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2403. Minimum Time to Kill All Monsters πŸ”’

Description

You are given an integer array power where power[i] is the power of the ith monster.

You start with 0 mana points, and each day you increase your mana points by gain where gain initially is equal to 1.

Each day, after gaining gain mana, you can defeat a monster if your mana points are greater than or equal to the power of that monster. When you defeat a monster:

  • your mana points will be reset to 0, and
  • the value of gain increases by 1.

Return the minimum number of days needed to defeat all the monsters.

 

Example 1:

Input: power = [3,1,4]
Output: 4
Explanation: The optimal way to beat all the monsters is to:
- Day 1: Gain 1 mana point to get a total of 1 mana point. Spend all mana points to kill the 2nd monster.
- Day 2: Gain 2 mana points to get a total of 2 mana points.
- Day 3: Gain 2 mana points to get a total of 4 mana points. Spend all mana points to kill the 3rd monster.
- Day 4: Gain 3 mana points to get a total of 3 mana points. Spend all mana points to kill the 1st monster.
It can be proven that 4 is the minimum number of days needed.

Example 2:

Input: power = [1,1,4]
Output: 4
Explanation: The optimal way to beat all the monsters is to:
- Day 1: Gain 1 mana point to get a total of 1 mana point. Spend all mana points to kill the 1st monster.
- Day 2: Gain 2 mana points to get a total of 2 mana points. Spend all mana points to kill the 2nd monster.
- Day 3: Gain 3 mana points to get a total of 3 mana points.
- Day 4: Gain 3 mana points to get a total of 6 mana points. Spend all mana points to kill the 3rd monster.
It can be proven that 4 is the minimum number of days needed.

Example 3:

Input: power = [1,2,4,9]
Output: 6
Explanation: The optimal way to beat all the monsters is to:
- Day 1: Gain 1 mana point to get a total of 1 mana point. Spend all mana points to kill the 1st monster.
- Day 2: Gain 2 mana points to get a total of 2 mana points. Spend all mana points to kill the 2nd monster.
- Day 3: Gain 3 mana points to get a total of 3 mana points.
- Day 4: Gain 3 mana points to get a total of 6 mana points.
- Day 5: Gain 3 mana points to get a total of 9 mana points. Spend all mana points to kill the 4th monster.
- Day 6: Gain 4 mana points to get a total of 4 mana points. Spend all mana points to kill the 3rd monster.
It can be proven that 6 is the minimum number of days needed.

 

Constraints:

  • 1 <= power.length <= 17
  • 1 <= power[i] <= 109

Solutions

We note that the number of monsters is at most \(17\), which means we can use a 17-bit binary number to represent the state of the monsters. The \(i\)-th bit being \(1\) indicates that the \(i\)-th monster is still alive, and \(0\) indicates that the \(i\)-th monster has been defeated.

We design a function \(\textit{dfs}(\textit{mask})\) to represent the minimum number of days needed to defeat all monsters when the current state of the monsters is \(\textit{mask}\). The answer is \(\textit{dfs}(2^n - 1)\), where \(n\) is the number of monsters.

The calculation of the function \(\textit{dfs}(\textit{mask})\) is as follows:

  • If \(\textit{mask} = 0\), it means all monsters have been defeated, return \(0\);
  • Otherwise, we enumerate each monster \(i\). If the \(i\)-th monster is still alive, we can choose to defeat the \(i\)-th monster, then recursively calculate \(\textit{dfs}(\textit{mask} \oplus 2^i)\), and update the answer to \(\textit{ans} = \min(\textit{ans}, \textit{dfs}(\textit{mask} \oplus 2^i) + \lceil \frac{x}{\textit{gain}} \rceil)\), where \(x\) is the strength of the \(i\)-th monster, and \(\textit{gain} = 1 + (n - \textit{mask}.\textit{bit\_count}())\) represents the current daily mana gain.

Finally, we return \(\textit{dfs}(2^n - 1)\).

