LeetCode #2087 — MEDIUM

Minimum Cost Homecoming of a Robot in a Grid

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

The Problem

Problem Statement

There is an m x n grid, where (0, 0) is the top-left cell and (m - 1, n - 1) is the bottom-right cell. You are given an integer array startPos where startPos = [start_row, start_col] indicates that initially, a robot is at the cell (start_row, start_col). You are also given an integer array homePos where homePos = [home_row, home_col] indicates that its home is at the cell (home_row, home_col).

The robot needs to go to its home. It can move one cell in four directions: left, right, up, or down, and it can not move outside the boundary. Every move incurs some cost. You are further given two 0-indexed integer arrays: rowCosts of length m and colCosts of length n.

If the robot moves up or down into a cell whose row is r, then this move costs rowCosts[r].
If the robot moves left or right into a cell whose column is c, then this move costs colCosts[c].

Return the minimum total cost for this robot to return home.

Example 1:

Input: startPos = [1, 0], homePos = [2, 3], rowCosts = [5, 4, 3], colCosts = [8, 2, 6, 7]
Output: 18
Explanation: One optimal path is that:
Starting from (1, 0)
-> It goes down to (2, 0). This move costs rowCosts[2] = 3.
-> It goes right to (2, 1). This move costs colCosts[1] = 2.
-> It goes right to (2, 2). This move costs colCosts[2] = 6.
-> It goes right to (2, 3). This move costs colCosts[3] = 7.
The total cost is 3 + 2 + 6 + 7 = 18

Example 2:

Input: startPos = [0, 0], homePos = [0, 0], rowCosts = [5], colCosts = [26]
Output: 0
Explanation: The robot is already at its home. Since no moves occur, the total cost is 0.

Constraints:

m == rowCosts.length
n == colCosts.length
1 <= m, n <= 10⁵
0 <= rowCosts[r], colCosts[c] <= 10⁴
startPos.length == 2
homePos.length == 2
0 <= start_row, home_row < m
0 <= start_col, home_col < n

Step 01

Brute Force Baseline

Problem summary: There is an m x n grid, where (0, 0) is the top-left cell and (m - 1, n - 1) is the bottom-right cell. You are given an integer array startPos where startPos = [startrow, startcol] indicates that initially, a robot is at the cell (startrow, startcol). You are also given an integer array homePos where homePos = [homerow, homecol] indicates that its home is at the cell (homerow, homecol). The robot needs to go to its home. It can move one cell in four directions: left, right, up, or down, and it can not move outside the boundary. Every move incurs some cost. You are further given two 0-indexed integer arrays: rowCosts of length m and colCosts of length n. If the robot moves up or down into a cell whose row is r, then this move costs rowCosts[r]. If the robot moves left or right into a cell whose column is c, then this move costs colCosts[c]. Return the minimum total cost for this robot to

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[1,0]
[2,3]
[5,4,3]
[8,2,6,7]

Example 2

[0,0]
[0,0]
[5]
[26]

Core Insight

What unlocks the optimal approach

Irrespective of what path the robot takes, it will have to traverse all the rows between startRow and homeRow and all the columns between startCol and homeCol.
Hence, making any other move other than traversing the required rows and columns will potentially incur more cost which can be avoided.

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 #2087: Minimum Cost Homecoming of a Robot in a Grid
class Solution {
    public int minCost(int[] startPos, int[] homePos, int[] rowCosts, int[] colCosts) {
        int i = startPos[0], j = startPos[1];
        int x = homePos[0], y = homePos[1];
        int ans = 0;
        if (i < x) {
            for (int k = i + 1; k <= x; ++k) {
                ans += rowCosts[k];
            }
        } else {
            for (int k = x; k < i; ++k) {
                ans += rowCosts[k];
            }
        }
        if (j < y) {
            for (int k = j + 1; k <= y; ++k) {
                ans += colCosts[k];
            }
        } else {
            for (int k = y; k < j; ++k) {
                ans += colCosts[k];
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2087: Minimum Cost Homecoming of a Robot in a Grid
func minCost(startPos []int, homePos []int, rowCosts []int, colCosts []int) (ans int) {
	i, j := startPos[0], startPos[1]
	x, y := homePos[0], homePos[1]
	if i < x {
		ans += sum(rowCosts, i+1, x+1)
	} else {
		ans += sum(rowCosts, x, i)
	}
	if j < y {
		ans += sum(colCosts, j+1, y+1)
	} else {
		ans += sum(colCosts, y, j)
	}
	return
}

func sum(nums []int, i, j int) (s int) {
	for k := i; k < j; k++ {
		s += nums[k]
	}
	return
}

