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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Cavity Map
  5. Discussions

Cavity Map

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  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/BweZoGPs08M

    vector<string> cavityMap(vector<string> grid) {
    for(int i = 1; i < grid.size() - 1; i++){
        for(int j = 1; j < grid[i].size()-1; j++){
            if(grid[i][j] > grid[i][j-1] && grid[i][j] > grid[i][j+1])
                if(grid[i][j] > grid[i-1][j] && grid[i][j] > grid[i+1][j])
                grid[i][j] = 'X'; 
        }
    }
    return grid;
}

  • patrickgao2020
     + 0 comments

    Here is my Python code utilizing list comprehensions! However, this does result in a line being incredibly long.

    def cavityMap(grid):
    changes = [(i + 1, num + 1) for i in range(len(grid) - 2) for num in range(len(grid[i + 1]) - 2) if grid[i + 1][num] < grid[i + 1][num + 1] and grid[i + 1][num + 1] > grid[i + 1][num + 2] and grid[i][num + 1] < grid[i + 1][num + 1] and grid[i + 2][num + 1] < grid[i + 1][num + 1]]
    for change in changes:
        grid[change[0]] = grid[change[0]][:change[1]] + "X" + grid[change[0]][change[1] + 1:]
    return grid

  • lukasz_jozefowi1
     + 0 comments

    Java

    public static List<String> cavityMap(List<String> grid) {
        for (int i = 1; i < grid.size() - 1; i++) {
            for (int j = 1; j < grid.get(i).length() - 1; j++) {
                int currentDigit = Character.getNumericValue(grid.get(i).charAt(j));
                int upDigit = Character.getNumericValue(grid.get(i-1).charAt(j));
                int downDigit = Character.getNumericValue(grid.get(i+1).charAt(j));
                int leftDigit = Character.getNumericValue(grid.get(i).charAt(j-1));
                int rightDigit = Character.getNumericValue(grid.get(i).charAt(j+1));
                if (currentDigit > upDigit 
                    && currentDigit > downDigit 
                    && currentDigit > leftDigit
                    && currentDigit > rightDigit) {
                        StringBuilder newRow = new StringBuilder(grid.get(i));
                        newRow.setCharAt(j, 'X');
                        grid.set(i, newRow.toString());
                }
            }
        }
        return grid;
    }

  • medyk_pawel
     + 0 comments

    Python3, straightforward, nothing smart; Python is not the best tool here, as its for loops cost a lot and replacing single string means recreating whole row

    def is_cavity(grid, i, j):
    max_adjacent = max(
        grid[i-1][j],
        grid[i+1][j],
        grid[i][j-1],
        grid[i][j+1]
    )
    return grid[i][j] > max_adjacent

def mark_cavity(grid, i, j):
    row: str = grid[i]
    grid[i] = row[:j] + 'X' + row[j+1:]

def cavityMap(grid):
    n = len(grid)
    cavity_mapped_grid = [row[:] for row in grid] 
    for i in range(1, n-1):
        for j in range(1, n-1):
            if is_cavity(grid, i, j):
                mark_cavity(cavity_mapped_grid, i, j)
    return cavity_mapped_grid

  • kunalkawate424
     + 0 comments

    Solution from my side, i know that it is not well oprimized. but the logic of this solution is in the understandable format.

    def cavityMap(grid):
    data = []
    result = []
    indexData = []

    for item in grid:
        data.append(list(item))

    for i in range(1,len(data)-1):
        for j in range(1,len(data)-1):
            if(data[i][j+1]<data[i][j]> data[i][j-1] and data[i+1][j]<data[i][j]> data[i-1][j]):
                indexData.append([i,j])

    for item01 in indexData:
        data[item01[0]][item01[1]] = 'X'

    for item03 in data:
        result.append(''.join(item03))
        
    return result