- Choose the invariant that makes a window valid or useful.
- Advance the right boundary and add the new element.
- Move the left boundary only as needed while maintaining the invariant and updating the answer.
Code notes
- 51 lines of Python from the credited upstream file minimum-operations-to-make-all-grid-elements-equal.py.
- The implementation visibly relies on sequence storage.
- No explicit loop blocks detected.
Complexity
Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.
Check the problem constraints before deciding whether this complexity will pass.
Use this to learn the idea, then write your own version.
123 45class Solution(object):6 def minOperations(self, grid, k):7 """8 :type grid: List[List[int]]9 :type k: int10 :rtype: int11 """12 c, target, mn = 0, 0, float("-inf")13 found = False14 lookup = [[0]*len(grid[0]) for _ in xrange(k)]15 cnt = [0]*len(grid[0])16 for i in xrange(len(grid)):17 total = 018 for j in xrange(len(grid[0])):19 total += cnt[j]20 diff = -(grid[i][j]+total) 21 if i+k-1 < len(grid) and j+k-1 < len(grid[0]):22 lookup[i%k][j] = diff23 cnt[j] += diff24 total += diff25 c += diff26 if i%k == 0 and j%k == 0: 27 mn = max(mn, -diff)28 elif not diff >= 0: 29 return -130 else:31 if (ik+1)*k <= len(grid) and (jk+1)*k <= len(grid[0]):32 if diff:33 return -134 else:35 if not found:36 found = True37 target = -diff38 elif target != -diff:39 return -140 if j-k+1 >= 0:41 total -= cnt[j-k+1]42 if i-k+1 >= 0:43 for j in xrange(len(grid[0])):44 cnt[j] -= lookup[(i-k+1)%k][j]45 lookup[(i-k+1)%k][j] = 046 if not found:47 target = mn48 elif target < mn:49 return -150 return c+target*((len(grid)k)*(len(grid[0])k))51