LeetCode 1240: Tiling a Rectangle with the Fewest Squares Solution

Master LeetCode problem 1240 (Tiling a Rectangle with the Fewest Squares), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1240. Tiling a Rectangle with the Fewest Squares

Hard

Backtracking

Problem Explanation

Explanation

To solve this problem, we can use a dynamic programming approach. We can create a 2D array dp where dp[i][j] represents the minimum number of squares needed to cover a rectangle of size i x j. We initialize the array with a large value and then iteratively update the values based on the minimum number of squares required.

We can iterate over all possible sizes of squares that can fit in the rectangle and update the minimum number of squares needed accordingly.

The time complexity of this approach is O(n * m * min(n, m)) and the space complexity is O(n * m).

Solution Code

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++) {
                dp[i][j] = Integer.MAX_VALUE;
                for (int k = 1; k <= Math.min(i, j); k++) {
                    dp[i][j] = Math.min(dp[i][j], 1 + dp[i - k][j] + dp[i][j - k]);
                }
                for (int a = 1; a <= i; a++) {
                    for (int b = 1; b <= j; b++) {
                        for (int x = 1; x <= a; x++) {
                            for (int y = 1; y <= b; y++) {
                                dp[i][j] = Math.min(dp[i][j], 1 + dp[x][y] + dp[a - x][b] + dp[i - a][j - b]);
                            }
                        }
                    }
                }
            }
        }
        
        return dp[n][m];
    }
}

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