LeetCode #403 — HARD

Frog Jump

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

The Problem

Problem Statement

A frog is crossing a river. The river is divided into some number of units, and at each unit, there may or may not exist a stone. The frog can jump on a stone, but it must not jump into the water.

Given a list of stones positions (in units) in sorted ascending order, determine if the frog can cross the river by landing on the last stone. Initially, the frog is on the first stone and assumes the first jump must be 1 unit.

If the frog's last jump was k units, its next jump must be either k - 1, k, or k + 1 units. The frog can only jump in the forward direction.

Example 1:

Input: stones = [0,1,3,5,6,8,12,17]
Output: true
Explanation: The frog can jump to the last stone by jumping 1 unit to the 2nd stone, then 2 units to the 3rd stone, then 2 units to the 4th stone, then 3 units to the 6th stone, 4 units to the 7th stone, and 5 units to the 8th stone.

Example 2:

Input: stones = [0,1,2,3,4,8,9,11]
Output: false
Explanation: There is no way to jump to the last stone as the gap between the 5th and 6th stone is too large.

Constraints:

2 <= stones.length <= 2000
0 <= stones[i] <= 2³¹ - 1
stones[0] == 0
stones is sorted in a strictly increasing order.

Step 01

Brute Force Baseline

Problem summary: A frog is crossing a river. The river is divided into some number of units, and at each unit, there may or may not exist a stone. The frog can jump on a stone, but it must not jump into the water. Given a list of stones positions (in units) in sorted ascending order, determine if the frog can cross the river by landing on the last stone. Initially, the frog is on the first stone and assumes the first jump must be 1 unit. If the frog's last jump was k units, its next jump must be either k - 1, k, or k + 1 units. The frog can only jump in the forward direction.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming

Example 1

[0,1,3,5,6,8,12,17]

Example 2

[0,1,2,3,4,8,9,11]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #403: Frog Jump
class Solution {
    private Boolean[][] f;
    private Map<Integer, Integer> pos = new HashMap<>();
    private int[] stones;
    private int n;

    public boolean canCross(int[] stones) {
        n = stones.length;
        f = new Boolean[n][n];
        this.stones = stones;
        for (int i = 0; i < n; ++i) {
            pos.put(stones[i], i);
        }
        return dfs(0, 0);
    }

    private boolean dfs(int i, int k) {
        if (i == n - 1) {
            return true;
        }
        if (f[i][k] != null) {
            return f[i][k];
        }
        for (int j = k - 1; j <= k + 1; ++j) {
            if (j > 0) {
                int h = stones[i] + j;
                if (pos.containsKey(h) && dfs(pos.get(h), j)) {
                    return f[i][k] = true;
                }
            }
        }
        return f[i][k] = false;
    }
}

// Accepted solution for LeetCode #403: Frog Jump
func canCross(stones []int) bool {
	n := len(stones)
	f := make([][]int, n)
	pos := map[int]int{}
	for i := range f {
		pos[stones[i]] = i
		f[i] = make([]int, n)
		for j := range f[i] {
			f[i][j] = -1
		}
	}
	var dfs func(int, int) bool
	dfs = func(i, k int) bool {
		if i == n-1 {
			return true
		}
		if f[i][k] != -1 {
			return f[i][k] == 1
		}
		for j := k - 1; j <= k+1; j++ {
			if j > 0 {
				if p, ok := pos[stones[i]+j]; ok {
					if dfs(p, j) {
						f[i][k] = 1
						return true
					}
				}
			}
		}
		f[i][k] = 0
		return false
	}
	return dfs(0, 0)
}

# Accepted solution for LeetCode #403: Frog Jump
class Solution:
    def canCross(self, stones: List[int]) -> bool:
        @cache
        def dfs(i, k):
            if i == n - 1:
                return True
            for j in range(k - 1, k + 2):
                if j > 0 and stones[i] + j in pos and dfs(pos[stones[i] + j], j):
                    return True
            return False

        n = len(stones)
        pos = {s: i for i, s in enumerate(stones)}
        return dfs(0, 0)

// Accepted solution for LeetCode #403: Frog Jump
use std::collections::HashMap;

impl Solution {
    #[allow(dead_code)]
    pub fn can_cross(stones: Vec<i32>) -> bool {
        let n = stones.len();
        let mut record = vec![vec![-1; n]; n];
        let mut pos = HashMap::new();
        for (i, &s) in stones.iter().enumerate() {
            pos.insert(s, i);
        }

        Self::dfs(&mut record, 0, 0, n, &pos, &stones)
    }

    #[allow(dead_code)]
    fn dfs(
        record: &mut Vec<Vec<i32>>,
        i: usize,
        k: usize,
        n: usize,
        pos: &HashMap<i32, usize>,
        stones: &Vec<i32>,
    ) -> bool {
        if i == n - 1 {
            return true;
        }

        if record[i][k] != -1 {
            return record[i][k] == 1;
        }

        let k = k as i32;
        for j in k - 1..=k + 1 {
            if j > 0
                && pos.contains_key(&(stones[i] + j))
                && Self::dfs(record, pos[&(stones[i] + j)], j as usize, n, pos, stones)
            {
                record[i][k as usize] = 1;
                return true;
            }
        }

        record[i][k as usize] = 0;
        false
    }
}

// Accepted solution for LeetCode #403: Frog Jump
function canCross(stones: number[]): boolean {
    const n = stones.length;
    const pos: Map<number, number> = new Map();
    for (let i = 0; i < n; ++i) {
        pos.set(stones[i], i);
    }
    const f: number[][] = new Array(n).fill(0).map(() => new Array(n).fill(-1));
    const dfs = (i: number, k: number): boolean => {
        if (i === n - 1) {
            return true;
        }
        if (f[i][k] !== -1) {
            return f[i][k] === 1;
        }
        for (let j = k - 1; j <= k + 1; ++j) {
            if (j > 0 && pos.has(stones[i] + j)) {
                if (dfs(pos.get(stones[i] + j)!, j)) {
                    f[i][k] = 1;
                    return true;
                }
            }
        }
        f[i][k] = 0;
        return false;
    };
    return dfs(0, 0);
}

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^2)

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.