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.

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