Press "Enter" to skip to content

Posts tagged as “matrix”

花花酱 LeetCode 1632. Rank Transform of a Matrix

Given an m x n matrix, return a new matrix answer where answer[row][col] is the rank of matrix[row][col].

The rank is an integer that represents how large an element is compared to other elements. It is calculated using the following rules:

  • The rank is an integer starting from 1.
  • If two elements p and q are in the same row or column, then:
    • If p < q then rank(p) < rank(q)
    • If p == q then rank(p) == rank(q)
    • If p > q then rank(p) > rank(q)
  • The rank should be as small as possible.

It is guaranteed that answer is unique under the given rules.

Example 1:

Input: matrix = [[1,2],[3,4]]
Output: [[1,2],[2,3]]
Explanation:
The rank of matrix[0][0] is 1 because it is the smallest integer in its row and column.
The rank of matrix[0][1] is 2 because matrix[0][1] > matrix[0][0] and matrix[0][0] is rank 1.
The rank of matrix[1][0] is 2 because matrix[1][0] > matrix[0][0] and matrix[0][0] is rank 1.
The rank of matrix[1][1] is 3 because matrix[1][1] > matrix[0][1], matrix[1][1] > matrix[1][0], and both matrix[0][1] and matrix[1][0] are rank 2.

Example 2:

Input: matrix = [[7,7],[7,7]]
Output: [[1,1],[1,1]]

Example 3:

Input: matrix = [[20,-21,14],[-19,4,19],[22,-47,24],[-19,4,19]]
Output: [[4,2,3],[1,3,4],[5,1,6],[1,3,4]]

Example 4:

Input: matrix = [[7,3,6],[1,4,5],[9,8,2]]
Output: [[5,1,4],[1,2,3],[6,3,1]]

Constraints:

  • m == matrix.length
  • n == matrix[i].length
  • 1 <= m, n <= 500
  • -109 <= matrix[row][col] <= 109

Solution: Union Find

Group cells by their values, process groups (cells that have the same value) in ascending order (smaller number has smaller rank).

For cells that are in the same row and same cols union them using union find, they should have the same rank which equals to max(max_rank_x[cols], max_rank_y[rows]) + 1.

Time complexity: O(m*n*(m+n))
Space complexity: O(m*n)

C++

花花酱 LeetCode 1605. Find Valid Matrix Given Row and Column Sums

You are given two arrays rowSum and colSum of non-negative integers where rowSum[i] is the sum of the elements in the ith row and colSum[j] is the sum of the elements of the jth column of a 2D matrix. In other words, you do not know the elements of the matrix, but you do know the sums of each row and column.

Find any matrix of non-negative integers of size rowSum.length x colSum.length that satisfies the rowSum and colSum requirements.

Return a 2D array representing any matrix that fulfills the requirements. It’s guaranteed that at least one matrix that fulfills the requirements exists.

Example 1:

Input: rowSum = [3,8], colSum = [4,7]
Output: [[3,0],
         [1,7]]
Explanation:
0th row: 3 + 0 = 0 == rowSum[0]
1st row: 1 + 7 = 8 == rowSum[1]
0th column: 3 + 1 = 4 == colSum[0]
1st column: 0 + 7 = 7 == colSum[1]
The row and column sums match, and all matrix elements are non-negative.
Another possible matrix is: [[1,2],
                             [3,5]]

Example 2:

Input: rowSum = [5,7,10], colSum = [8,6,8]
Output: [[0,5,0],
         [6,1,0],
         [2,0,8]]

Example 3:

Input: rowSum = [14,9], colSum = [6,9,8]
Output: [[0,9,5],
         [6,0,3]]

Example 4:

Input: rowSum = [1,0], colSum = [1]
Output: [[1],
         [0]]

Example 5:

Input: rowSum = [0], colSum = [0]
Output: [[0]]

Constraints:

  • 1 <= rowSum.length, colSum.length <= 500
  • 0 <= rowSum[i], colSum[i] <= 108
  • sum(rows) == sum(columns)

Solution: Greedy

Let a = min(row[i], col[j]), m[i][j] = a, row[i] -= a, col[j] -=a

Time complexity: O(m*n)
Space complexity: O(m*n)

C++

花花酱 LeetCode 1594. Maximum Non Negative Product in a Matrix

You are given a rows x cols matrix grid. Initially, you are located at the top-left corner (0, 0), and in each step, you can only move right or down in the matrix.

