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  1. Rotten Oranges. Difficulty: Medium Accuracy: 46.02% Submissions: 152K+ Points: 4. Given a matrix where each cell in the matrix 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 time after which all the oranges are rotten.

  2. Sep 17, 2024 · Given a matrix of dimension M * N, where each cell in the matrix 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. The task is to the minimum time required so that all the oranges become rotten.

    • 17 min
  3. May 27, 2024 · 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. Examples: Input: arr[][C] = { {2, 1, 0, 2, 1}, {1, 0, 1, 2, 1}, {1, 0, 0, 2, 1}}; Output: 2. Explanation: At 0th time frame: {2, 1, 0, 2, 1}

  4. 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.

  5. Sep 13, 2020 · Problem statement. You have been given a grid containing some oranges. Each cell of this grid has one of the three integers values: Value 0 - representing an empty cell. Value 1 - representing a fresh orange. Value 2 - representing a rotten orange. Every second, any fresh orange that is adjacent (4-directionally) to a rotten orange becomes rotten.

  6. We have to determine what is the minimum time required to rot all oranges. A rotten orange at index [i,j] can rot other fresh orange at indexes [i-1,j], [i+1,j], [i,j-1], [i,j+1] ( up , down , left and right ) in unit time.

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  8. 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.