2661. First Completely Painted Row or Column

Explanation

To solve this problem, we will iterate through the elements in the arr array and keep track of how many cells have been painted in each row and each column. We will use two arrays, rows and cols, to store the number of painted cells in each row and each column respectively. Additionally, we will use two sets, paintedRows and paintedCols, to store the rows and columns that are completely painted.

As we iterate through the arr array, we will increment the count of painted cells in the corresponding row and column. If the count reaches the total number of cells in a row or a column, we will add that row or column to the respective set of completely painted rows or columns. We will return the index at which either a row or a column is completely painted first.

Time complexity: O(m*n) where m is the number of rows and n is the number of columns in the matrix. Space complexity: O(m + n) for the row and column arrays.

View Leetcode Problem

class Solution {
    public int paintMatrix(int[] arr, int[][] mat) {
        int m = mat.length;
        int n = mat[0].length;
        int[] rows = new int[m];
        int[] cols = new int[n];
        Set<Integer> paintedRows = new HashSet<>();
        Set<Integer> paintedCols = new HashSet();

        for (int i = 0; i < arr.length; i++) {
            int num = arr[i];
            int r = (num - 1) / n;
            int c = (num - 1) % n;
            rows[r]++;
            cols[c]++;
            if (rows[r] == n) paintedRows.add(r);
            if (cols[c] == m) paintedCols.add(c);
            if (paintedRows.contains(r) || paintedCols.contains(c)) {
                return i + 1;
            }
        }
        
        return -1;
    }
}

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