LeetCode #1240 — HARD

Tiling a Rectangle with the Fewest Squares

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

Example 1:

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:

Input: n = 5, m = 8
Output: 5

Example 3:

Input: n = 11, m = 13
Output: 6

Constraints:

1 <= n, m <= 13

Step 01

Brute Force Baseline

Problem summary: Given a rectangle of size n x m, return the minimum number of integer-sided squares that tile the rectangle.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Backtracking

Example 1

2
3

Example 2

5
8

Example 3

11
13

Core Insight

What unlocks the optimal approach

Can you use backtracking to solve this problem ?.
Suppose you've placed a bunch of squares. Where is the natural spot to place the next square ?.
The maximum number of squares to be placed will be ≤ max(n,m).

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1240: Tiling a Rectangle with the Fewest Squares
class Solution {
    private int n;
    private int m;
    private int[] filled;
    private int ans;

    public int tilingRectangle(int n, int m) {
        this.n = n;
        this.m = m;
        ans = n * m;
        filled = new int[n];
        dfs(0, 0, 0);
        return ans;
    }

    private 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) == 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) == 1) {
                    break;
                }
                ++r;
            }
            for (int k = j; k < m; ++k) {
                if ((filled[i] >> k & 1) == 1) {
                    break;
                }
                ++c;
            }
            int mx = Math.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;
                }
            }
        }
    }
}

// Accepted solution for LeetCode #1240: Tiling a Rectangle with the Fewest Squares
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
}

# Accepted solution for LeetCode #1240: Tiling a Rectangle with the Fewest Squares
class Solution:
    def tilingRectangle(self, n: int, m: int) -> int:
        def dfs(i: int, j: int, t: int):
            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

// Accepted solution for LeetCode #1240: Tiling a Rectangle with the Fewest Squares
struct Solution;

use std::collections::HashMap;

impl Solution {
    fn tiling_rectangle(n: i32, m: i32) -> i32 {
        let mut memo: HashMap<(i32, i32), i32> = HashMap::new();
        Self::dp(n, m, &mut memo)
    }
    fn dp(n: i32, m: i32, memo: &mut HashMap<(i32, i32), i32>) -> i32 {
        if n == m {
            1
        } else {
            if let Some(&res) = memo.get(&(n, m)) {
                return res;
            }
            let mut res = std::i32::MAX;

            for i in 1..n {
                res = res.min(Self::dp(i, m, memo) + Self::dp(n - i, m, memo));
            }

            for j in 1..m {
                res = res.min(Self::dp(n, j, memo) + Self::dp(n, m - j, memo));
            }

            for i in 1..n - 1 {
                for j in 1..m - 1 {
                    let a = Self::dp(i, j + 1, memo);
                    let b = Self::dp(i + 1, m - j - 1, memo);
                    let c = Self::dp(n - i, j, memo);
                    let d = Self::dp(n - i - 1, m - j, memo);
                    res = res.min(a + b + c + d + 1);
                }
            }
            memo.insert((n, m), res);
            res
        }
    }
}

#[test]
fn test() {
    let n = 2;
    let m = 3;
    let res = 3;
    assert_eq!(Solution::tiling_rectangle(n, m), res);
    let n = 5;
    let m = 8;
    let res = 5;
    assert_eq!(Solution::tiling_rectangle(n, m), res);
    let n = 11;
    let m = 13;
    let res = 6;
    assert_eq!(Solution::tiling_rectangle(n, m), res);
    let n = 7;
    let m = 6;
    let res = 5;
    assert_eq!(Solution::tiling_rectangle(n, m), res);
    let n = 10;
    let m = 9;
    let res = 6;
    assert_eq!(Solution::tiling_rectangle(n, m), res);
}

// Accepted solution for LeetCode #1240: Tiling a Rectangle with the Fewest Squares
function tilingRectangle(n: number, m: number): number {
    let ans = n * m;
    const filled: number[] = new Array(n).fill(0);
    const dfs = (i: number, j: number, t: number) => {
        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) {
            let [r, c] = [0, 0];
            for (let k = i; k < n; ++k) {
                if ((filled[k] >> j) & 1) {
                    break;
                }
                ++r;
            }
            for (let k = j; k < m; ++k) {
                if ((filled[i] >> k) & 1) {
                    break;
                }
                ++c;
            }
            const mx = Math.min(r, c);
            for (let w = 1; w <= mx; ++w) {
                for (let 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 (let x = i; x < i + mx; ++x) {
                for (let y = j; y < j + mx; ++y) {
                    filled[x] ^= 1 << y;
                }
            }
        }
    };
    dfs(0, 0, 0);
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n!)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.