chapter 1 sorted matrix search v1
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# 10.9. Sorted matrix search
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## Given an M x N matrix in which each row and each column is sorted in ascending order, write a method to find an element
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> example
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```python
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A = [[1,2,4,5,7],
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[2,3,5,6,8],
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[3,6,7,8,10],
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[4,7,8,9,11]]
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num = 3
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```
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## First idea
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If A[0][3] < num and A[3][0] < num, if A[3][3] < num, return -1. Otherwise it must be in the matrix.
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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.
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```python
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def search_sorted_matrix(M, N, A, num):
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if num < A[0][0] or num > A[M][N]:
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return 0
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if num == A[0][0] or num == A[M][N]:
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return 1
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if num < A[0][N]:
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initidx = 0
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maxidx = N
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while True:
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i = maxidx // 2
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if A[0][i] == num:
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return 1
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elif A[0][i] > num:
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maxidx = i
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elif A[0][i] < num:
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initidx = i
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if A[0][initidx] < num and A[0][initidx+1] > num:
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break
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```
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import unittest
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M = 5 - 1
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N = 5 - 1
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def search_sorted_matrix(A, num):
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if num < A[0][0] or num > A[M][N]:
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return 0
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if num == A[0][0] or num == A[M][N]:
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return 1
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if num < A[0][N]:
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initidx = 0
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maxidx = N
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while True:
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i = maxidx // 2
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if A[0][i] == num:
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return 1
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elif A[0][i] > num:
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maxidx = i
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elif A[0][i] < num:
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initidx = i
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if A[0][initidx] < num and A[0][initidx+1] > num:
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break
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return 0
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class Test(unittest.TestCase):
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data = [
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([
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[0, 1, 3, 4, 6],
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[6, 7, 8, 9, 10],
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[11, 12, 13, 14, 15],
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[16, 17, 18, 19, 20],
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[21, 22, 23, 24, 25]
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], 2, 1)
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]
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def test_rotate_matrix(self):
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for test_matrix, num, expected in self.data:
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actual = search_sorted_matrix(test_matrix, num)
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self.assertEqual(actual, expected)
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if __name__ == "__main__":
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unittest.main()
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