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diagonal_traverse.py
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diagonal_traverse.py
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"""
498. Diagonal Traverse
Medium
Add to List
Share
Given an m x n matrix mat, return an array of all the elements of the array in a diagonal order.
Example 1:
Input: mat = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,4,7,5,3,6,8,9]
Example 2:
Input: mat = [[1,2],[3,4]]
Output: [1,2,3,4]
Constraints:
m == mat.length
n == mat[i].length
1 <= m, n <= 104
1 <= m * n <= 104
-105 <= mat[i][j] <= 105
"""
# V0
# IDEA : while loop + boundary conditions
### NOTE : the "directions" trick
class Solution(object):
def findDiagonalOrder(self, matrix):
if not matrix or not matrix[0]: return []
### NOTE this trick
directions = [(-1, 1), (1, -1)]
count = 0
res = []
i, j = 0, 0
M, N = len(matrix), len(matrix[0])
while len(res) < M * N:
if 0 <= i < M and 0 <= j < N:
res.append(matrix[i][j])
direct = directions[count % 2]
i, j = i + direct[0], j + direct[1]
continue
elif i < 0 and 0 <= j < N:
i += 1
elif 0 <= i < M and j < 0:
j += 1
elif i < M and j >= N:
i += 2
j -= 1
elif i >= M and j < N:
j += 2
i -= 1
count += 1
return res
# V0'
# IDEA : Diagonal Iteration and Reversal
# NOTE !!! : for "odd" diagoal traversal, we just need to go over it and REVERSE it before append tp res
class Solution:
def findDiagonalOrder(self, matrix):
# Check for empty matrices
if not matrix or not matrix[0]:
return []
# Variables to track the size of the matrix
N, M = len(matrix), len(matrix[0])
# The two arrays as explained in the algorithm
result, intermediate = [], []
# We have to go over all the elements in the first
# row and the last column to cover all possible diagonals
for d in range(N + M - 1):
# Clear the intermediate array everytime we start
# to process another diagonal
intermediate = []
# We need to figure out the "head" of this diagonal
# The elements in the first row and the last column
# are the respective heads.
# r : row idx
# c : col idx
#r, c = 0 if d < M else d - M + 1, d if d < M else M - 1
if d < M:
r = 0
c = d
else:
r = d - M + 1
c = M - 1
# Iterate until one of the indices goes out of scope
# Take note of the index math to go down the diagonal
while r < N and c > -1:
intermediate.append(matrix[r][c])
r += 1
c -= 1
# Reverse even numbered diagonals. The
# article says we have to reverse odd
# numbered articles but here, the numbering
# is starting from 0 :P
if d % 2 == 0:
result.extend(intermediate[::-1])
else:
result.extend(intermediate)
return result
# V1
# IDEA : Diagonal Iteration and Reversal
# https://leetcode.com/problems/diagonal-traverse/solution/
class Solution:
def findDiagonalOrder(self, matrix: List[List[int]]) -> List[int]:
# Check for empty matrices
if not matrix or not matrix[0]:
return []
# Variables to track the size of the matrix
N, M = len(matrix), len(matrix[0])
# The two arrays as explained in the algorithm
result, intermediate = [], []
# We have to go over all the elements in the first
# row and the last column to cover all possible diagonals
for d in range(N + M - 1):
# Clear the intermediate array everytime we start
# to process another diagonal
intermediate.clear()
# We need to figure out the "head" of this diagonal
# The elements in the first row and the last column
# are the respective heads.
r, c = 0 if d < M else d - M + 1, d if d < M else M - 1
# Iterate until one of the indices goes out of scope
# Take note of the index math to go down the diagonal
while r < N and c > -1:
intermediate.append(matrix[r][c])
r += 1
c -= 1
# Reverse even numbered diagonals. The
# article says we have to reverse odd
# numbered articles but here, the numbering
# is starting from 0 :P
if d % 2 == 0:
result.extend(intermediate[::-1])
else:
result.extend(intermediate)
return result
# V1
# IDEA : Simulation
# https://leetcode.com/problems/diagonal-traverse/solution/
class Solution:
def findDiagonalOrder(self, matrix: List[List[int]]) -> List[int]:
# Check for an empty matrix
if not matrix or not matrix[0]:
return []
# The dimensions of the matrix
N, M = len(matrix), len(matrix[0])
# Incides that will help us progress through
# the matrix, one element at a time.
row, column = 0, 0
# As explained in the article, this is the variable
# that helps us keep track of what direction we are
# processing the current diaonal
direction = 1
# Final result array that will contain all the elements
# of the matrix
result = []
# The uber while loop which will help us iterate over all
# the elements in the array.
while row < N and column < M:
# First and foremost, add the current element to
# the result matrix.
result.append(matrix[row][column])
# Move along in the current diagonal depending upon
# the current direction.[i, j] -> [i - 1, j + 1] if
# going up and [i, j] -> [i + 1][j - 1] if going down.
new_row = row + (-1 if direction == 1 else 1)
new_column = column + (1 if direction == 1 else -1)
# Checking if the next element in the diagonal is within the
# bounds of the matrix or not. If it's not within the bounds,
# we have to find the next head.
if new_row < 0 or new_row == N or new_column < 0 or new_column == M:
# If the current diagonal was going in the upwards
# direction.
if direction:
# For an upwards going diagonal having [i, j] as its tail
# If [i, j + 1] is within bounds, then it becomes
# the next head. Otherwise, the element directly below
# i.e. the element [i + 1, j] becomes the next head
row += (column == M - 1)
column += (column < M - 1)
else:
# For a downwards going diagonal having [i, j] as its tail
# if [i + 1, j] is within bounds, then it becomes
# the next head. Otherwise, the element directly below
# i.e. the element [i, j + 1] becomes the next head
column += (row == N - 1)
row += (row < N - 1)
# Flip the direction
direction = 1 - direction
else:
row = new_row
column = new_column
return result
# V1'
# https://blog.csdn.net/fuxuemingzhu/article/details/82528226
class Solution(object):
def findDiagonalOrder(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: List[int]
"""
if not matrix or not matrix[0]: return []
directions = [(-1, 1), (1, -1)]
count = 0
res = []
i, j = 0, 0
M, N = len(matrix), len(matrix[0])
while len(res) < M * N:
if 0 <= i < M and 0 <= j < N:
res.append(matrix[i][j])
direct = directions[count % 2]
i, j = i + direct[0], j + direct[1]
continue
elif i < 0 and 0 <= j < N:
i += 1
elif 0 <= i < M and j < 0:
j += 1
elif i < M and j >= N:
i += 2
j -= 1
elif i >= M and j < N:
j += 2
i -= 1
count += 1
return res
# V1''
# http://bookshadow.com/weblog/2017/02/05/leetcode-diagonal-traverse/
class Solution(object):
def findDiagonalOrder(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: List[int]
"""
if not matrix: return []
i, j, k = 0, 0, 1
w, h = len(matrix), len(matrix[0])
ans = []
for x in range(w * h):
ans.append(matrix[i][j])
if k > 0:
di, dj = i - 1, j + 1
else:
di, dj = i + 1, j - 1
if 0 <= di < w and 0 <= dj < h:
i, j = di, dj
else:
if k > 0:
if j + 1 < h:
j += 1
else:
i += 1
else:
if i + 1 < w:
i += 1
else:
j += 1
k *= -1
return ans
# V2