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Leetcode 650. 2 Keys Keyboard

Problem Statement

You have a text editor with only two operations:

  1. “Copy All”: You can copy all the text present on the screen (copy operation does not delete the text that’s already on the screen).
  2. “Paste”: You can paste the text which was last copied.

Initially, you have only one character ‘A’ on the screen. You need to get exactly n ‘A’s on the screen by performing the minimum number of operations.

Write a function minSteps that takes an integer n and returns the minimum number of operations to get n ‘A’s on the screen.

Clarifying Questions

  1. What is the range of n?
    • The problem constraints usually ensure that n is a positive integer. For this problem, 1 <= n <= 1000.
  2. Is there any initial number of operations or state?
    • The only character initially on the screen is one ‘A’.

Strategy

The problem can be approached using dynamic programming (DP) or prime factorization technique to minimize the operations.

Dynamic Programming Approach:

  1. Define DP array:
    • Create a DP array dp of size n+1, where dp[i] represents the minimum number of operations needed to get exactly i ‘A’s.
  2. Initial State:
    • dp[1] = 0, since we already have one ‘A’ on the screen.
  3. Transition:
    • For each i from 2 to n, determine the minimum operations by checking for all possible j (1 <= j < i) such that i is divisible by j. This means we can generate i from j by copying j and pasting (i/j)-1 times.
  4. Formula:
    • dp[i] = dp[j] + (i/j) for every j that divides i.

Time Complexity:

Code

Here’s how the solution can be implemented in C++:

#include <vector>

class Solution {
public:
    int minSteps(int n) {
        if (n == 1) return 0;
        
        std::vector<int> dp(n + 1, 0);
        
        for (int i = 2; i <= n; ++i) {
            dp[i] = i; // Initialize with maximum operations possibly required.
            for (int j = 1; j <= i / 2; ++j) {
                if (i % j == 0) {
                    dp[i] = std::min(dp[i], dp[j] + i / j);
                }
            }
        }
        
        return dp[n];
    }
};

Explanation:

  1. Initialize a DP array dp with n+1 elements, all set to 0.
  2. Iterate from 2 to n.
  3. For each i, initialize dp[i] with i assuming the worst case (only pasting each step).
  4. For each i, check all possible j from 1 to i/2 to see if i is divisible by j.
  5. Update dp[i] with the minimum value found by performing the necessary operations.

This solution efficiently determines the minimum steps required to achieve exactly n ‘A’s on the screen.

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