Leetcode 1536. Minimum Swaps to Arrange a Binary Grid

Seawolf

1. Given an n x n binary grid, in one step you can choose two adjacent rows of the grid and swap them.
   
2. 
   
3. A grid is said to be valid if all the cells above the main diagonal are zeros.
   
4. 
   
5. Return the minimum number of steps needed to make the grid valid, or -1 if the grid cannot be valid.
   
6. 
   
7. The main diagonal of a grid is the diagonal that starts at cell (1, 1) and ends at cell (n, n).
   
8. 
   
9. 
   
10. 
    
11. Example 1:
    
12. 
    
13. 
    
14. Input: grid = [[0,0,1],[1,1,0],[1,0,0]]
    
15. Output: 3
    
16. Example 2:
    
17. 
    
18. 
    
19. Input: grid = [[0,1,1,0],[0,1,1,0],[0,1,1,0],[0,1,1,0]]
    
20. Output: -1
    
21. Explanation: All rows are similar, swaps have no effect on the grid.
    
22. Example 3:
    
23. 
    
24. 
    
25. Input: grid = [[1,0,0],[1,1,0],[1,1,1]]
    
26. Output: 0
    
27. 
    
28. 
    
29. Constraints:
    
30. 
    
31. n == grid.length
    
32. n == grid[i].length
    
33. 1 <= n <= 200
    
34. grid[i][j] is 0 or 1

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1. class Solution:
   
2.     def minSwaps(self, grid: List[List[int]]) -> int:
   
3.         zeros = [0 for _ in grid]
   
4.         result = 0
   
5.         for i in range(len(grid)):
   
6.             count = 0
   
7.             for j in range(len(grid[i]) - 1, -1, -1):
   
8.                 if grid[i][j] != 0:
   
9.                     break
   
10.                 count += 1
    
11.             zeros[i] = count
    
12.         
    
13.         n = len(grid)
    
14.         for i in range(len(grid)):
    
15.             target = n - i - 1
    
16.             for j in range(i, len(grid) + 1):
    
17.                
    
18.                 if j < n and zeros[j] >= target:
    
19.                     break
    
20.             if j == n:
    
21.                 return -1
    
22.             result += j - i
    
23.             temp = zeros[j]
    
24.             for k in range(j, i, -1):
    
25.                 zeros[k] = zeros[k - 1]
    
26.             zeros[i] = temp
    
27.         return result

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