The time complexity is \(O(2^n \times n)\), and the space complexity is \(O(2^n)\). Here, \(n\) is the number of monsters.

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class Solution:
    def minimumTime(self, power: List[int]) -> int:
        @cache
        def dfs(mask: int) -> int:
            if mask == 0:
                return 0
            ans = inf
            gain = 1 + (n - mask.bit_count())
            for i, x in enumerate(power):
                if mask >> i & 1:
                    ans = min(ans, dfs(mask ^ (1 << i)) + (x + gain - 1) // gain)
            return ans

        n = len(power)
        return dfs((1 << n) - 1)

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class Solution {
    private int n;
    private int[] power;
    private Long[] f;

    public long minimumTime(int[] power) {
        n = power.length;
        this.power = power;
        f = new Long[1 << n];
        return dfs((1 << n) - 1);
    }

    private long dfs(int mask) {
        if (mask == 0) {
            return 0;
        }
        if (f[mask] != null) {
            return f[mask];
        }
        f[mask] = Long.MAX_VALUE;
        int gain = 1 + (n - Integer.bitCount(mask));
        for (int i = 0; i < n; ++i) {
            if ((mask >> i & 1) == 1) {
                f[mask] = Math.min(f[mask], dfs(mask ^ 1 << i) + (power[i] + gain - 1) / gain);
            }
        }
        return f[mask];
    }
}

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class Solution {
public:
    long long minimumTime(vector<int>& power) {
        int n = power.size();
        long long f[1 << n];
        memset(f, -1, sizeof(f));
        auto dfs = [&](this auto&& dfs, int mask) -> long long {
            if (mask == 0) {
                return 0;
            }
            if (f[mask] != -1) {
                return f[mask];
            }
            f[mask] = LLONG_MAX;
            int gain = 1 + (n - __builtin_popcount(mask));
            for (int i = 0; i < n; ++i) {
                if (mask >> i & 1) {
                    f[mask] = min(f[mask], dfs(mask ^ (1 << i)) + (power[i] + gain - 1) / gain);
                }
            }
            return f[mask];
        };
        return dfs((1 << n) - 1);
    }
};

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func minimumTime(power []int) int64 {
    n := len(power)
    f := make([]int64, 1<<n)
    for i := range f {
        f[i] = -1
    }
    var dfs func(mask int) int64
    dfs = func(mask int) int64 {
        if mask == 0 {
            return 0
        }
        if f[mask] != -1 {
            return f[mask]
        }
        f[mask] = 1e18
        gain := 1 + (n - bits.OnesCount(uint(mask)))
        for i, x := range power {
            if mask>>i&1 == 1 {
                f[mask] = min(f[mask], dfs(mask^(1<<i))+int64(x+gain-1)/int64(gain))
            }
        }
        return f[mask]
    }
    return dfs(1<<n - 1)
}

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function minimumTime(power: number[]): number {
    const n = power.length;
    const f: number[] = Array(1 << n).fill(-1);
    const dfs = (mask: number): number => {
        if (mask === 0) {
            return 0;
        }
        if (f[mask] !== -1) {
            return f[mask];
        }
        f[mask] = Infinity;
        const gain = 1 + (n - bitCount(mask));
        for (let i = 0; i < n; ++i) {
            if ((mask >> i) & 1) {
                f[mask] = Math.min(f[mask], dfs(mask ^ (1 << i)) + Math.ceil(power[i] / gain));
            }
        }
        return f[mask];
    };
    return dfs((1 << n) - 1);
}

function bitCount(i: number): number {
    i = i - ((i >>> 1) & 0x55555555);
    i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
    i = (i + (i >>> 4)) & 0x0f0f0f0f;
    i = i + (i >>> 8);
    i = i + (i >>> 16);
    return i & 0x3f;
}

Solution 2: State Compression + Dynamic Programming

We can convert the memoization search in Solution 1 to dynamic programming. Define \(f[\textit{mask}]\) to represent the minimum number of days needed to defeat all monsters when the current state of the monsters is \(\textit{mask}\). Here, \(\textit{mask}\) is an \(n\)-bit binary number, where the \(i\)-th bit being \(1\) indicates that the \(i\)-th monster has been defeated, and \(0\) indicates that the \(i\)-th monster is still alive. Initially, \(f[0] = 0\), and the rest \(f[\textit{mask}] = +\infty\). The answer is \(f[2^n - 1]\).

