = j + 2  # Copy all + Ctrl+C
                pastes = (i - steps_needed)  # Remaining steps for paste
                if pastes >= 1:
                    dp[i] = max(dp[i], dp[j] * (pastes + 1))
                else:
                    break  # No paste possible

        # Step 3: Return maximum A's for n
        return dp[n]

• Lines 4-6: Init: dp[i] = i for small i (typing 'A').
• Lines 9-18: Fill:
• Type 'A' as baseline.
• Try j as last copy point, compute pastes, update max.
• Line 21: Return dp[n].
• Time Complexity: O(n²)—n states, O(n) inner loop.
• Space Complexity: O(n)—DP array.

This is like an 'A' maximizer—plan and type!

Alternative Solution: Greedy with Iteration

Why an Alternative Approach?

The greedy with iteration approach approximates the max 'A's—O(n) time, O(1) space. It’s simpler, using a heuristic to balance typing and pasting, but not guaranteed optimal without exhaustive checks. It’s like guessing the best copy-paste rhythm!

How It Works

Picture this as a rhythm finder:

• Step 1: For small n, type 'A' (n steps).
• Step 2: For larger n, iterate paste counts:
• Assume copy at k, paste m times, total steps = k + m + 1.
• 'A's = k * (m + 1).
• Maximize within n steps.
• Step 3: Return max 'A's.

It’s like a paste guesser—try and max!

Step-by-Step Example

Example: n = 7

• Step 1: n ≤ 6, try typing: 7 'A's.
• Step 2: Try sequences:
• k=2, m=3: 2+3+1=6, 2*4=8.
• k=3, m=2: 3+2+1=6, 3*3=9.
• k=4, m=1: 4+1+1=6, 4*2=8.
• Max within 7: 9 (adjust k=3, m=3 invalid, use k=3, m=2).
• Step 3: Return 9.
• Output: 9.

Code for Greedy Approach

Copy codeclass Solution:
    def maxA(self, n: int) -> int:
        # Step 1: Small n, just type A
        if n <= 6:
            return n

        # Step 2: Greedy iteration
        max_a = n
        for k in range(1, n - 2):  # k = initial A's
            remaining = n - k - 1  # Steps left after copy
            if remaining <= 0:
                break
            pastes = remaining
            curr_a = k * (pastes + 1)
            max_a = max(max_a, curr_a)

        # Step 3: Return maximum A's
        return max_a

• Lines 4-5: Small n: type 'A'.
• Lines 8-14: Iterate:
• Try k 'A's, compute pastes, maximize 'A's.
• Line 17: Return max.
• Time Complexity: O(n)—single loop.
• Space Complexity: O(1)—few variables.

It’s a greedy typer—guess and max!

Comparing the Two Solutions

• DP (Best):
• Pros: O(n²) time, O(n) space, guaranteed optimal.
• Cons: More complex.
• Greedy (Alternative):
• Pros: O(n) time, O(1) space, simple.
• Cons: Heuristic, may miss optimal.

DP wins for accuracy.

Additional Examples and Edge Cases

• Input: n = 1
• Output: 1.
• Input: n = 9
• Output: 12 (AA, Ctrl+A, Ctrl+C, Ctrl+V*4).

DP handles these precisely.

Complexity Breakdown

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

DP is optimal for correctness.

Key Takeaways

• DP: Planned precision—smart!
• Greedy: Quick guess—clear!
• Keys: Typing is fun.
• Python Tip: DP rocks—see Python Basics.

Final Thoughts: Max Those 'A's

LeetCode 651: 4 Keys Keyboard in Python is a fun optimization challenge. DP with optimization ensures the max 'A's, while greedy offers a quick heuristic. Want more? Try LeetCode 650: 2 Keys Keyboard or LeetCode 279: Perfect Squares. Ready to type? Head to Solve LeetCode 651 on LeetCode and maximize those 'A's today!