LeetCode #3197 — HARD

Find the Minimum Area to Cover All Ones II

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

The Problem

Problem Statement

You are given a 2D binary array grid. You need to find 3 non-overlapping rectangles having non-zero areas with horizontal and vertical sides such that all the 1's in grid lie inside these rectangles.

Return the minimum possible sum of the area of these rectangles.

Note that the rectangles are allowed to touch.

Example 1:

Input: grid = [[1,0,1],[1,1,1]]

Output: 5

Explanation:

The 1's at (0, 0) and (1, 0) are covered by a rectangle of area 2.
The 1's at (0, 2) and (1, 2) are covered by a rectangle of area 2.
The 1 at (1, 1) is covered by a rectangle of area 1.

Example 2:

Input: grid = [[1,0,1,0],[0,1,0,1]]

Output: 5

Explanation:

The 1's at (0, 0) and (0, 2) are covered by a rectangle of area 3.
The 1 at (1, 1) is covered by a rectangle of area 1.
The 1 at (1, 3) is covered by a rectangle of area 1.

Constraints:

1 <= grid.length, grid[i].length <= 30
grid[i][j] is either 0 or 1.
The input is generated such that there are at least three 1's in grid.

Step 01

Brute Force Baseline

Problem summary: You are given a 2D binary array grid. You need to find 3 non-overlapping rectangles having non-zero areas with horizontal and vertical sides such that all the 1's in grid lie inside these rectangles. Return the minimum possible sum of the area of these rectangles. Note that the rectangles are allowed to touch.

Baseline thinking

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

Pattern signal: Array

Example 1

[[1,0,1],[1,1,1]]

Example 2

[[1,0,1,0],[0,1,0,1]]

Core Insight

What unlocks the optimal approach

Consider covering using 2 rectangles. As the rectangles don’t overlap, one of the rectangles must either be vertically above or horizontally left to the other.
To find the minimum area, check all possible vertical and horizontal splits.
For 3 rectangles, extend the idea to first covering using one rectangle, and then try splitting leftover ones both horizontally and vertically.

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 #3197: Find the Minimum Area to Cover All Ones II
class Solution {
    private final int inf = 1 << 30;
    private int[][] grid;

    public int minimumSum(int[][] grid) {
        this.grid = grid;
        int m = grid.length;
        int n = grid[0].length;
        int ans = m * n;

        for (int i1 = 0; i1 < m - 1; i1++) {
            for (int i2 = i1 + 1; i2 < m - 1; i2++) {
                ans = Math.min(
                    ans, f(0, 0, i1, n - 1) + f(i1 + 1, 0, i2, n - 1) + f(i2 + 1, 0, m - 1, n - 1));
            }
        }

        for (int j1 = 0; j1 < n - 1; j1++) {
            for (int j2 = j1 + 1; j2 < n - 1; j2++) {
                ans = Math.min(
                    ans, f(0, 0, m - 1, j1) + f(0, j1 + 1, m - 1, j2) + f(0, j2 + 1, m - 1, n - 1));
            }
        }

        for (int i = 0; i < m - 1; i++) {
            for (int j = 0; j < n - 1; j++) {
                ans = Math.min(
                    ans, f(0, 0, i, j) + f(0, j + 1, i, n - 1) + f(i + 1, 0, m - 1, n - 1));
                ans = Math.min(
                    ans, f(0, 0, i, n - 1) + f(i + 1, 0, m - 1, j) + f(i + 1, j + 1, m - 1, n - 1));

                ans = Math.min(
                    ans, f(0, 0, i, j) + f(i + 1, 0, m - 1, j) + f(0, j + 1, m - 1, n - 1));
                ans = Math.min(
                    ans, f(0, 0, m - 1, j) + f(0, j + 1, i, n - 1) + f(i + 1, j + 1, m - 1, n - 1));
            }
        }
        return ans;
    }

    private int f(int i1, int j1, int i2, int j2) {
        int x1 = inf, y1 = inf;
        int x2 = -inf, y2 = -inf;
        for (int i = i1; i <= i2; i++) {
            for (int j = j1; j <= j2; j++) {
                if (grid[i][j] == 1) {
                    x1 = Math.min(x1, i);
                    y1 = Math.min(y1, j);
                    x2 = Math.max(x2, i);
                    y2 = Math.max(y2, j);
                }
            }
        }
        return (x2 - x1 + 1) * (y2 - y1 + 1);
    }
}