We enumerate \(\textit{mask}\) in the range \([1, 2^n - 1]\). For each \(\textit{mask}\), we enumerate each monster \(i\). If the \(i\)-th monster is defeated, it can be transferred from the previous state \(\textit{mask} \oplus 2^i\), with a transfer cost of \((\textit{power}[i] + \textit{gain} - 1) / \textit{gain}\), where \(\textit{gain} = \textit{mask}.\textit{bitCount}()\).

Finally, return \(f[2^n - 1]\).

The time complexity is \(O(2^n \times n)\), and the space complexity is \(O(2^n)\). Here, \(n\) is the number of monsters.

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class Solution:
    def minimumTime(self, power: List[int]) -> int:
        n = len(power)
        f = [inf] * (1 << n)
        f[0] = 0
        for mask in range(1, 1 << n):
            gain = mask.bit_count()
            for i, x in enumerate(power):
                if mask >> i & 1:
                    f[mask] = min(f[mask], f[mask ^ (1 << i)] + (x + gain - 1) // gain)
        return f[-1]

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class Solution {
    public long minimumTime(int[] power) {
        int n = power.length;
        long[] f = new long[1 << n];
        Arrays.fill(f, Long.MAX_VALUE);
        f[0] = 0;
        for (int mask = 1; mask < 1 << n; ++mask) {
            int gain = Integer.bitCount(mask);
            for (int i = 0; i < n; ++i) {
                if ((mask >> i & 1) == 1) {
                    f[mask] = Math.min(f[mask], f[mask ^ 1 << i] + (power[i] + gain - 1) / gain);
                }
            }
        }
        return f[(1 << n) - 1];
    }
}

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class Solution {
public:
    long long minimumTime(vector<int>& power) {
        int n = power.size();
        long long f[1 << n];
        memset(f, 0x3f, sizeof(f));
        f[0] = 0;
        for (int mask = 1; mask < 1 << n; ++mask) {
            int gain = __builtin_popcount(mask);
            for (int i = 0; i < n; ++i) {
                if (mask >> i & 1) {
                    f[mask] = min(f[mask], f[mask ^ (1 << i)] + (power[i] + gain - 1) / gain);
                }
            }
        }
        return f[(1 << n) - 1];
    }
};

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func minimumTime(power []int) int64 {
    n := len(power)
    f := make([]int64, 1<<n)
    for i := range f {
        f[i] = 1e18
    }
    f[0] = 0
    for mask := 1; mask < 1<<n; mask++ {
        gain := bits.OnesCount(uint(mask))
        for i, x := range power {
            if mask>>i&1 == 1 {
                f[mask] = min(f[mask], f[mask^(1<<i)]+int64(x+gain-1)/int64(gain))
            }
        }
    }
    return f[1<<n-1]
}

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function minimumTime(power: number[]): number {
    const n = power.length;
    const f: number[] = Array(1 << n).fill(Infinity);
    f[0] = 0;
    for (let mask = 1; mask < 1 << n; ++mask) {
        const gain = bitCount(mask);
        for (let i = 0; i < n; ++i) {
            if ((mask >> i) & 1) {
                f[mask] = Math.min(f[mask], f[mask ^ (1 << i)] + Math.ceil(power[i] / gain));
            }
        }
    }
    return f.at(-1)!;
}

function bitCount(i: number): number {
    i = i - ((i >>> 1) & 0x55555555);
    i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
    i = (i + (i >>> 4)) & 0x0f0f0f0f;
    i = i + (i >>> 8);
    i = i + (i >>> 16);
    return i & 0x3f;
}

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