LeetCode #3651 — HARD

Minimum Cost Path with Teleportations

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

The Problem

Problem Statement

You are given a m x n 2D integer array grid and an integer k. You start at the top-left cell (0, 0) and your goal is to reach the bottom‐right cell (m - 1, n - 1).

There are two types of moves available:

Normal move: You can move right or down from your current cell (i, j), i.e. you can move to (i, j + 1) (right) or (i + 1, j) (down). The cost is the value of the destination cell.
Teleportation: You can teleport from any cell (i, j), to any cell (x, y) such that grid[x][y] <= grid[i][j]; the cost of this move is 0. You may teleport at most k times.

Return the minimum total cost to reach cell (m - 1, n - 1) from (0, 0).

Example 1:

Input: grid = [[1,3,3],[2,5,4],[4,3,5]], k = 2

Output: 7

Explanation:

Initially we are at (0, 0) and cost is 0.

Current Position	Move	New Position	Total Cost
`(0, 0)`	Move Down	`(1, 0)`	`0 + 2 = 2`
`(1, 0)`	Move Right	`(1, 1)`	`2 + 5 = 7`
`(1, 1)`	Teleport to `(2, 2)`	`(2, 2)`	`7 + 0 = 7`

The minimum cost to reach bottom-right cell is 7.

Example 2:

Input: grid = [[1,2],[2,3],[3,4]], k = 1

Output: 9

Explanation:

Initially we are at (0, 0) and cost is 0.

Current Position	Move	New Position	Total Cost
`(0, 0)`	Move Down	`(1, 0)`	`0 + 2 = 2`
`(1, 0)`	Move Right	`(1, 1)`	`2 + 3 = 5`
`(1, 1)`	Move Down	`(2, 1)`	`5 + 4 = 9`

The minimum cost to reach bottom-right cell is 9.

Constraints:

2 <= m, n <= 80
m == grid.length
n == grid[i].length
0 <= grid[i][j] <= 10⁴
0 <= k <= 10

Step 01

Brute Force Baseline

Problem summary: You are given a m x n 2D integer array grid and an integer k. You start at the top-left cell (0, 0) and your goal is to reach the bottom‐right cell (m - 1, n - 1). There are two types of moves available: Normal move: You can move right or down from your current cell (i, j), i.e. you can move to (i, j + 1) (right) or (i + 1, j) (down). The cost is the value of the destination cell. Teleportation: You can teleport from any cell (i, j), to any cell (x, y) such that grid[x][y] <= grid[i][j]; the cost of this move is 0. You may teleport at most k times. Return the minimum total cost to reach cell (m - 1, n - 1) from (0, 0).

Baseline thinking

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

Pattern signal: Array · Dynamic Programming

Example 1

[[1,3,3],[2,5,4],[4,3,5]]
2

Example 2

[[1,2],[2,3],[3,4]]
1

Step 02

Core Insight

What unlocks the optimal approach

Use dynamic programming to solve the problem efficiently.
Think of the solution in terms of up to <code>k</code> teleportation steps. At each step, compute the minimum cost to reach each cell, either through a normal move or a teleportation from the previous step.

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 #3651: Minimum Cost Path with Teleportations
class Solution {
    public int minCost(int[][] grid, int k) {
        int m = grid.length, n = grid[0].length;
        int inf = Integer.MAX_VALUE / 2;

        int[][][] f = new int[k + 1][m][n];
        for (int t = 0; t <= k; t++) {
            for (int i = 0; i < m; i++) {
                Arrays.fill(f[t][i], inf);
            }
        }

        f[0][0][0] = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i > 0) {
                    f[0][i][j] = Math.min(f[0][i][j], f[0][i - 1][j] + grid[i][j]);
                }
                if (j > 0) {
                    f[0][i][j] = Math.min(f[0][i][j], f[0][i][j - 1] + grid[i][j]);
                }
            }
        }

        Map<Integer, List<int[]>> g = new HashMap<>();
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int x = grid[i][j];
                g.computeIfAbsent(x, z -> new ArrayList<>()).add(new int[] {i, j});
            }
        }

