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chapter 1 sorted matrix search

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anebz 2019-04-09 23:51:12 +02:00
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223
Chapter 10 Sorting and searching/10.9. sorted_matrix_search.md
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@ -20,16 +20,16 @@ If A[0][3] < num and A[3][0] < num, if A[3][3] < num, return -1. Otherwise it mu
If A[0][3] => num or A[3][0] => num, Iterate through the last column, find first row where `A[row][3] > num`. Then iterate that row from the back. If not found, iterate through the last row, do the same. This should take O(N+M) in the worst case.
```python
def search_sorted_matrix(M, N, A, num):
def search_sorted_matrix(A, num):
if num < A[0][0] or num > A[M][N]:
return 0
if num == A[0][0] or num == A[M][N]:
return 1
if num < A[0][N]:
if num < A[0][N]: # case for small numbers, first row
initidx = 0
maxidx = N
while True:
i = maxidx // 2
i = (maxidx + initidx) // 2
if A[0][i] == num:
return 1
elif A[0][i] > num:
@ -40,5 +40,222 @@ def search_sorted_matrix(M, N, A, num):
if A[0][initidx] < num and A[0][initidx+1] > num:
break
# we need to look in column initidx
for i in range(initidx+1, 0, -1):
for j in range(N):
if A[i][j] == num:
return 1
if A[i][j] > num:
break
elif num < A[M][0]: # case for small numbers, first column
initidx = 0
maxidx = M
while True:
i = (maxidx + initidx) // 2
if A[i][0] == num:
return 1
elif A[i][0] > num:
maxidx = i
elif A[i][0] < num:
initidx = i
if A[initidx][0] < num and A[initidx+1][0] > num:
break
# we need to look in column initidx
for i in range(initidx+1, 0, -1):
for j in range(M):
if A[j][i] == num:
return 1
if A[j][i] > num:
break
elif num > A[M][0]:
initidx = 0
maxidx = N
while True:
i = (maxidx + initidx) // 2
if A[M][i] == num:
return 1
elif A[M][i] > num:
maxidx = i
elif A[M][i] < num:
initidx = i
if A[M][initidx] < num and A[M][initidx+1] > num:
initidx += 1
break
# we need to look in column initidx
for j in range(initidx, N):
for i in range(M-1, 1, -1):
if A[i][j] == num:
return 1
if A[i][j] < num:
break
return 0
```
Complexity is O(logN + ) or O(logM + ), bffI don't know exactly. O(1) space.
We only want to check rows where num > A[i][0] and num < A[i][M-1]. If it's equal to the top or botton fow, return 1. For example, only rows 1,2 qualify. We check the columns, and maybe 3,4 qualify. Then we only have to check the mini-matrix of the rows and cols.
> example:
```python
data = [
([
[0, 1, 3, 5, 6],
[1, 2, 4, 9, 10],
[3, 12, 13, 14, 15],
[6, 17, 18, 25, 26],
[8, 22, 23, 28, 30]
], 4, 0)
]
minimatrix = [[0, 1, 3],
[1, 2, 4],
[3, 12, 13]]
```
This is recursive, until we either find it or end up with a minimatrix of size 1 and A[i][j] != num, return 0
```python
def search_sorted_matrix(A, num):
if num < A[0][0] or num > A[M-1][N-1]:
return 0
if num == A[0][0] or num == A[M-1][N-1]:
return 1
rows = [0, M-1]
cols = [0, N-1]
while True:
new_rows = []
new_cols = []
flag = False
# check bounded rows
for i in range(rows[0], rows[1]+1):
if A[i][cols[0]] == num or A[i][cols[1]] == num:
return 1
if A[i][cols[0]] > num:
new_rows.append(i - 1)
flag = False
break
if i == rows[1]:
new_rows.append(i)
flag = False
break
if A[i][cols[0]] < num and A[i][cols[1]] > num:
if not flag:
new_rows.append(i)
flag = True
# check bounded cols
for j in range(cols[0], cols[1]+1):
if A[rows[0]][j] == num or A[rows[1]][j] == num:
return 1
if A[rows[0]][j] > num:
new_cols.append(j - 1)
flag = False
break
if j == cols[1]:
new_cols.append(j)
flag = False
break
if A[rows[0]][j] < num and A[rows[1]][j] > num:
if not flag:
new_cols.append(j)
flag = True
rows = new_rows
cols = new_cols
return 0
```
The matrix is recursively getting smaller, need to check for false test cases:
```python
def search_sorted_matrix(A, num):
if num < A[0][0] or num > A[M-1][N-1]:
return 0
if num == A[0][0] or num == A[M-1][N-1]:
return 1
rows = [0, M-1]
cols = [0, N-1]
while True:
new_rows = []
new_cols = []
flag = False
if rows[0] == rows[1]:
for j in range(cols[0], cols[1]+1):
if A[rows[0]][j] == num:
return 1
return 0
if cols[0] == cols[1]:
for i in range(rows[0], rows[1]+1):
if A[i][cols[0]] == num:
return 1
return 0
# check bounded rows
for i in range(rows[0], rows[1]+1):
if A[i][cols[0]] == num or A[i][cols[1]] == num:
return 1
if A[i][cols[0]] > num:
new_rows.append(i - 1)
flag = False
break
if i == rows[1]:
new_rows.append(i)
flag = False
break
if A[i][cols[0]] < num and A[i][cols[1]] > num:
if not flag:
new_rows.append(i)
flag = True
# check bounded cols
for j in range(cols[0], cols[1]+1):
if A[rows[0]][j] == num or A[rows[1]][j] == num:
return 1
if A[rows[0]][j] > num:
new_cols.append(j - 1)
flag = False
break
if j == cols[1]:
new_cols.append(j)
flag = False
break
if A[rows[0]][j] < num and A[rows[1]][j] > num:
if not flag:
new_cols.append(j)
flag = True
if len(new_rows) == len(new_cols) == 1 or new_rows[0] == new_rows[1] and new_cols[0] == new_cols[1]:
return 0
rows = new_rows
cols = new_cols
return 0
```
This works for occurring numbers and missing numbers also.
## Hints and solution
We can do binary search in each column, but that takes O(Mlog(N)). one binary search per row.
Keep track of possible rows and cols, where num > first elem, and num < last elem, and keep track of this using arrays, move up, down, left and right around the rows and the columns.
If we compare `num` to the center number in the matrix, we can eliminate roughly one quarter. hm.
p410, I don't really understand it. We can discard quarters of the matrix by checking the middle values in the matrix and recursively check the lower left quadrant and the upper right quadrant.

