LeetCode 650: 2 Keys Keyboard Solution

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming. We maintain a dp array where dp[i] represents the minimum number of operations to get 'A' exactly i times on the screen. At each step, we can either copy all characters (which increments the number of characters by the current number of characters on the screen) or paste the characters (which increments the number of characters by the characters copied in the previous step). We iterate from 2 to n and update dp[i] based on the minimum of dp[i/2] + 2 (copy all and paste) or dp[j] + i/j (copy j characters and paste i/j times).

Time complexity: O(n^2)
Space complexity: O(n)

Solution Code

class Solution {
    public int minSteps(int n) {
        int[] dp = new int[n + 1];
        for (int i = 2; i <= n; i++) {
            dp[i] = i;
            for (int j = i / 2; j >= 2; j--) {
                if (i % j == 0) {
                    dp[i] = dp[j] + i / j;
                    break;
                }
            }
        }
        return dp[n];
    }
}

Frequently Asked Questions

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The time complexity for LeetCode 650 (2 Keys Keyboard) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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650. 2 Keys Keyboard