Among all possible paths starting from the top-left corner (0, 0) and ending in the bottom-right corner (rows - 1, cols - 1), find the path with the maximum non-negative product. The product of a path is the product of all integers in the grid cells visited along the path.

Return the maximum non-negative product modulo 109 + 7If the maximum product is negative return -1.

Notice that the modulo is performed after getting the maximum product.

Example 1:

Input: grid = [[-1,-2,-3],
               [-2,-3,-3],
               [-3,-3,-2]]
Output: -1
Explanation: It's not possible to get non-negative product in the path from (0, 0) to (2, 2), so return -1.

Example 2:

Input: grid = [[1,-2,1],
               [1,-2,1],
               [3,-4,1]]
Output: 8
Explanation: Maximum non-negative product is in bold (1 * 1 * -2 * -4 * 1 = 8).

Example 3:

Input: grid = [[1, 3],
               [0,-4]]
Output: 0
Explanation: Maximum non-negative product is in bold (1 * 0 * -4 = 0).

Example 4:

Input: grid = [[ 1, 4,4,0],
               [-2, 0,0,1],
               [ 1,-1,1,1]]
Output: 2
Explanation: Maximum non-negative product is in bold (1 * -2 * 1 * -1 * 1 * 1 = 2).

Constraints:

  • 1 <= rows, cols <= 15
  • -4 <= grid[i][j] <= 4

Solution: DP

Use two dp arrays,

dp_max[i][j] := max product of matrix[0~i][0~j]
dp_min[i][j] := min product of matrix[0~i][0~j]

Time complexity: O(m*n)
Space complexity: O(m*n)

C++

花花酱 LeetCode 1582. Special Positions in a Binary Matrix

Given a rows x cols matrix mat, where mat[i][j] is either 0 or 1, return the number of special positions in mat.

A position (i,j) is called special if mat[i][j] == 1 and all other elements in row i and column j are 0 (rows and columns are 0-indexed).

Example 1:

Input: mat = [[1,0,0],
              [0,0,1],
              [1,0,0]]
Output: 1
Explanation: (1,2) is a special position because mat[1][2] == 1 and all other elements in row 1 and column 2 are 0.

Example 2:

Input: mat = [[1,0,0],
              [0,1,0],
              [0,0,1]]
Output: 3
Explanation: (0,0), (1,1) and (2,2) are special positions. 

Example 3:

Input: mat = [[0,0,0,1],
              [1,0,0,0],
              [0,1,1,0],
              [0,0,0,0]]
Output: 2

Example 4:

Input: mat = [[0,0,0,0,0],
              [1,0,0,0,0],
              [0,1,0,0,0],
              [0,0,1,0,0],
              [0,0,0,1,1]]
Output: 3

Constraints:

  • rows == mat.length
  • cols == mat[i].length
  • 1 <= rows, cols <= 100
  • mat[i][j] is 0 or 1.

Solution: Sum for each row and column

Brute force:
Time complexity: O(R*C*(R+C))
Space complexity: O(1)

We can pre-compute the sums for each row and each column, ans = sum(mat[r][c] == 1 and rsum[r] == 1 and csum[c] == 1)

Time complexity: O(R*C)
Space complexity: O(R+C)

C++

花花酱 LeetCode 1572. Matrix Diagonal Sum

Given a square matrix mat, return the sum of the matrix diagonals.

Only include the sum of all the elements on the primary diagonal and all the elements on the secondary diagonal that are not part of the primary diagonal.

Example 1:

Input: mat = [[1,2,3],
              [4,5,6],
              [7,8,9]]
Output: 25
Explanation: Diagonals sum: 1 + 5 + 9 + 3 + 7 = 25
Notice that element mat[1][1] = 5 is counted only once.

Example 2:

Input: mat = [[1,1,1,1],
              [1,1,1,1],
              [1,1,1,1],
              [1,1,1,1]]
Output: 8

Example 3:

Input: mat = [[5]]
Output: 5

Constraints:

  • n == mat.length == mat[i].length
  • 1 <= n <= 100
  • 1 <= mat[i][j] <= 100

Solution: Brute Force

Note: if n is odd, be careful not to double count the center one.

Time complexity: O(n)
Space complexity: O(1)

C++