LeetCode-in-Java

2661. First Completely Painted Row or Column

Medium

You are given a 0-indexed integer array arr, and an m x n integer matrix mat. arr and mat both contain all the integers in the range [1, m * n].

Go through each index i in arr starting from index 0 and paint the cell in mat containing the integer arr[i].

Return the smallest index i at which either a row or a column will be completely painted in mat.

Example 1:

Input: arr = [1,3,4,2], mat = [[1,4],[2,3]]

Output: 2

Explanation: The moves are shown in order, and both the first row and second column of the matrix become fully painted at arr[2].

Example 2:

Input: arr = [2,8,7,4,1,3,5,6,9], mat = [[3,2,5],[1,4,6],[8,7,9]]

Output: 3

Explanation: The second column becomes fully painted at arr[3].

Constraints:

• m == mat.length
• n = mat[i].length
• arr.length == m * n
• 1 <= m, n <= 105
• 1 <= m * n <= 105
• 1 <= arr[i], mat[r][c] <= m * n
• All the integers of arr are unique.
• All the integers of mat are unique.

Solution

import java.util.HashMap;

public class Solution {
    public int firstCompleteIndex(int[] arr, int[][] mat) {
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < arr.length; i++) {
            map.put(arr[i], i);
        }

        for (int i = 0; i < mat.length; i++) {
            for (int j = 0; j < mat[0].length; j++) {
                mat[i][j] = map.get(mat[i][j]);
            }
        }

        int ans = Integer.MAX_VALUE;
        for (int[] ints : mat) {
            int max = 0;
            for (int j = 0; j < mat[0].length; j++) {
                max = Math.max(max, ints[j]);
            }
            ans = Math.min(ans, max);
        }

        for (int i = 0; i < mat[0].length; i++) {
            int max = 0;
            for (int[] ints : mat) {
                max = Math.max(max, ints[i]);
            }
            ans = Math.min(ans, max);
        }

        return ans;
    }
}

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