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Apr 25, 2024 · 2: Cells have rotten oranges; The task is to the minimum time required so that all the oranges become rotten. A rotten orange at index (i,j ) can rot other fresh oranges which are its neighbors (up, down, left, and right). If it is impossible to rot every orange then simply return -1.
Rotting Oranges - You are given an m x n grid where each cell can have one of three values: * 0 representing an empty cell, * 1 representing a fresh orange, or * 2 representing a rotten orange. Every minute, any fresh orange that is 4-directionally adjacent to a rotten orange becomes rotten.
Given a grid of dimension nxm where each cell in the grid can have values 0, 1 or 2 which has the following meaning:0 : Empty cell 1 : Cells have fresh oranges 2 : Cells have rotten oranges . We have to determine what is the earliest ti.
Nov 11, 2021 · The key observation is that fresh oranges adjacent to rotten oranges are rotten on day 1, those adjacent to those oranges are rotten on day 2, and so on. The phenomenon is similar to a level order traversal on a graph, where all the initial rotten oranges act as root nodes.
Can you solve this real interview question? Rotting Oranges - Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.
Rotting Oranges - In-Depth Explanation. Leetcode Link. You are given an m x n grid where each cell can have one of three values: 0 representing an empty cell, 1 representing a fresh orange, or. 2 representing a rotten orange. Every minute, any fresh orange that is 4-directionally adjacent to a rotten orange becomes rotten.
Jul 24, 2021 · In this tutorial, i am going to discuss a very interesting problem rotting oranges (minimum time required to rot all fresh oranges). Given an m*n grid where each cell in the matrix can have values 0, 1 or 2. 0 represents an empty cell. 1 represents cells have fresh oranges. 2 represents cells have rotten oranges.
Apr 26, 2024 · Embark on a journey to optimize the process of rotting oranges efficiently. In this tutorial, we delve into the intriguing world of rotting oranges and explore algorithms to minimize the time required for all oranges to become rotten.
Aug 20, 2018 · Rotting Oranges. Description. You are given an m x n grid where each cell can have one of three values: 0 representing an empty cell, 1 representing a fresh orange, or. 2 representing a rotten orange. Every minute, any fresh orange that is 4-directionally adjacent to a rotten orange becomes rotten.
class Solution {public: int orangesRotting (vector < vector < int >>& grid) {constexpr int dirs [4][2] = {{0, 1}, {1, 0}, {0,-1}, {-1, 0}}; const int m = grid. size (); const int n = grid [0]. size (); auto isNeighborRotten = [&](int i, int j, const vector < vector < int >>& grid) {for (const auto & [dx, dy]: dirs) {const int r = i + dx; const ...
