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Grid Challenge

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207 Discussions

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  • t_vashchenko_dev
     + 0 comments

    Java 15

    public static String gridChallenge(List<String> grid) {
        final String YES = "YES";
        final String NO = "NO";
        int columnCount = grid.get(0).length();
        int rowCount = grid.size();
        
        var sortedRowsGrid = grid.stream()
            .map(row -> 
                row.chars().sorted().mapToObj(num -> (char) num).collect(toList())
            ).collect(toList());
        
        for (int j = 0; j < columnCount; j++) {
            for (int i = 1; i < rowCount; i++) {
                char currentLetter = sortedRowsGrid.get(i).get(j);
                char previousLetter = sortedRowsGrid.get(i - 1).get(j);
                
                if (currentLetter < previousLetter) {
                    return NO;
                }  
            } 
        }
        
        return YES;
    }

  • ankitraj311
     + 0 comments

    Easiest approch I found on CPP:

    string gridChallenge(vector grid) { bool ok = true; int n = grid.size();

    for (int i = 0; i < n; i++) {
    sort(grid[i].begin(), grid[i].end());    //Sorting taking care of row's
}

for (int i = 0; i < n; i++) {
    for (int j = 0; j < grid[i].size()-1; j++) {
        if(grid[j][i] > grid[j+1][i]) //Taking care of coloumb
        {
            return "NO";
        }
    }
}

return "YES";

    }

  • daniel_alvarado2
     + 0 comments

    C#

    public static string gridChallenge(List grid) {

    List<string> arrangedGrid = new List<string>();

    foreach(string s in grid)
    {
        arrangedGrid.Add(new string(s.OrderBy(x=>x).ToArray()));

    }               

    for(int row = 0; row < arrangedGrid.Count - 1; row++)
    {
        for (int column = 0; column < arrangedGrid[0].Count(); column++)
        {
            if(arrangedGrid[row][column] > arrangedGrid[row + 1][column])
            {
                return "NO";
            }
        }
    }

    return "YES";

}

  • micheldejesus
     + 0 comments

    Java 

        public static String gridChallenge(List<String> grid) {
        int[][] values = new int[grid.size()][grid.get(0).length()];
        for (int i=0 ; i<grid.size() ; i++) {
            char[] chars = grid.get(i).toCharArray();
            Arrays.sort(chars);
            
            for (int j=0 ; j<chars.length ; j++) {
                values[i][j] = (int) chars[j];
                if (i > 0 && values[i-1][j] > values[i][j]) {
                     return "NO";
                }
            }     
        }
        return "YES";
    }

  • gillrodrigues_
     + 0 comments

    There is no way to run out of O(n²) because of the column's checking, but follow a O(n) way to sort each row's string, 

    function gridChallenge(grid) {
    for (let i = 0; i < grid.length; i++) {
        grid[i] = sortString(grid[i]);
    }
    let answer = 'YES';
    for (let idr = 1; idr < grid.length; idr++) {
        for (let idc = 0; idc < grid[0].length; idc++) {
            if (grid[idr][idc] < grid[idr-1][idc]) {
                answer = 'NO';
                break;
            }
        }
    }
    return answer;
}

function sortString(str) {
    const ALPHABET_SIZE = 26;
    const charCount = Array(ALPHABET_SIZE).fill(0);
    const asciiStartLowercase = "a".charCodeAt(0);

    for (const char of str) {
        charCount[char.charCodeAt(0) - asciiStartLowercase]++;
    }

    let sortedStr = "";
    for (let i = 0; i < charCount.length; i++) {
        if (charCount[i] > 0) {
            sortedStr += String.fromCharCode(i + asciiStartLowercase).repeat(charCount[i]);
        }
    }
    return sortedStr;
}