Question
Formatted question description: https://leetcode.ca/all/174.html
174 Dungeon Game
The demons had captured the princess (P) and imprisoned her in the bottomright corner of a dungeon.
The dungeon consists of M x N rooms laid out in a 2D grid.
Our valiant knight (K) was initially positioned in the topleft room
and must fight his way through the 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, so the knight loses health (negative integers) upon entering these rooms;
other rooms are either empty (0's) or contain magic orbs that increase the knight's health (positive integers).
In order to reach the princess as quickly as possible,
the knight decides to move only rightward or downward in each step.
Write a function to determine the knight's minimum initial health so that he is able to rescue the princess.
For example, given the dungeon below, the initial health of the knight must be at least 7
if he follows the optimal path RIGHT> RIGHT > DOWN > DOWN.
2(K) 3 3
5 10 1
10 30 5 (P)
Note:
The knight's health has no upper bound.
Any room can contain threats or powerups,
even the first room the knight enters and the bottomright room where the princess is imprisoned.
@tagdp
Algorithm
Dynamic Programming is used to create a twodimensional array dp, where dp[i][j] is used to represent the initial HP from the current position (i, j), and the first thing to process is the initial HP of the room where the princess is, and then slowly spread to the first room, continuously getting the best health value in each location. Reverse engineering is the essence of this question.
Formula: dp[i][j] = max(1, min(dp[i+1][j], dp[i][j+1])  dungeon[i][j])
Code
Java

/* test case: 1 (K) 3 3 0 2 0 3 3 3 (P) */ public class Dungeon_Game { class Solution { public int calculateMinimumHP(int[][] dungeon) { if (dungeon == null  dungeon[0].length == 0) { return 0; } int m = dungeon.length; int n = dungeon[0].length; int[][] dp = new int[m][n]; // dp[i][j] is used to indicate the starting HP from the current position (i, j) dp[m  1][n  1] = Math.max(1, 1  dungeon[m  1][n  1]); // Initialization the last column for (int i = m  2; i >= 0; i) { dp[i][n  1] = Math.max(1, dp[i + 1][n  1]  dungeon[i][n  1]); } // Initialization the last row for (int i = n  2; i >= 0; i) { dp[m  1][i] = Math.max(1, dp[m  1][i + 1]  dungeon[m  1][i]); } for (int i = m  2; i >= 0; i) { for (int j = n  2; j >= 0; j) { dp[i][j] = Math.max( 1, Math.min(dp[i + 1][j], dp[i][j + 1])  dungeon[i][j] ); } } return dp[0][0]; } } }

// OJ: https://leetcode.com/problems/dungeongame // Time: O(MN) // Space: O(1) class Solution { public: int calculateMinimumHP(vector<vector<int>>& A) { int M = A.size(), N = A[0].size(); for (int i = M  1; i >= 0; i) { for (int j = N  1; j >= 0; j) { int need = min(i + 1 < M ? A[i + 1][j] : INT_MAX, j + 1 < N ? A[i][j + 1] : INT_MAX); if (need == INT_MAX) need = 1; A[i][j] = max(1, need  A[i][j]); } } return A[0][0]; } };

class Solution(object): def calculateMinimumHP(self, dungeon): """ :type dungeon: List[List[int]] :rtype: int """ n = len(dungeon[0]) need = [2 ** 31] * (n  1) + [1] for row in dungeon[::1]: for j in range(n)[::1]: need[j] = max(min(need[j:j + 2])  row[j], 1) return need[0]