// Accepted solution for LeetCode #3197: Find the Minimum Area to Cover All Ones II
func minimumSum(grid [][]int) int {
	m := len(grid)
	n := len(grid[0])
	ans := m * n
	inf := math.MaxInt32

	f := func(i1, j1, i2, j2 int) int {
		x1, y1 := inf, inf
		x2, y2 := -inf, -inf
		for i := i1; i <= i2; i++ {
			for j := j1; j <= j2; j++ {
				if grid[i][j] == 1 {
					x1 = min(x1, i)
					y1 = min(y1, j)
					x2 = max(x2, i)
					y2 = max(y2, j)
				}
			}
		}
		if x1 == inf {
			return 0
		}
		return (x2 - x1 + 1) * (y2 - y1 + 1)
	}

	for i1 := 0; i1 < m-1; i1++ {
		for i2 := i1 + 1; i2 < m-1; i2++ {
			ans = min(ans, f(0, 0, i1, n-1)+f(i1+1, 0, i2, n-1)+f(i2+1, 0, m-1, n-1))
		}
	}

	for j1 := 0; j1 < n-1; j1++ {
		for j2 := j1 + 1; j2 < n-1; j2++ {
			ans = min(ans, f(0, 0, m-1, j1)+f(0, j1+1, m-1, j2)+f(0, j2+1, m-1, n-1))
		}
	}

	for i := 0; i < m-1; i++ {
		for j := 0; j < n-1; j++ {
			ans = min(ans, f(0, 0, i, j)+f(0, j+1, i, n-1)+f(i+1, 0, m-1, n-1))
			ans = min(ans, f(0, 0, i, n-1)+f(i+1, 0, m-1, j)+f(i+1, j+1, m-1, n-1))
			ans = min(ans, f(0, 0, i, j)+f(i+1, 0, m-1, j)+f(0, j+1, m-1, n-1))
			ans = min(ans, f(0, 0, m-1, j)+f(0, j+1, i, n-1)+f(i+1, j+1, m-1, n-1))
		}
	}

	return ans
}

# Accepted solution for LeetCode #3197: Find the Minimum Area to Cover All Ones II
class Solution:
    def minimumSum(self, grid: List[List[int]]) -> int:
        def f(i1: int, j1: int, i2: int, j2: int) -> int:
            x1 = y1 = inf
            x2 = y2 = -inf
            for i in range(i1, i2 + 1):
                for j in range(j1, j2 + 1):
                    if grid[i][j] == 1:
                        x1 = min(x1, i)
                        y1 = min(y1, j)
                        x2 = max(x2, i)
                        y2 = max(y2, j)
            return (x2 - x1 + 1) * (y2 - y1 + 1)

        m, n = len(grid), len(grid[0])
        ans = m * n
        for i1 in range(m - 1):
            for i2 in range(i1 + 1, m - 1):
                ans = min(
                    ans,
                    f(0, 0, i1, n - 1)
                    + f(i1 + 1, 0, i2, n - 1)
                    + f(i2 + 1, 0, m - 1, n - 1),
                )
        for j1 in range(n - 1):
            for j2 in range(j1 + 1, n - 1):
                ans = min(
                    ans,
                    f(0, 0, m - 1, j1)
                    + f(0, j1 + 1, m - 1, j2)
                    + f(0, j2 + 1, m - 1, n - 1),
                )
        for i in range(m - 1):
            for j in range(n - 1):
                ans = min(
                    ans,
                    f(0, 0, i, j) + f(0, j + 1, i, n - 1) + f(i + 1, 0, m - 1, n - 1),
                )
                ans = min(
                    ans,
                    f(0, 0, i, n - 1)
                    + f(i + 1, 0, m - 1, j)
                    + f(i + 1, j + 1, m - 1, n - 1),
                )

                ans = min(
                    ans,
                    f(0, 0, i, j) + f(i + 1, 0, m - 1, j) + f(0, j + 1, m - 1, n - 1),
                )
                ans = min(
                    ans,
                    f(0, 0, m - 1, j)
                    + f(0, j + 1, i, n - 1)
                    + f(i + 1, j + 1, m - 1, n - 1),
                )
        return ans

