Dungeon Game

The demons had captured the princess (P) 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 (K) was initially positioned in the top-left 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 power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned.

 class Solution {
    public int calculateMinimumHP(int[][] dungeon) {
    int m = dungeon.length;
    int n = dungeon[0].length;

    int[][] dp = new int[m][n];
 
    dp[m - 1][n - 1] = Math.max(1 - dungeon[m - 1][n - 1], 1); //step 1
 
    for (int i = m - 2; i >= 0; i--) {  //step 2
        dp[i][n - 1] = Math.max(dp[i + 1][n - 1] - dungeon[i][n - 1], 1);
    }
 
    for (int j = n - 2; j >= 0; j--) {  //step 3
        dp[m - 1][j] = Math.max(dp[m - 1][j + 1] - dungeon[m - 1][j], 1);
    }
 
    for (int i = m - 2; i >= 0; i--) {   //step 4
        for (int j = n - 2; j >= 0; j--) {
            int down = Math.max(dp[i + 1][j] - dungeon[i][j], 1);
            int right = Math.max(dp[i][j + 1] - dungeon[i][j], 1);
            dp[i][j] = Math.min(right, down);
        }
    }
 
    return dp[0][0];
}
}


Here, we are going to solve using bottom up approach. While solving take care of all important points given in problem.

Below are steps to be followed for solving this problem..,(please follow step number as mentioned in code)
1) Fill last cell with max value such that it should be positive, if it is below 1 (value should not be negative) set 1 as value.
2) Fill last column and update dp array.
3) Fill last row and update dp array.
4) Finally, do iteration for remaining elements, until it reaches dp[0][0].

Return dp[0][0], which will be the health of knight.

[Example]

Consider taking above given table as a input for a problem, then the updated dp array will be..,

7             5             2
6            11            5
1              1            6


Click Follow button to receive updates from us instantly.

Please let us know about your views in comment section.

Happy Coding...😍

The intention behind this blog is to help all my fellow programmers. Also to provide solutions for various coding platform problems with solution and explanation. I will add many tutorials in future. I am not a professional blogger but whenever time permits I will post. I hope you’ll enjoy the content and find it useful! Check out my Writer blog for a more valuable quotes. For any queries or suggestions, please feel free to reach out to me. Happy coding!

Comments

Post a Comment

Popular posts from this blog

First Unique Character in a String

Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1. Examples: s = "leetcode" return 0. s = "loveleetcode", return 2. Note: You may assume the string contain only lowercase letters. class Solution {     public int firstUniqChar(String s) {         int length = s.length();         String str;         for(int i=0;i<length;i++){             if(i==length-1){                 str = s.substring(0, i);             }             else{                 str = s.substring(0,i)+s.substring(i+1);             }             if(!str.contains(s.charAt(i)+""))                 return i;         }         return -1;     } } Try it on Leetcode Here, we are going to use substring method. 1) If character is at end, then take substring from 0 till before last character of string. 2) Else make a temporary string, by removing exactly that particular character using index. [e.g..
Read more

Balanced Binary Tree

Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as: a binary tree in which the left and right subtrees of  every  node differ in height by no more than 1. Example 1: Input: root = [3,9,20,null,null,15,7] Output: true Example 2: Input: root = [1,2,2,3,3,null,null,4,4] Output: false Example 3: Input: root = [] Output: true   Constraints: The number of nodes in the tree is in the range  [0, 5000] . -10 4  <= Node.val <= 10 4 Solution: /**  * Definition for a binary tree node.  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode() {}  *     TreeNode(int val) { this.val = val; }  *     TreeNode(int val, TreeNode left, TreeNode right) {  *         this.val = val;  *         this.left = left;  *         this.right = right;  *     }
Read more

Majority Element

Given an array of size n , find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. You may assume that the array is non-empty and the majority element always exist in the array. Example 1: Input: [3,2,3] Output: 3 Example 2: Input: [2,2,1,1,1,2,2] Output: 2   class Solution { public int majorityElement(int[] nums) { Arrays.sort(nums); return nums[nums.length/2]; } } Here, we are sorting the entire array using sort() of Arrays class. Then, number at index position n/2 would be a majority element. There is no chance of wrong values in odd or even length.   Below is another method using HashMap. METHOD 2: class Solution { public int majorityElement(int[] nums) { HashMap<Integer, Integer>list = new HashMap<Integer, Integer>(); int repeatLen
Read more

Smallest Range II

Given an array   A   of integers, for each integer   A[i]   we need to choose   either  x = -K  or  x = K , and add   x   to   A[i]  (only once) . After this process, we have some array  B . Return the smallest possible difference between the maximum value of  B  and the minimum value of  B . Example 1: Input: A = [1] , K = 0 Output: 0 Explanation : B = [1] Example 2: Input: A = [0,10] , K = 2 Output: 6 Explanation : B = [2,8] Example 3: Input: A = [1,3,6] , K = 3 Output: 3 Explanation : B = [4,6,3] Note: 1 <= A.length <= 10000 0 <= A[i] <= 10000 0 <= K <= 10000 Solution: class Solution {     public int smallestRangeII(int[] A, int K) {         int i, min=0, max=0, min1, min2, max1, max2, length = A.length, diff
Read more