LeetCode #1536 — MEDIUM

Minimum Swaps to Arrange a Binary Grid

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

Given an n x n binary grid, in one step you can choose two adjacent rows of the grid and swap them.

A grid is said to be valid if all the cells above the main diagonal are zeros.

Return the minimum number of steps needed to make the grid valid, or -1 if the grid cannot be valid.

The main diagonal of a grid is the diagonal that starts at cell (1, 1) and ends at cell (n, n).

Example 1:

Input: grid = [[0,0,1],[1,1,0],[1,0,0]]
Output: 3

Example 2:

Input: grid = [[0,1,1,0],[0,1,1,0],[0,1,1,0],[0,1,1,0]]
Output: -1
Explanation: All rows are similar, swaps have no effect on the grid.

Example 3:

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

Constraints:

n == grid.length == grid[i].length
1 <= n <= 200
grid[i][j] is either 0 or 1

Step 01

Brute Force Baseline

Problem summary: Given an n x n binary grid, in one step you can choose two adjacent rows of the grid and swap them. A grid is said to be valid if all the cells above the main diagonal are zeros. Return the minimum number of steps needed to make the grid valid, or -1 if the grid cannot be valid. The main diagonal of a grid is the diagonal that starts at cell (1, 1) and ends at cell (n, n).

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

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

Example 2

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

Example 3

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

Step 02

Core Insight

What unlocks the optimal approach

For each row of the grid calculate the most right 1 in the grid in the array maxRight.
To check if there exist answer, sort maxRight and check if maxRight[i] ≤ i for all possible i's.
If there exist an answer, simulate the swaps.

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

Upper-end input sizes

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 #1536: Minimum Swaps to Arrange a Binary Grid
class Solution {
    public int minSwaps(int[][] grid) {
        int n = grid.length;
        int[] pos = new int[n];
        Arrays.fill(pos, -1);
        for (int i = 0; i < n; ++i) {
            for (int j = n - 1; j >= 0; --j) {
                if (grid[i][j] == 1) {
                    pos[i] = j;
                    break;
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            int k = -1;
            for (int j = i; j < n; ++j) {
                if (pos[j] <= i) {
                    ans += j - i;
                    k = j;
                    break;
                }
            }
            if (k == -1) {
                return -1;
            }
            for (; k > i; --k) {
                int t = pos[k];
                pos[k] = pos[k - 1];
                pos[k - 1] = t;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1536: Minimum Swaps to Arrange a Binary Grid
func minSwaps(grid [][]int) (ans int) {
	n := len(grid)
	pos := make([]int, n)
	for i := range pos {
		pos[i] = -1
	}
	for i := 0; i < n; i++ {
		for j := n - 1; j >= 0; j-- {
			if grid[i][j] == 1 {
				pos[i] = j
				break
			}
		}
	}
	for i := 0; i < n; i++ {
		k := -1
		for j := i; j < n; j++ {
			if pos[j] <= i {
				ans += j - i
				k = j
				break
			}
		}
		if k == -1 {
			return -1
		}
		for ; k > i; k-- {
			pos[k], pos[k-1] = pos[k-1], pos[k]
		}
	}
	return
}

# Accepted solution for LeetCode #1536: Minimum Swaps to Arrange a Binary Grid
class Solution:
    def minSwaps(self, grid: List[List[int]]) -> int:
        n = len(grid)
        pos = [-1] * n
        for i in range(n):
            for j in range(n - 1, -1, -1):
                if grid[i][j] == 1:
                    pos[i] = j
                    break
        ans = 0
        for i in range(n):
            k = -1
            for j in range(i, n):
                if pos[j] <= i:
                    ans += j - i
                    k = j
                    break
            if k == -1:
                return -1
            while k > i:
                pos[k], pos[k - 1] = pos[k - 1], pos[k]
                k -= 1
        return ans

// Accepted solution for LeetCode #1536: Minimum Swaps to Arrange a Binary Grid
struct Solution;

impl Solution {
    fn min_swaps(mut grid: Vec<Vec<i32>>) -> i32 {
        let n = grid.len();
        let mut res = 0;
        'outer: for i in 0..n {
            for j in i..n {
                if grid[j][i + 1..].iter().all(|&x| x == 0) {
                    let mut k = j;
                    while k > i {
                        grid.swap(k, k - 1);
                        res += 1;
                        k -= 1;
                    }
                    continue 'outer;
                }
            }
            return -1;
        }
        res as i32
    }
}

#[test]
fn test() {
    let grid = vec_vec_i32![[0, 0, 1], [1, 1, 0], [1, 0, 0]];
    let res = 3;
    assert_eq!(Solution::min_swaps(grid), res);
    let grid = vec_vec_i32![
        [1, 0, 0, 0, 0, 0],
        [0, 0, 0, 1, 0, 0],
        [0, 0, 0, 1, 0, 0],
        [0, 1, 0, 0, 0, 0],
        [0, 0, 1, 0, 0, 0],
        [0, 0, 0, 0, 0, 1]
    ];
    let res = 4;
    assert_eq!(Solution::min_swaps(grid), res);
    let grid = vec_vec_i32![[0, 1, 1, 0], [0, 1, 1, 0], [0, 1, 1, 0], [0, 1, 1, 0]];
    let res = -1;
    assert_eq!(Solution::min_swaps(grid), res);
    let grid = vec_vec_i32![[1, 0, 0], [1, 1, 0], [1, 1, 1]];
    let res = 0;
    assert_eq!(Solution::min_swaps(grid), res);
}

// Accepted solution for LeetCode #1536: Minimum Swaps to Arrange a Binary Grid
function minSwaps(grid: number[][]): number {
    const n = grid.length;
    const pos: number[] = Array(n).fill(-1);
    for (let i = 0; i < n; ++i) {
        for (let j = n - 1; ~j; --j) {
            if (grid[i][j] === 1) {
                pos[i] = j;
                break;
            }
        }
    }
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        let k = -1;
        for (let j = i; j < n; ++j) {
            if (pos[j] <= i) {
                ans += j - i;
                k = j;
                break;
            }
        }
        if (k === -1) {
            return -1;
        }
        for (; k > i; --k) {
            [pos[k], pos[k - 1]] = [pos[k - 1], pos[k]];
        }
    }
    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^2)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.