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650. 2 Keys Keyboard

DifficultyMedium

Description

There is only one character 'A' on the screen of a notepad. You can perform one of two operations on this notepad for each step:

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

Given an integer n, return the minimum number of operations to get the character 'A' exactly n times on the screen.

 

Example 1:

Input: n = 3
Output: 3
Explanation: Initially, 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'.

Example 2:

Input: n = 1
Output: 0

 

Constraints:

  • 1 <= n <= 1000

Solutions

Thinking

Start from one A and reach \(n\) by copy and paste. The search tree of sequences is wide.

If the last step pastes a block of length \(n/j\) into \(j\) copies, \(dfs(n)=\min(dfs(n/j)+j)\). Memoize over factors; \(n=1\) costs \(0\).

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class Solution:
    def minSteps(self, n: int) -> int:
        @cache
        def dfs(n):
            if n == 1:
                return 0
            i, ans = 2, n
            while i * i <= n:
                if n % i == 0:
                    ans = min(ans, dfs(n // i) + i)
                i += 1
            return ans

        return dfs(n)

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class Solution {
    private int[] f;

    public int minSteps(int n) {
        f = new int[n + 1];
        Arrays.fill(f, -1);
        return dfs(n);
    }

    private int dfs(int n) {
        if (n == 1) {
            return 0;
        }
        if (f[n] != -1) {
            return f[n];
        }
        int ans = n;
        for (int i = 2; i * i <= n; ++i) {
            if (n % i == 0) {
                ans = Math.min(ans, dfs(n / i) + i);
            }
        }
        f[n] = ans;
        return ans;
    }
}

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class Solution {
public:
    vector<int> f;

    int minSteps(int n) {
        f.assign(n + 1, -1);
        return dfs(n);
    }

    int dfs(int n) {
        if (n == 1) return 0;
        if (f[n] != -1) return f[n];
        int ans = n;
        for (int i = 2; i * i <= n; ++i) {
            if (n % i == 0) {
                ans = min(ans, dfs(n / i) + i);
            }
        }
        f[n] = ans;
        return ans;
    }
};

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func minSteps(n int) int {
    f := make([]int, n+1)
    for i := range f {
        f[i] = -1
    }
    var dfs func(int) int
    dfs = func(n int) int {
        if n == 1 {
            return 0
        }
        if f[n] != -1 {
            return f[n]
        }
        ans := n
        for i := 2; i*i <= n; i++ {
            if n%i == 0 {
                ans = min(ans, dfs(n/i)+i)
            }
        }
        return ans
    }
    return dfs(n)
}

Solution 2: Dynamic Programming

Thinking

Memoization is top-down. The same recurrences fill \(dp[i]\) bottom-up over factors, without recursion.

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class Solution:
    def minSteps(self, n: int) -> int:
        dp = list(range(n + 1))
        dp[1] = 0
        for i in range(2, n + 1):
            j = 2
            while j * j <= i:
                if i % j == 0:
                    dp[i] = min(dp[i], dp[i // j] + j)
                j += 1
        return dp[-1]

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class Solution {
    public int minSteps(int n) {
        int[] dp = new int[n + 1];
        for (int i = 0; i < n + 1; ++i) {
            dp[i] = i;
        }
        dp[1] = 0;
        for (int i = 2; i < n + 1; ++i) {
            for (int j = 2; j * j <= i; ++j) {
                if (i % j == 0) {
                    dp[i] = Math.min(dp[i], dp[i / j] + j);
                }
            }
        }
        return dp[n];
    }
}

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class Solution {
public:
    int minSteps(int n) {
        vector<int> dp(n + 1);
        iota(dp.begin(), dp.end(), 0);
        dp[1] = 0;
        for (int i = 2; i < n + 1; ++i) {
            for (int j = 2; j * j <= i; ++j) {
                if (i % j == 0) {
                    dp[i] = min(dp[i], dp[i / j] + j);
                }
            }
        }
        return dp[n];
    }
};

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func minSteps(n int) int {
    dp := make([]int, n+1)
    for i := range dp {
        dp[i] = i
    }
    dp[1] = 0
    for i := 2; i < n+1; i++ {
        for j := 2; j*j <= i; j++ {
            if i%j == 0 {
                dp[i] = min(dp[i], dp[i/j]+j)
            }
        }
    }
    return dp[n]
}

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function minSteps(n: number): number {
    const dp = Array(n + 1).fill(1000);
    dp[1] = 0;

    for (let i = 2; i <= n; i++) {
        for (let j = 1, half = i / 2; j <= half; j++) {
            if (i % j === 0) {
                dp[i] = Math.min(dp[i], dp[j] + i / j);
            }
        }
    }

    return dp[n];
}

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/**
 * @param {number} n
 * @return {number}
 */
var minSteps = function (n) {
    const dp = Array(n + 1).fill(1000);
    dp[1] = 0;

    for (let i = 2; i <= n; i++) {
        for (let j = 1, half = i / 2; j <= half; j++) {
            if (i % j === 0) {
                dp[i] = Math.min(dp[i], dp[j] + i / j);
            }
        }
    }

    return dp[n];
};

Solution 3: Math

Thinking

The optimum equals the sum of prime factors of \(n\): factor \(i\) is one copy plus \(i-1\) pastes. Factorize \(n\) and skip the DP table.

Factorize \(n\); each prime factor \(i\) costs \(i\) operations.

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class Solution {
    public int minSteps(int n) {
        int res = 0;
        for (int i = 2; n > 1; ++i) {
            while (n % i == 0) {
                res += i;
                n /= i;
            }
        }
        return res;
    }
}

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