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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:

https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2600-2699/2661.First%20Completely%20Painted%20Row%20or%20Column/images/image explanation for 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:

image explanation for 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.

  • 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;
                }
            }
        }
    };
    
  • 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
    
    
  • 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[] = new Array(m).fill(0);
        const col: number[] = new 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
        }
    }
    
    

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