89
Chapter 10 Sorting and searching/10.9. sorted_matrix_search.py
View File

@ -1,38 +1,85 @@
import unittest
M = 5 - 1
N = 5 - 1
M = 5
N = 5
def search_sorted_matrix(A, num):
if num < A[0][0] or num > A[M][N]:
if num < A[0][0] or num > A[M-1][N-1]:
return 0
if num == A[0][0] or num == A[M][N]:
if num == A[0][0] or num == A[M-1][N-1]:
return 1
if num < A[0][N]:
initidx = 0
maxidx = N
while True:
i = maxidx // 2
if A[0][i] == num:
return 1
elif A[0][i] > num:
maxidx = i
elif A[0][i] < num:
initidx = i
if A[0][initidx] < num and A[0][initidx+1] > num:
rows = [0, M-1]
cols = [0, N-1]
while True:
new_rows = []
new_cols = []
flag = False
if rows[0] == rows[1]:
for j in range(cols[0], cols[1]+1):
if A[rows[0]][j] == num:
return 1
return 0
if cols[0] == cols[1]:
for i in range(rows[0], rows[1]+1):
if A[i][cols[0]] == num:
return 1
return 0
# check bounded rows
for i in range(rows[0], rows[1]+1):
if A[i][cols[0]] == num or A[i][cols[1]] == num:
return 1
if A[i][cols[0]] > num:
new_rows.append(i - 1)
flag = False
break
if i == rows[1]:
new_rows.append(i)
flag = False
break
if A[i][cols[0]] < num and A[i][cols[1]] > num:
if not flag:
new_rows.append(i)
flag = True
# check bounded cols
for j in range(cols[0], cols[1]+1):
if A[rows[0]][j] == num or A[rows[1]][j] == num:
return 1
if A[rows[0]][j] > num:
new_cols.append(j - 1)
flag = False
break
if j == cols[1]:
new_cols.append(j)
flag = False
break
if A[rows[0]][j] < num and A[rows[1]][j] > num:
if not flag:
new_cols.append(j)
flag = True
if len(new_rows) == len(new_cols) == 1 or new_rows[0] == new_rows[1] and new_cols[0] == new_cols[1]:
return 0
rows = new_rows
cols = new_cols
return 0
class Test(unittest.TestCase):
data = [
([
[0, 1, 3, 4, 6],
[6, 7, 8, 9, 10],
[11, 12, 13, 14, 15],
[16, 17, 18, 19, 20],
[21, 22, 23, 24, 25]
], 2, 1)
[1, 3, 4, 7, 14],
[3, 12, 13, 14, 15],
[6, 17, 18, 25, 26],
[8, 22, 23, 28, 30]
], 2, 0)
]
def test_rotate_matrix(self):