// Accepted solution for LeetCode #3197: Find the Minimum Area to Cover All Ones II
impl Solution {
    pub fn minimum_sum(grid: Vec<Vec<i32>>) -> i32 {
        let m = grid.len();
        let n = grid[0].len();
        let mut ans = (m * n) as i32;
        let inf = i32::MAX / 4;

        let f = |i1: usize, j1: usize, i2: usize, j2: usize| -> i32 {
            let mut x1 = inf;
            let mut y1 = inf;
            let mut x2 = -inf;
            let mut y2 = -inf;

            for i in i1..=i2 {
                for j in j1..=j2 {
                    if grid[i][j] == 1 {
                        x1 = x1.min(i as i32);
                        y1 = y1.min(j as i32);
                        x2 = x2.max(i as i32);
                        y2 = y2.max(j as i32);
                    }
                }
            }
            if x1 > x2 || y1 > y2 {
                inf
            } else {
                (x2 - x1 + 1) * (y2 - y1 + 1)
            }
        };

        for i1 in 0..m - 1 {
            for i2 in i1 + 1..m - 1 {
                ans = ans
                    .min(f(0, 0, i1, n - 1) + f(i1 + 1, 0, i2, n - 1) + f(i2 + 1, 0, m - 1, n - 1));
            }
        }

        for j1 in 0..n - 1 {
            for j2 in j1 + 1..n - 1 {
                ans = ans
                    .min(f(0, 0, m - 1, j1) + f(0, j1 + 1, m - 1, j2) + f(0, j2 + 1, m - 1, n - 1));
            }
        }

        for i in 0..m - 1 {
            for j in 0..n - 1 {
                ans = ans.min(f(0, 0, i, j) + f(0, j + 1, i, n - 1) + f(i + 1, 0, m - 1, n - 1));
                ans = ans
                    .min(f(0, 0, i, n - 1) + f(i + 1, 0, m - 1, j) + f(i + 1, j + 1, m - 1, n - 1));
                ans = ans.min(f(0, 0, i, j) + f(i + 1, 0, m - 1, j) + f(0, j + 1, m - 1, n - 1));
                ans = ans
                    .min(f(0, 0, m - 1, j) + f(0, j + 1, i, n - 1) + f(i + 1, j + 1, m - 1, n - 1));
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #3197: Find the Minimum Area to Cover All Ones II
function minimumSum(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    let ans = m * n;
    const inf = Number.MAX_SAFE_INTEGER;
    const f = (i1: number, j1: number, i2: number, j2: number): number => {
        let [x1, y1] = [inf, inf];
        let [x2, y2] = [-inf, -inf];
        for (let i = i1; i <= i2; i++) {
            for (let j = j1; j <= j2; j++) {
                if (grid[i][j] === 1) {
                    x1 = Math.min(x1, i);
                    y1 = Math.min(y1, j);
                    x2 = Math.max(x2, i);
                    y2 = Math.max(y2, j);
                }
            }
        }
        return x1 === inf ? 0 : (x2 - x1 + 1) * (y2 - y1 + 1);
    };

    for (let i1 = 0; i1 < m - 1; i1++) {
        for (let i2 = i1 + 1; i2 < m - 1; i2++) {
            ans = Math.min(
                ans,
                f(0, 0, i1, n - 1) + f(i1 + 1, 0, i2, n - 1) + f(i2 + 1, 0, m - 1, n - 1),
            );
        }
    }

    for (let j1 = 0; j1 < n - 1; j1++) {
        for (let j2 = j1 + 1; j2 < n - 1; j2++) {
            ans = Math.min(
                ans,
                f(0, 0, m - 1, j1) + f(0, j1 + 1, m - 1, j2) + f(0, j2 + 1, m - 1, n - 1),
            );
        }
    }

    for (let i = 0; i < m - 1; i++) {
        for (let j = 0; j < n - 1; j++) {
            ans = Math.min(ans, f(0, 0, i, j) + f(0, j + 1, i, n - 1) + f(i + 1, 0, m - 1, n - 1));
            ans = Math.min(
                ans,
                f(0, 0, i, n - 1) + f(i + 1, 0, m - 1, j) + f(i + 1, j + 1, m - 1, n - 1),
            );
            ans = Math.min(ans, f(0, 0, i, j) + f(i + 1, 0, m - 1, j) + f(0, j + 1, m - 1, n - 1));
            ans = Math.min(
                ans,
                f(0, 0, m - 1, j) + f(0, j + 1, i, n - 1) + f(i + 1, j + 1, m - 1, n - 1),
            );
        }
    }

    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(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.