        List<Integer> keys = new ArrayList<>(g.keySet());
        keys.sort(Collections.reverseOrder());

        for (int t = 1; t <= k; t++) {
            int mn = inf;
            for (int key : keys) {
                List<int[]> pos = g.get(key);
                for (int[] p : pos) {
                    mn = Math.min(mn, f[t - 1][p[0]][p[1]]);
                }
                for (int[] p : pos) {
                    f[t][p[0]][p[1]] = mn;
                }
            }
            for (int i = 0; i < m; i++) {
                for (int j = 0; j < n; j++) {
                    if (i > 0) {
                        f[t][i][j] = Math.min(f[t][i][j], f[t][i - 1][j] + grid[i][j]);
                    }
                    if (j > 0) {
                        f[t][i][j] = Math.min(f[t][i][j], f[t][i][j - 1] + grid[i][j]);
                    }
                }
            }
        }

        int ans = inf;
        for (int t = 0; t <= k; t++) {
            ans = Math.min(ans, f[t][m - 1][n - 1]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3651: Minimum Cost Path with Teleportations
func minCost(grid [][]int, k int) int {
	m, n := len(grid), len(grid[0])
	inf := int(^uint(0)>>1) / 4

	f := make([][][]int, k+1)
	for t := 0; t <= k; t++ {
		f[t] = make([][]int, m)
		for i := 0; i < m; i++ {
			f[t][i] = make([]int, n)
			for j := 0; j < n; j++ {
				f[t][i][j] = inf
			}
		}
	}

	f[0][0][0] = 0
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if i > 0 {
				f[0][i][j] = min(f[0][i][j], f[0][i-1][j]+grid[i][j])
			}
			if j > 0 {
				f[0][i][j] = min(f[0][i][j], f[0][i][j-1]+grid[i][j])
			}
		}
	}

	g := make(map[int][][]int)
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			x := grid[i][j]
			g[x] = append(g[x], []int{i, j})
		}
	}

	keys := make([]int, 0, len(g))
	for key := range g {
		keys = append(keys, key)
	}
	sort.Sort(sort.Reverse(sort.IntSlice(keys)))

	for t := 1; t <= k; t++ {
		mn := inf
		for _, key := range keys {
			pos := g[key]
			for _, p := range pos {
				mn = min(mn, f[t-1][p[0]][p[1]])
			}
			for _, p := range pos {
				f[t][p[0]][p[1]] = mn
			}
		}
		for i := 0; i < m; i++ {
			for j := 0; j < n; j++ {
				if i > 0 {
					f[t][i][j] = min(f[t][i][j], f[t][i-1][j]+grid[i][j])
				}
				if j > 0 {
					f[t][i][j] = min(f[t][i][j], f[t][i][j-1]+grid[i][j])
				}
			}
		}
	}

	ans := inf
	for t := 0; t <= k; t++ {
		ans = min(ans, f[t][m-1][n-1])
	}
	return ans
}

# Accepted solution for LeetCode #3651: Minimum Cost Path with Teleportations
class Solution:
    def minCost(self, grid: List[List[int]], k: int) -> int:
        m, n = len(grid), len(grid[0])
        f = [[[inf] * n for _ in range(m)] for _ in range(k + 1)]
        f[0][0][0] = 0
        for i in range(m):
            for j in range(n):
                if i:
                    f[0][i][j] = min(f[0][i][j], f[0][i - 1][j] + grid[i][j])
                if j:
                    f[0][i][j] = min(f[0][i][j], f[0][i][j - 1] + grid[i][j])
        g = defaultdict(list)
        for i, row in enumerate(grid):
            for j, x in enumerate(row):
                g[x].append((i, j))
        keys = sorted(g, reverse=True)
        for t in range(1, k + 1):
            mn = inf
            for key in keys:
                pos = g[key]
                for i, j in pos:
                    mn = min(mn, f[t - 1][i][j])
                for i, j in pos:
                    f[t][i][j] = mn
            for i in range(m):
                for j in range(n):
                    if i:
                        f[t][i][j] = min(f[t][i][j], f[t][i - 1][j] + grid[i][j])
                    if j:
                        f[t][i][j] = min(f[t][i][j], f[t][i][j - 1] + grid[i][j])
        return min(f[t][m - 1][n - 1] for t in range(k + 1))

