g0101_0200.s0200_number_of_islands.Solution.rs Maven / Gradle / Ivy

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104 LeetCode algorithm problem solutions

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// #Medium #Top_100_Liked_Questions #Top_Interview_Questions #Array #Depth_First_Search
// #Breadth_First_Search #Matrix #Union_Find
// #Algorithm_II_Day_6_Breadth_First_Search_Depth_First_Search
// #Graph_Theory_I_Day_1_Matrix_Related_Problems #Level_1_Day_9_Graph/BFS/DFS #Udemy_Graph
// #Big_O_Time_O(M*N)_Space_O(M*N) #2024_09_10_Time_7_ms_(86.79%)_Space_9.1_MB_(42.14%)

impl Solution {
    pub fn num_islands(grid: Vec>) -> i32 {
        let mut num_of_islands: i32 = 0;
        let mut current_index: i32 = 1;
        let height = grid.len();
        let width = grid[0].len();
        let mut part_of_island = vec![0; width * height];
        let mut pos;

        for current in 0..height * width {
            let x = current / width;
            let y = current % width;

            if grid[x][y] == '1' {
                let left = if y > 0 {
                    part_of_island[current - 1]
                } else {
                    0
                };

                let up = if x > 0 {
                    part_of_island[current - width]
                } else {
                    0
                };

                if up > 1 {
                    if left > 1 && left != up {
                        for i in 0..width {
                            pos = part_of_island[current - i];
                            if pos == left {
                                part_of_island[current - i] = up;
                            }
                        }
                        num_of_islands -= 1;
                    }
                    part_of_island[current] = up;
                    continue;
                } else if left > 1 {
                    part_of_island[current] = left;
                    continue;
                }

                current_index += 1;
                num_of_islands += 1;
                part_of_island[current] = current_index;
            }
        }

        num_of_islands
    }
}