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650. 2 Keys Keyboard

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

Solution 1

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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

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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]
}

Solution 3

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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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