650

650. 2 Keys Keyboard (Medium)

Initially on a notepad only one character 'A' is present. You can perform two operations on this notepad for each step:

1. Copy All: You can copy all the characters present on the notepad (partial copy is not allowed).
2. Paste: You can paste the characters which are copied last time.

 

Given a number n. You have to get exactly n 'A' on the notepad by performing the minimum number of steps permitted. Output the minimum number of steps to get n 'A'.

Example 1:

Input: 3
Output: 3
Explanation:
Intitally, we have one character 'A'.
In step 1, we use Copy All operation.
In step 2, we use Paste operation to get 'AA'.
In step 3, we use Paste operation to get 'AAA'.

 

Note:

1. The n will be in the range [1, 1000].

 

Related Topics:
Dynamic Programming

Similar Questions:

• 4 Keys Keyboard (Medium)
• Broken Calculator (Medium)

Solution 1. DP

Let f(n) be the answer given input n.

f(1) = 0
f(2) = 2 (cp [1*2])
f(3) = 3 (cpp [1*3])
f(4) = 4 (cpcp [2*2]; cppp [1*4])
f(5) = 5 (cpppp [1*5])
f(6) = 5 (cppppp [1*6]; cpcpp [2*3]; cppcp [3*2])
f(7) = 7 (cpppppp [1*7])

So the pattern is that, if n = a * b, then f(a) + b is one option of f(n). We can get all combinations of a and b and f(n) = min(f(a) + b | n = a * b).

f(8) = 6 (0+8 [1*8]; 2+4 [2*4]; 4+2 [4*2])
f(9) = 6 (0+9 [1*9]; 3+3 [3*3])
f(10) = 7 (0+10 [1*10]; 2+5 [2*5]; 5+2 [5*2])
f(11) = 11 (0+11 [1*11])
f(12) = 7 (0+12 [1*12]; 2+6 [2*6]; 4+3[3*4]; 4+3 [4*3]; 5+2 [6*2])

// OJ: https://leetcode.com/problems/2-keys-keyboard/
// Author: github.com/lzl124631x
// Time: O(N^(3/2))
// Space: O(N)
class Solution {
public:
    int minSteps(int n) {
        vector<int> dp(n + 1, INT_MAX);
        dp[1] = 0;
        for (int i = 2; i <= n; ++i) {
            for (int j = 1, end = sqrt(i); j <= end; ++j) {
                if (i % j) continue;
                dp[i] = min({ dp[i], dp[j] + i / j, j == 1 ? INT_MAX : (dp[i / j] + j) });
            }
        }
        return dp[n];
    }
};

Solution 2. DP

For n >= 2, f(n) is at most n, i.e. cppp...ppp (1 c and n - 1 ps)

By observation, if a != 1 is the smallest factor of n, then f(n) = f(n / a) + a.

// OJ: https://leetcode.com/problems/2-keys-keyboard/
// Author: github.com/lzl124631x
// Time: O(N^(3/2))
// Space: O(N)
class Solution {
public:
    int minSteps(int n) {
        vector<int> dp(n + 1, INT_MAX);
        dp[1] = 0;
        for (int i = 2; i <= n; ++i) {
            dp[i] = i;
            for (int j = 2, end = sqrt(i); j <= end; ++j) {
                if (i % j) continue;
                dp[i] = min(dp[i], dp[i / j] + j);
            }
        }
        return dp[n];
    }
};

Solution 3. Greedy (Integer Factorization)

// OJ: https://leetcode.com/problems/2-keys-keyboard/
// Author: github.com/lzl124631x
// Time: O(sqrt(N))
// Space: O(1)
// Ref: https://leetcode.com/problems/2-keys-keyboard/solution/
class Solution {
public:
    int minSteps(int n) {
        int ans = 0, d = 2;
        while (n > 1) {
            while (n % d == 0) {
                ans += d;
                n /= d;
            }
            ++d;
        }
        return ans;
    }
};