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Formatted question description: https://leetcode.ca/all/1240.html

1240. Tiling a Rectangle with the Fewest Squares

Level

Hard

Description

Given a rectangle of size n x m, find the minimum number of integer-sided squares that tile the rectangle.

Example 1:

Image text

Input: n = 2, m = 3

Output: 3

Explanation: 3 squares are necessary to cover the rectangle.

2 (squares of 1x1)

1 (square of 2x2)

Example 2:

Image text

Input: n = 5, m = 8

Output: 5

Example 3:

Image text

Input: n = 11, m = 13

Output: 6

Constraints:

  • 1 <= n <= 13
  • 1 <= m <= 13

Solution

Use dynamic programming. Create a 2D array dp of n + 1 rows and m + 1 columns, where dp[i][j] represents the minimum number of squares required to tile the i * j rectangle.

Obviously, if i == j, dp[i][j] = 1. Otherwise, the rectangle can be split by a horizontal line or a vertical line, which leads to two smaller rectangles, or a square may be put in the middle. For each case, calculate the minimum number of squares needed.

Finally, return dp[n][m].

  • class Solution {
    public int tilingRectangle(int n, int m) {
        int[][] dp = new int[n + 1][m + 1];
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                if (i == j)
                    dp[i][j] = 1;
                else {
                    dp[i][j] = Integer.MAX_VALUE;
                    int rowStart = 1, rowEnd = i / 2;
                    for (int row = rowStart; row <= rowEnd; row++)
                        dp[i][j] = Math.min(dp[i][j], dp[row][j] + dp[i - row][j]);
                    int columnStart = 1, columnEnd = j / 2;
                    for (int column = columnStart; column <= columnEnd; column++)
                        dp[i][j] = Math.min(dp[i][j], dp[i][column] + dp[i][j - column]);
                    for (int row = 1; row <= i; row++) {
                        for (int column = 1; column <= j; column++) {
                            for (int side = Math.min(row, column) - 1; side > 0; side--)
                                dp[i][j] = Math.min(dp[i][j], 1 + dp[row - side][column] + dp[i - row + side][column - side] + dp[i - row][j - column + side] + dp[row][j - column]);
                        }
                    }
                }
            }
        }
        return dp[n][m];
    }
}

  • class Solution {
public:
    int tilingRectangle(int n, int m) {
        memset(filled, 0, sizeof(filled));
        this->n = n;
        this->m = m;
        ans = n * m;
        dfs(0, 0, 0);
        return ans;
    }

private:
    int filled[13];
    int n, m;
    int ans;

    void dfs(int i, int j, int t) {
        if (j == m) {
            ++i;
            j = 0;
        }
        if (i == n) {
            ans = t;
            return;
        }
        if (filled[i] >> j & 1) {
            dfs(i, j + 1, t);
        } else if (t + 1 < ans) {
            int r = 0, c = 0;
            for (int k = i; k < n; ++k) {
                if (filled[k] >> j & 1) {
                    break;
                }
                ++r;
            }
            for (int k = j; k < m; ++k) {
                if (filled[i] >> k & 1) {
                    break;
                }
                ++c;
            }
            int mx = min(r, c);
            for (int w = 1; w <= mx; ++w) {
                for (int k = 0; k < w; ++k) {
                    filled[i + w - 1] |= 1 << (j + k);
                    filled[i + k] |= 1 << (j + w - 1);
                }
                dfs(i, j + w, t + 1);
            }
            for (int x = i; x < i + mx; ++x) {
                for (int y = j; y < j + mx; ++y) {
                    filled[x] ^= 1 << y;
                }
            }
        }
    }
};

  • class Solution:
    def tilingRectangle(self, n: int, m: int) -> int:
        def dfs(i, j, t):
            nonlocal ans
            if j == m:
                i += 1
                j = 0
            if i == n:
                ans = t
                return
            if filled[i] >> j & 1:
                dfs(i, j + 1, t)
            elif t + 1 < ans:
                r = c = 0
                for k in range(i, n):
                    if filled[k] >> j & 1:
                        break
                    r += 1
                for k in range(j, m):
                    if filled[i] >> k & 1:
                        break
                    c += 1
                mx = r if r < c else c
                for w in range(1, mx + 1):
                    for k in range(w):
                        filled[i + w - 1] |= 1 << (j + k)
                        filled[i + k] |= 1 << (j + w - 1)
                    dfs(i, j + w, t + 1)
                for x in range(i, i + mx):
                    for y in range(j, j + mx):
                        filled[x] ^= 1 << y

        ans = n * m
        filled = [0] * n
        dfs(0, 0, 0)
        return ans


  • func tilingRectangle(n int, m int) int {
	ans := n * m
	filled := make([]int, n)
	var dfs func(i, j, t int)
	dfs = func(i, j, t int) {
		if j == m {
			i++
			j = 0
		}
		if i == n {
			ans = t
			return
		}
		if filled[i]>>j&1 == 1 {
			dfs(i, j+1, t)
		} else if t+1 < ans {
			var r, c int
			for k := i; k < n; k++ {
				if filled[k]>>j&1 == 1 {
					break
				}
				r++
			}
			for k := j; k < m; k++ {
				if filled[i]>>k&1 == 1 {
					break
				}
				c++
			}
			mx := min(r, c)
			for w := 1; w <= mx; w++ {
				for k := 0; k < w; k++ {
					filled[i+w-1] |= 1 << (j + k)
					filled[i+k] |= 1 << (j + w - 1)
				}
				dfs(i, j+w, t+1)
			}
			for x := i; x < i+mx; x++ {
				for y := j; y < j+mx; y++ {
					filled[x] ^= 1 << y
				}
			}
		}
	}
	dfs(0, 0, 0)
	return ans
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

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