2661. First Completely Painted Row or Column

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.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.

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.

Python3JavaC++GoTypeScriptRust

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

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;
            }
        }
    }
}

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;
            }
        }
    }
};

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
        }
    }
}

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;
        }
    }
}

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;
            }
        }

        -1
    }
}

2661. First Completely Painted Row or Column

Description

Solutions

Solution 1: Hash Table + Array Counting

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