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.

中文翻譯：

給你一個(gè)下標(biāo)從 0 開始的整數(shù)數(shù)組 arr 和一個(gè) m x n 的整數(shù) 矩陣 mat 。arr 和 mat 都包含范圍 [1，m * n] 內(nèi)的 所有 整數(shù)。

從下標(biāo) 0 開始遍歷 arr 中的每個(gè)下標(biāo) i ，并將包含整數(shù) arr[i] 的 mat 單元格涂色。

請(qǐng)你找出 arr 中在 mat 的某一行或某一列上都被涂色且下標(biāo)最小的元素，并返回其下標(biāo) i 。

多用hash；

下面是代碼：

Runtime:?90 ms, faster than?11.53%?of?Java?online submissions for?First Completely Painted Row or Column.

Memory Usage:?80.3 MB, less than?7.42%?of?Java?online submissions for?First Completely Painted Row or Column.