Skip to main content

Dynamic Programming - Medium Level - Question 2

Dynamic Programming - Medium Level - Question 2


Leetcode 174 Dungeon Game

The demons had captured the princess and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of m x n rooms laid out in a 2D grid. Our valiant knight was initially positioned in the top-left room and must fight his way through dungeon to rescue the princess.

The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.

Some of the rooms are guarded by demons (represented by negative integers), so the knight loses health upon entering these rooms; other rooms are either empty (represented as 0) or contain magic orbs that increase the knight's health (represented by positive integers).

To reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.

Return the knight's minimum initial health so that he can rescue the princess.

Note that any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned.

Constraints:

m == dungeon.length

n == dungeon[i].length

1 <= m, n <= 200

-1000 <= dungeon[i][j] <= 1000


Analysis:

If we start from the left top corner to think about this question, it seems not easy... In this kind of situation, what about starting from the last step?

At the last step, without knowing all the previous data, we can calculated the minimum health point when the knight reaches this position.

If the dungeon[m-1][n-1] >= 0, then the minimum value is 1; if negative, it should be 1 - dungeon[m-1][n-1].

This can be the starting point for this question.

If we define f(i, j) is the minimum health point when reaching at location (i, j), then f(i, j) can be derived from f(i+1, j) and f(i, j+1).

int val = min(f(i+1, j), f(i, j+1)) - dungeon[i][j]

If val is positive, (for example when dungeon[i][j] is negative, or smaller than min(f(i+1, j), f(i, j+1)), then f(i, j) should be at least be val; otherwise, f(i, j) can be set as 1, since dungeon[i][j] is large enough for the requirement of next step, f(i+1, j) or f(i, j+1). Thus,

f(i, j) = val <= 0 ? 1 : val

f(0, 0) would be the final result wanted.


See the code below:

class Solution {
public:
    int calculateMinimumHP(vector<vector<int>>& dungeon) {
        if(!dungeon.size() || !dungeon.front().size()) return 0;
        int m = dungeon.size(), n = dungeon.front().size();
        vector<vector<int>> dp(m+1, vector<int>(n+1, INT_MAX));
        dp[m-1][n-1] = dungeon[m-1][n-1] >= 0 ? 1 : 1 - dungeon[m-1][n-1]; 
        for(int i=m-1; i>=0; --i) {
            for(int j=n-1; j>=0; --j) {
                if(i==m-1 && j==n-1) continue;
                int val = min(dp[i+1][j], dp[i][j+1]) - dungeon[i][j];
                dp[i][j] = (val <= 0 ? 1 : val);
            }
        }
        return dp[0][0];
    }
};


Based on the "rolling array" trick, the space complexity can be reduced to O(n). 

See the code below:


class Solution {
public:
    int calculateMinimumHP(vector<vector<int>>& dungeon) {
        if(!dungeon.size() || !dungeon.front().size()) return 0;
        int m = dungeon.size(), n = dungeon.front().size();
        vector<int> dp(n+1, INT_MAX);
        dp[n-1] = dungeon[m-1][n-1] >= 0 ? 1 : 1 - dungeon[m-1][n-1]; 
        for(int i=m-1; i>=0; --i) {
            for(int j=n-1; j>=0; --j) {
                if(i==m-1 && j==n-1) continue;
                int val = min(dp[j], dp[j+1]) - dungeon[i][j];
                dp[j] = (val <= 0 ? 1 : val);
            }
        }
        return dp[0];
    }
};


If still want to start from the left top corner, a different algorithm can be applied: binary search.

The basic idea is that we first make a guess of the initial value: mid. Then we do a check to see whether the knight can reach the end point or not with mid health point. If no, the search range becomes the larger half; if yes, the search range is narrowed down to the smaller half.

See the code below:


class Solution {
public:
    int calculateMinimumHP(vector<vector<int>>& dungeon) {
        if(!dungeon.size() || !dungeon.front().size()) return 1;
        int left = 1, right = INT_MAX;
        while(left < right) {
            int mid = left + (right - left) /2;
            if(!isV(dungeon, mid)) left = mid + 1;
            else right = mid;
        }
        return left;
    }
private:
    bool isV(vector<vector<int>>& vs, int val) {
        int m = vs.size(), n = vs.front().size();
        vector<vector<int>> dp(m+1, vector<int>(n+1, 0));
        dp[1][1] = vs[0][0] + val;
        if(dp[1][1] <= 0) return false;
        for(int i=1; i<=m; ++i) {
            for(int j=1; j<=n; ++j) {
                if(i==1 && j==1) continue;
                int t = max(dp[i][j-1], dp[i-1][j]);
                dp[i][j] = t <= 0 ? -1 : (t + vs[i-1][j-1]);
            }
        }
        return dp[m][n] > 0;
    }
};

Note:

The overall time complexity is O(m*n*log(INT_MAX)), so it is a factor of log(INT_MAX) larger than the dp method above. But this number is smaller, which is ~ 31 or 32.


Upper Layer

Comments

Popular posts from this blog

Brute Force - Question 2

2105. Watering Plants II Alice and Bob want to water n plants in their garden. The plants are arranged in a row and are labeled from 0 to n - 1 from left to right where the ith plant is located at x = i. Each plant needs a specific amount of water. Alice and Bob have a watering can each, initially full. They water the plants in the following way: Alice waters the plants in order from left to right, starting from the 0th plant. Bob waters the plants in order from right to left, starting from the (n - 1)th plant. They begin watering the plants simultaneously. It takes the same amount of time to water each plant regardless of how much water it needs. Alice/Bob must water the plant if they have enough in their can to fully water it. Otherwise, they first refill their can (instantaneously) then water the plant. In case both Alice and Bob reach the same plant, the one with more water currently in his/her watering can should water this plant. If they have the same amount of water, then Alice ...

Sweep Line

Sweep (or scanning) line algorithm is very efficient for some specific questions involving discrete intervals. The intervals could be the lasting time of events, or the width of a building or an abstract square, etc. In the scanning line algorithm, we usually need to distinguish the start and the end of an interval. After the labeling of the starts and ends, we can sort them together based on the values of the starts and ends. Thus, if there are N intervals in total, we will have 2*N data points (since each interval will contribute 2). The sorting becomes the most time-consuming step, which is O(2N*log(2N) ~ O(N*logN). After the sorting, we usually can run a linear sweep for all the data points. If the data point is labeled as a starting point, it means a new interval is in the processing; when an ending time is reached, it means one of the interval has ended. In such direct way, we can easily figure out how many intervals are in the processes. Other related information can also be obt...

Dynamic Programming - Easy Level - Question 1

Dynamic Programming - Easy Level - Question 1 Leetcode 1646  Get Maximum in Generated Array You are given an integer n. An array nums of length n + 1 is generated in the following way: nums[0] = 0 nums[1] = 1 nums[2 * i] = nums[i] when 2 <= 2 * i <= n nums[2 * i + 1] = nums[i] + nums[i + 1] when 2 <= 2 * i + 1 <= n Return the maximum integer in the array nums​​​. Constraints: 0 <= n <= 100 Analysis: This question is quick straightforward: the state and transitional formula are given; the initialization is also given. So we can just ready the code to iterate all the states and find the maximum. See the code below: class Solution { public: int getMaximumGenerated(int n) { int res = 0; if(n<2) return n; vector<int> f(n+1, 0); f[1] = 1; for(int i=2; i<=n; ++i) { if(i&1) f[i] = f[i/2] + f[i/2+1]; else f[i] = f[i/2]; // cout<<i<<" "<<f[i]<<endl; ...