Description#
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.lengthn = mat[i].lengtharr.length == m * n1 <= m, n <= 1051 <= m * n <= 1051 <= arr[i], mat[r][c] <= m * n- All the integers of
arr are unique. - All the integers of
mat are unique.
Solutions#
Solution 1: Hash Table + Array Counting#
We use a hash table $idx$ to record the position of each element in the matrix $mat$, that is $idx[mat[i][j]] = (i, j)$, and define two arrays $row$ and $col$ to record the number of colored elements in each row and each column respectively.
Traverse the array $arr$. For each element $arr[k]$, we find its position $(i, j)$ in the matrix $mat$, and then add $row[i]$ and $col[j]$ by one. If $row[i] = n$ or $col[j] = m$, it means that the $i$-th row or the $j$-th column has been colored, so $arr[k]$ is the element we are looking for, and we return $k$.
The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Here $m$ and $n$ are the number of rows and columns of the matrix $mat$ respectively.
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| class Solution:
def firstCompleteIndex(self, arr: List[int], mat: List[List[int]]) -> int:
m, n = len(mat), len(mat[0])
idx = {}
for i in range(m):
for j in range(n):
idx[mat[i][j]] = (i, j)
row = [0] * m
col = [0] * n
for k in range(len(arr)):
i, j = idx[arr[k]]
row[i] += 1
col[j] += 1
if row[i] == n or col[j] == m:
return k
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| class Solution {
public int firstCompleteIndex(int[] arr, int[][] mat) {
int m = mat.length, n = mat[0].length;
Map<Integer, int[]> idx = new HashMap<>();
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
idx.put(mat[i][j], new int[] {i, j});
}
}
int[] row = new int[m];
int[] col = new int[n];
for (int k = 0;; ++k) {
var x = idx.get(arr[k]);
int i = x[0], j = x[1];
++row[i];
++col[j];
if (row[i] == n || col[j] == m) {
return k;
}
}
}
}
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| class Solution {
public:
int firstCompleteIndex(vector<int>& arr, vector<vector<int>>& mat) {
int m = mat.size(), n = mat[0].size();
unordered_map<int, pair<int, int>> idx;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
idx[mat[i][j]] = {i, j};
}
}
vector<int> row(m), col(n);
for (int k = 0;; ++k) {
auto [i, j] = idx[arr[k]];
++row[i];
++col[j];
if (row[i] == n || col[j] == m) {
return k;
}
}
}
};
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| func firstCompleteIndex(arr []int, mat [][]int) int {
m, n := len(mat), len(mat[0])
idx := map[int][2]int{}
for i := range mat {
for j := range mat[i] {
idx[mat[i][j]] = [2]int{i, j}
}
}
row := make([]int, m)
col := make([]int, n)
for k := 0; ; k++ {
x := idx[arr[k]]
i, j := x[0], x[1]
row[i]++
col[j]++
if row[i] == n || col[j] == m {
return k
}
}
}
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| function firstCompleteIndex(arr: number[], mat: number[][]): number {
const m = mat.length;
const n = mat[0].length;
const idx: Map<number, number[]> = new Map();
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
idx.set(mat[i][j], [i, j]);
}
}
const row: number[] = Array(m).fill(0);
const col: number[] = Array(n).fill(0);
for (let k = 0; ; ++k) {
const [i, j] = idx.get(arr[k])!;
++row[i];
++col[j];
if (row[i] === n || col[j] === m) {
return k;
}
}
}
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| use std::collections::HashMap;
impl Solution {
pub fn first_complete_index(arr: Vec<i32>, mat: Vec<Vec<i32>>) -> i32 {
let m = mat.len();
let n = mat[0].len();
let mut idx = HashMap::new();
for i in 0..m {
for j in 0..n {
idx.insert(mat[i][j], [i, j]);
}
}
let mut row = vec![0; m];
let mut col = vec![0; n];
for k in 0..arr.len() {
let x = idx.get(&arr[k]).unwrap();
let i = x[0];
let j = x[1];
row[i] += 1;
col[j] += 1;
if row[i] == n || col[j] == m {
return k as i32;
}
}
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}
}
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