Leetcode #782: Transform to Chessboard
In this guide, we solve Leetcode #782 Transform to Chessboard in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given an n x n binary grid board. In each move, you can swap any two rows with each other, or any two columns with each other.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Bit Manipulation, Array, Math, Matrix
Intuition
The problem structure lets us track state with bitwise operations.
Bit operations are constant time and avoid extra memory.
Approach
Apply XOR/AND/OR and shifts to maintain the required invariant.
Aggregate the result in a single pass.
Steps:
- Identify a bitwise invariant.
- Combine values with bit operations.
- Return the aggregated result.
Example
Input: board = [[0,1,1,0],[0,1,1,0],[1,0,0,1],[1,0,0,1]]
Output: 2
Explanation: One potential sequence of moves is shown.
The first move swaps the first and second column.
The second move swaps the second and third row.
Python Solution
class Solution:
def movesToChessboard(self, board: List[List[int]]) -> int:
def f(mask, cnt):
ones = mask.bit_count()
if n & 1:
if abs(n - 2 * ones) != 1 or abs(n - 2 * cnt) != 1:
return -1
if ones == n // 2:
return n // 2 - (mask & 0xAAAAAAAA).bit_count()
return (n + 1) // 2 - (mask & 0x55555555).bit_count()
else:
if ones != n // 2 or cnt != n // 2:
return -1
cnt0 = n // 2 - (mask & 0xAAAAAAAA).bit_count()
cnt1 = n // 2 - (mask & 0x55555555).bit_count()
return min(cnt0, cnt1)
n = len(board)
mask = (1 << n) - 1
rowMask = colMask = 0
for i in range(n):
rowMask |= board[0][i] << i
colMask |= board[i][0] << i
revRowMask = mask ^ rowMask
revColMask = mask ^ colMask
sameRow = sameCol = 0
for i in range(n):
curRowMask = curColMask = 0
for j in range(n):
curRowMask |= board[i][j] << j
curColMask |= board[j][i] << j
if curRowMask not in (rowMask, revRowMask) or curColMask not in (
colMask,
revColMask,
):
return -1
sameRow += curRowMask == rowMask
sameCol += curColMask == colMask
t1 = f(rowMask, sameRow)
t2 = f(colMask, sameCol)
return -1 if t1 == -1 or t2 == -1 else t1 + t2
Complexity
The time complexity is , where is the size of the chessboard. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.