Python - Algorithm #47

Python Algorithm

LeetCode

1252. Cells with Odd Values in a Matrix

• There is an m x n matrix that is initialized to all 0's. There is also a 2D array [indices] where each indices[i] = [rᵢ, cᵢ] represents a 0-indexed location to perform some increment operations on the matrix.
• For each location indices[i], do both of the following:
• Increment all the cells on row rᵢ.
• Increment all the cells on column cᵢ.
• Given m, n, and [indices], return the number of odd-valued cells in the matrix after applying the increment to all locations in [indices].
• Constraints
  • 1 ≤ m, n ≤ 50
  • 1 ≤ indices.length ≤ 100
  • 0 ≤ ri < m
  • 0 ≤ ci < n

class Solution:
    def oddCells(self, m: int, n: int, indices: List[List[int]]) -> int:
        matrix = [[0 for x in range(n)] for y in range(m)]
        for row in matrix:
            print(row)
        answer = 0
        for row, col in indices:
            for i in range(n):
                # 특정 row의 값을 1씩 증가시켜 주기
                matrix[row][i] += 1
            for i in range(m):
                # 특정 column 값을 1씩 증가시켜 주기
                matrix[i][col] += 1
        # 전체 matrix 순회하면서 홀수 개수 세어서 리턴
        for i in range(m):
            for j in range(n):
                if matrix[i][j] % 2 == 1:
                    answer += 1
        return answer

import numpy as np

class Solution:
    def oddCells(self, m: int, n: int, indices: List[List[int]]) -> int:
        answer = 0
        mat = np.zeros((m, n))
        for r, c in indices:
            mat[r] += 1
            mat[:, c] += 1
        for i in range(m):
            for j in range(n):
                if mat[i][j] % 2 != 0:
                    answer += 1
        return answer

class Solution:
    def oddCells(self, m: int, n: int, indices: List[List[int]]) -> int:
        answer = 0
        mat = [[0 for _ in range(n)] for _ in range(m)]
        for r, c in indices:
            mat[r] = [x + 1 for x in mat[r]]
            for col in mat:
                col[c] += 1
        for i in range(m):
            for j in range(n):
                if mat[i][j] % 2 != 0:
                    answer += 1
        return answer

class Solution:
    def oddCells(self, m: int, n: int, indices: List[List[int]]) -> int:
        answer = 0
        mat = [[0 for _ in range(n)] for _ in range(m)]
        for r, c in indices:
            for i in range(n):
                mat[r][i] += 1
            for i in range(m):
                mat[i][c] += 1
        for i in range(m):
            for j in range(n):
                if mat[i][j] % 2 != 0:
                    answer += 1
        return answer

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1351. Count Negative Numbers in a Sorted Matrix

• Given a m x n matrix [grid] which is sorted in non-increasing order both row-wise and column-wise, return the number of negative numbers in [grid].
• Constraints
  
  • m == grid.length
  • n == grid[i].length
  • 1 ≤ m, n ≤ 100
  • -100 ≤ grid[i][j] ≤ 100

class Solution:
    def countNegatives(self, grid: List[List[int]]) -> int:
        return sum([1 for row in grid for num in row if num < 0])

class Solution:
    def countNegatives(self, grid: List[List[int]]) -> int:
        answer = 0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] < 0:
                    answer += 1
        return answer

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https://leetcode.com/problems/cells-with-odd-values-in-a-matrix/

https://leetcode.com/problems/count-negative-numbers-in-a-sorted-matrix/