LeetCode 651: 4 Keys Keyboard Solution
Master LeetCode problem 651 (4 Keys Keyboard), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
651. 4 Keys Keyboard
Problem Explanation
Explanation:
This problem asks us to find the maximum number of 'A's that can be printed on the screen using the following operations:
- 'A': Print one 'A' on the screen.
- Ctrl-A: Select the whole screen.
- Ctrl-C: Copy the selected text to the buffer.
- Ctrl-V: Print the content of the buffer on the screen.
To solve this problem, we can use dynamic programming. We can maintain a dp array where dp[i] represents the maximum number of 'A's that can be printed with i keystrokes. At each step, we can either press 'A' or use Ctrl-A, Ctrl-C, and Ctrl-V. We can iterate through the possible number of keystrokes and calculate the maximum number of 'A's that can be printed.
Algorithm Steps:
- Initialize a dp array of size n+1 to store the maximum number of 'A's that can be printed with i keystrokes.
- Iterate from i=1 to n and for each i:
- Initialize dp[i] = i (press 'A' i times).
- Iterate j from i-3 down to 1 (to consider the Ctrl-A, Ctrl-C, Ctrl-V operations):
- Update dp[i] to max(dp[i], dp[j] * (i-j-1)).
The final result will be stored in dp[n], which represents the maximum number of 'A's that can be printed with n keystrokes.
Time Complexity: The time complexity of this algorithm is O(n^2) where n is the given number of keystrokes.
Space Complexity: The space complexity is O(n) to store the dp array.
:
Solution Code
class Solution {
public int maxA(int n) {
int[] dp = new int[n + 1];
for (int i = 1; i <= n; i++) {
dp[i] = i;
for (int j = i - 3; j >= 1; j--) {
dp[i] = Math.max(dp[i], dp[j] * (i - j - 1));
}
}
return dp[n];
}
}Try It Yourself
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