// Accepted solution for LeetCode #3651: Minimum Cost Path with Teleportations
use std::collections::HashMap;

impl Solution {
    pub fn min_cost(grid: Vec<Vec<i32>>, k: i32) -> i32 {
        let m = grid.len();
        let n = grid[0].len();
        let k = k as usize;
        let inf: i32 = i32::MAX / 2;

        let mut f = vec![vec![vec![inf; n]; m]; k + 1];

        f[0][0][0] = 0;
        for i in 0..m {
            for j in 0..n {
                if i > 0 {
                    f[0][i][j] = f[0][i][j].min(f[0][i - 1][j] + grid[i][j]);
                }
                if j > 0 {
                    f[0][i][j] = f[0][i][j].min(f[0][i][j - 1] + grid[i][j]);
                }
            }
        }

        let mut g: HashMap<i32, Vec<(usize, usize)>> = HashMap::new();
        for i in 0..m {
            for j in 0..n {
                g.entry(grid[i][j]).or_default().push((i, j));
            }
        }

        let mut keys: Vec<i32> = g.keys().cloned().collect();
        keys.sort_by(|a, b| b.cmp(a));

        for t in 1..=k {
            let mut mn = inf;
            for &key in &keys {
                let pos = &g[&key];
                for &(i, j) in pos {
                    mn = mn.min(f[t - 1][i][j]);
                }
                for &(i, j) in pos {
                    f[t][i][j] = mn;
                }
            }
            for i in 0..m {
                for j in 0..n {
                    if i > 0 {
                        f[t][i][j] = f[t][i][j].min(f[t][i - 1][j] + grid[i][j]);
                    }
                    if j > 0 {
                        f[t][i][j] = f[t][i][j].min(f[t][i][j - 1] + grid[i][j]);
                    }
                }
            }
        }

        let mut ans = inf;
        for t in 0..=k {
            ans = ans.min(f[t][m - 1][n - 1]);
        }
        ans
    }
}

// Accepted solution for LeetCode #3651: Minimum Cost Path with Teleportations
function minCost(grid: number[][], k: number): number {
    const [m, n] = [grid.length, grid[0].length];
    const INF = 1e18;

    const f: number[][][] = Array.from({ length: k + 1 }, () =>
        Array.from({ length: m }, () => Array(n).fill(INF)),
    );

    f[0][0][0] = 0;
    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            if (i > 0) f[0][i][j] = Math.min(f[0][i][j], f[0][i - 1][j] + grid[i][j]);
            if (j > 0) f[0][i][j] = Math.min(f[0][i][j], f[0][i][j - 1] + grid[i][j]);
        }
    }

    const g = new Map<number, Array<[number, number]>>();
    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            const x = grid[i][j];
            if (!g.has(x)) {
                g.set(x, []);
            }
            g.get(x)!.push([i, j]);
        }
    }

    const keys = Array.from(g.keys()).sort((a, b) => b - a);

    for (let t = 1; t <= k; t++) {
        let mn = INF;

        for (const key of keys) {
            const pos = g.get(key)!;
            for (const [x, y] of pos) {
                mn = Math.min(mn, f[t - 1][x][y]);
            }
            for (const [x, y] of pos) {
                f[t][x][y] = mn;
            }
        }

        for (let i = 0; i < m; i++) {
            for (let j = 0; j < n; j++) {
                if (i > 0) {
                    f[t][i][j] = Math.min(f[t][i][j], f[t][i - 1][j] + grid[i][j]);
                }
                if (j > 0) {
                    f[t][i][j] = Math.min(f[t][i][j], f[t][i][j - 1] + grid[i][j]);
                }
            }
        }
    }

    let ans = INF;
    for (let t = 0; t <= k; t++) {
        ans = Math.min(ans, f[t][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((k + log mn)

Space

O(k × mn)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.