# Accepted solution for LeetCode #2087: Minimum Cost Homecoming of a Robot in a Grid
class Solution:
    def minCost(
        self,
        startPos: List[int],
        homePos: List[int],
        rowCosts: List[int],
        colCosts: List[int],
    ) -> int:
        i, j = startPos
        x, y = homePos
        ans = 0
        if i < x:
            ans += sum(rowCosts[i + 1 : x + 1])
        else:
            ans += sum(rowCosts[x:i])
        if j < y:
            ans += sum(colCosts[j + 1 : y + 1])
        else:
            ans += sum(colCosts[y:j])
        return ans

// Accepted solution for LeetCode #2087: Minimum Cost Homecoming of a Robot in a Grid
/**
 * [2087] Minimum Cost Homecoming of a Robot in a Grid
 *
 * There is an m x n grid, where (0, 0) is the top-left cell and (m - 1, n - 1) is the bottom-right cell. You are given an integer array startPos where startPos = [startrow, startcol] indicates that initially, a robot is at the cell (startrow, startcol). You are also given an integer array homePos where homePos = [homerow, homecol] indicates that its home is at the cell (homerow, homecol).
 * The robot needs to go to its home. It can move one cell in four directions: left, right, up, or down, and it can not move outside the boundary. Every move incurs some cost. You are further given two 0-indexed integer arrays: rowCosts of length m and colCosts of length n.
 *
 * 	If the robot moves up or down into a cell whose row is r, then this move costs rowCosts[r].
 * 	If the robot moves left or right into a cell whose column is c, then this move costs colCosts[c].
 *
 * Return the minimum total cost for this robot to return home.
 *  
 * Example 1:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/10/11/eg-1.png" style="width: 282px; height: 217px;" />
 * Input: startPos = [1, 0], homePos = [2, 3], rowCosts = [5, 4, 3], colCosts = [8, 2, 6, 7]
 * Output: 18
 * Explanation: One optimal path is that:
 * Starting from (1, 0)
 * -> It goes down to (<u>2</u>, 0). This move costs rowCosts[2] = 3.
 * -> It goes right to (2, <u>1</u>). This move costs colCosts[1] = 2.
 * -> It goes right to (2, <u>2</u>). This move costs colCosts[2] = 6.
 * -> It goes right to (2, <u>3</u>). This move costs colCosts[3] = 7.
 * The total cost is 3 + 2 + 6 + 7 = 18
 * Example 2:
 *
 * Input: startPos = [0, 0], homePos = [0, 0], rowCosts = [5], colCosts = [26]
 * Output: 0
 * Explanation: The robot is already at its home. Since no moves occur, the total cost is 0.
 *
 *  
 * Constraints:
 *
 * 	m == rowCosts.length
 * 	n == colCosts.length
 * 	1 <= m, n <= 10^5
 * 	0 <= rowCosts[r], colCosts[c] <= 10^4
 * 	startPos.length == 2
 * 	homePos.length == 2
 * 	0 <= startrow, homerow < m
 * 	0 <= startcol, homecol < n
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/minimum-cost-homecoming-of-a-robot-in-a-grid/
// discuss: https://leetcode.com/problems/minimum-cost-homecoming-of-a-robot-in-a-grid/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn min_cost(
        start_pos: Vec<i32>,
        home_pos: Vec<i32>,
        row_costs: Vec<i32>,
        col_costs: Vec<i32>,
    ) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2087_example_1() {
        let start_pos = vec![0, 0];
        let home_pos = vec![2, 3];
        let row_costs = vec![5, 4, 3];
        let col_costs = vec![8, 2, 6, 7];

        let result = 2;

        assert_eq!(
            Solution::min_cost(start_pos, home_pos, row_costs, col_costs),
            result
        );
    }

    #[test]
    #[ignore]
    fn test_2087_example_2() {
        let start_pos = vec![1, 0];
        let home_pos = vec![0, 0];
        let row_costs = vec![5];
        let col_costs = vec![26];

        let result = 18;

        assert_eq!(
            Solution::min_cost(start_pos, home_pos, row_costs, col_costs),
            result
        );
    }
}

// Accepted solution for LeetCode #2087: Minimum Cost Homecoming of a Robot in a Grid
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2087: Minimum Cost Homecoming of a Robot in a Grid
// class Solution {
//     public int minCost(int[] startPos, int[] homePos, int[] rowCosts, int[] colCosts) {
//         int i = startPos[0], j = startPos[1];
//         int x = homePos[0], y = homePos[1];
//         int ans = 0;
//         if (i < x) {
//             for (int k = i + 1; k <= x; ++k) {
//                 ans += rowCosts[k];
//             }
//         } else {
//             for (int k = x; k < i; ++k) {
//                 ans += rowCosts[k];
//             }
//         }
//         if (j < y) {
//             for (int k = j + 1; k <= y; ++k) {
//                 ans += colCosts[k];
//             }
//         } else {
//             for (int k = y; k < j; ++k) {
//                 ans += colCosts[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 log n)

Space

O(1)

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.