Skip to main content

Breadth-first Search - Middle Level - Question 1

Breadth-first Search - Middle Level - Question 1


Leetcode 1162  As Far from Land as Possible

Given an n x n grid containing only values 0 and 1, where 0 represents water and 1 represents land, find a water cell such that its distance to the nearest land cell is maximized, and return the distance. If no land or water exists in the grid, return -1.

The distance used in this problem is the Manhattan distance: the distance between two cells (x0, y0) and (x1, y1) is |x0 - x1| + |y0 - y1|.

Constraints:

n == grid.length

n == grid[i].length

1 <= n <= 100

grid[i][j] is 0 or 1


Analysis:

One way to solve this problem is that: for each water position, we run a bfs search to the find the shortest distance to the land. Then the get the maximum of these shortest distances. The time complexity is O(n^4), which is too large to this question.

Alternatively, we can start with all lands, then do a bsf search. The last find water should have the largest distance to the land. The time complexity is O(n^2). 


See the code below:


/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int maxDistance(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<vector<int>> visit (n, vector<int>(n, 0));
        queue<pair<int,int>> q;
        vector<int> dirs={-1, 0, 1, 0, -1};
        for(int i=0; i<n; ++i) {
            for(int j=0; j<n; ++j) {
                if(grid[i][j]==1) {
                    q.push({i,j});
                    visit[i][j]=1;
                }
            }
        }
        if(q.empty() || (int)q.size() == n*n) return -1;
        int res = 0;
        while (q.size()) {
            ++res;
            int t = q.size();
            // cout<<t<<endl;
            for(int i=0; i<t; ++i) {
                auto a = q.front();
                q.pop();
                for(int k=0; k+1<dirs.size(); ++k) {
                    int x= a.first + dirs[k], y=a.second + dirs[k+1];
                    if(x<0 || x>=n || y<0 || y>=n || grid[x][y] || visit[x][y]) continue;
                    q.push({x,y});
                    visit[x][y]=1;
                }
            }
        }
        return res - 1;
    }
};


Comments

Post a Comment

Popular posts from this blog

Segment Tree

Segment tree can be viewed as an abstract data structure which using some more space to trade for speed. For example, for a typical question with O(N^2) time complexity, the segment tree method can decrease it to O(N*log(N)).  To make it understandable, let us consider one example. Say we have an integer array of N size, and what we want is to query the maximum with a query range [idx1, idx2], where idx1 is the left indexes, and idx2 is the right indexes inclusive. If we only do this kind of query once, then we just need to scan through the array from idx1 to idx2 once, and record the maximum, done. The time complexity is O(N), which is decent enough in most cases even though it is not the optimal one (for example, with a segment tree built, the time complexity can decrease down to O(log(N))). However, how about we need to query the array N times? If we continue to use the naïve way above, then the time complexity is O(N^2), since for each query we need to scan the query range once...
Post a Comment
Read more

Bit Manipulation - Example

  Leetcode 136 Single Number Given a non-empty array of integers nums, every element appears twice except for one. Find that single one. You must implement a solution with a linear runtime complexity and use only constant extra space. Constraints: 1 <= nums.length <= 3 * 10^4 -3 * 10^4 <= nums[i] <= 3 * 10^4 Each element in the array appears twice except for one element which appears only once. Analysis: If there is no space limitation, this question can be solved by counting easily. But counting requires additional space. Here we can use xor (^) operation based on some interesting observations:  A^A = 0, here A is any number A^0 = A, here A is any number Since all the number appears twice except one, then all the number appear even numbers will be cancelled out, and only the number appears one time is left, which is what we want. See the code below: class Solution { public: int singleNumber(vector<int>& nums) { int res = 0; for(auto ...
Post a Comment
Read more

Rolling Hash

Rolling hash is one common trick used to increase efficiency of substring comparisons by compressing (or hashing) a string into a integer. After this step, we can compare two strings directly without comparing each chars. So the efficiency can be increased from O(N) to O(1). So how to implement the rolling hash? First we need to choose a base for the expansion and a modulo to mod. The basic formula is (suppose the window is n, and the rolling direction is from left to right), HashVal = (A1*p^(n-1) + A2*p^(n-2) + ... + An-1*p^1 + An*p^0)%mod where HashVal is the hash value, Ai is the ith element, p is the base, and mod is the modulo. To avoid collision as much as we can, p and modulo usually need to be large prime numbers. One corner case is that the base order in the above formula cannot be reversed. Or to be more clear, if the rolling direction is from left to right in an array, the first element should be in the highest order of the base, or times p^(n-1), and the last element times ...
Post a Comment
Read more