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Lachlan Jacob 5 年前
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共有 3 个文件被更改,包括 65 次插入0 次删除
  1. 32
    0
      problems/1252/main.py
  2. 30
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      problems/1252/problem.txt
  3. 3
    0
      problems/1252/run.sh

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problems/1252/main.py 查看文件

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class Solution:
def oddCells(self, n, m, indices):
# Initialise the matrix
matrix = []
for x in range(n):
new_row = []
for y in range(m):
new_row.append(0)
matrix.append(new_row)
# Now apply steps
for i in indices:
ri = i[0]
ci = i[1]
for n in range(len(matrix[ri])):
matrix[ri][n] += 1
for n in range(len(matrix)):
matrix[n][ci] += 1
# now count odd values (this could be done while changes are made, but this is easier to reason)
count = 0
for x in range(len(matrix)):
for y in range(len(matrix[x])):
if matrix[x][y] % 2 != 0:
count += 1
return count

s = Solution()
print("Expected: 6")
print("Got:", s.oddCells(2, 3, [[0, 1], [1, 1]]))

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problems/1252/problem.txt 查看文件

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Given n and m which are the dimensions of a matrix initialized by zeros and given an array indices where indices[i] = [ri, ci]. For each pair of [ri, ci] you have to increment all cells in row ri and column ci by 1.

Return the number of cells with odd values in the matrix after applying the increment to all indices.

(There was pictures here, but hopefully this is clear enough)

Example 1:

Input: n = 2, m = 3, indices = [[0,1],[1,1]]
Output: 6
Explanation: Initial matrix = [[0,0,0],[0,0,0]].
After applying first increment it becomes [[1,2,1],[0,1,0]].
The final matrix will be [[1,3,1],[1,3,1]] which contains 6 odd numbers.

Example 2:

Input: n = 2, m = 2, indices = [[1,1],[0,0]]
Output: 0
Explanation: Final matrix = [[2,2],[2,2]]. There is no odd number in the final matrix.


Constraints:

1 <= n <= 50
1 <= m <= 50
1 <= indices.length <= 100
0 <= indices[i][0] < n
0 <= indices[i][1] < m


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problems/1252/run.sh 查看文件

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#!/bin/bash

python3 main.py

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