The O(n) Club: Frog Jump: Can Your Recursion Swim, or Will It Croak?

The O(n) Club: Frog Jump — Can Your Recursion Swim, or Will It Croak?

⚡ TL;DR

Frog at stone 0 wants to land on the last stone, only jumping forward by k-1, k, or k+1 units (where k is his last jump). The first jump is always a modest 1. Brute force will drown your CPU; memoization is how your frog stays dry. Here’s a Java sketch to get the river water swirling:

// Brute force (don't blame me if it times out)
boolean canCross(int[] stones) {
    return jump(stones, 0, 0);
}
boolean jump(int[] stones, int idx, int k) {
    if (idx == stones.length - 1) return true;
    for (int next = idx + 1; next < stones.length; next++) {
        int gap = stones[next] - stones[idx];
        if (gap >= k - 1 && gap <= k + 1 && gap > 0) {
            if (jump(stones, next, gap)) return true;
        }
    }
    return false;
}

🧠 How Devs Usually Mess This Up

Common land mines:

• Ignoring that the very first jump must be exactly 1. No short cuts, no creative liberties.
• Letting the frog hop backwards (bonus points if you want a time machine, but not here).
• Blind recursion, no memoization—see also: “interviewer sighs, your stack overflows, and the frog silently drifts downstream.”
• Assuming jumps must change every time, when repeats are allowed if a stone obediently sits at the target spot.

🔍 What’s Actually Going On

This is Frogger with dignity: after each heroic leap (of k units), the next jump must bravely differ by at most 1 (k-1, k, or k+1 units). The frog never backtracks and only lands where a stone exists. Thus, you always track two features in parallel universe:

• Where’s the frog? (The location/stone position)
• How big was the last jump? (Jump size)

Repeat that state, repeat your doom—so cache it! When picking a jump, always check if there’s a stone at the landing spot, or this frog’s going for an unwanted swim. Good thing Sets turn river navigation into a lookup party.

🛠️ PseudoCode

• Build a HashSet of all stone positions.
• Write a recursion: canCross(position, jumpSize).
• Base case: If position == last stone, congrats—frog is home.
• For possibleJump in jumpSize-1, jumpSize, jumpSize+1:
  • Skip if possibleJump <= 0. (Only positive hops allowed—no reverse or hover mode!)
  • Calculate nextPosition = position + possibleJump
  • If nextPosition is a stone: canCross(nextPosition, possibleJump)
• Memoize (position, jumpSize) pairs where the frog fails, or it’ll keep trying the same damp magic trick.
• Remember: first jump must be exactly 1.

  // Start: canCross(stones[0], 0)
  // First move: try only jump size 1

💻 The Code

import java.util.*;
 public class FrogJump {
    public boolean canCross(int[] stones) {
        Set
<integer> stoneSet = new HashSet<>();
        for (int s : stones) stoneSet.add(s);
        Map<string boolean> memo = new HashMap<>();
        return dfs(stones[0], 0, stones[stones.length - 1], stoneSet, memo);
    }
     private boolean dfs(int pos, int jump, int last, Set
<integer> stoneSet, Map<string boolean> memo) {
        String key = pos + "," + jump;
        if (memo.containsKey(key)) return memo.get(key);
        if (pos == last) return true;
        for (int step = jump - 1; step <= jump + 1; step++) {
            if (step <= 0) continue;
            int next = pos + step;
            if (stoneSet.contains(next)) {
                if (dfs(next, step, last, stoneSet, memo)) {
                    memo.put(key, true);
                    return true;
                }
            }
        }
        memo.put(key, false);
        return false;
    }
}</string></integer></string></integer>

⚠️ Pitfalls, Traps & Runtime Smacks

• Stubbing your first jump: No stone at distance 1? Don’t start packing for Paris—the frog’s going nowhere.
• Negative or zero jumps: Only valid hops are >0. The rest is frog yoga, not coding.
• No memoization, exponential misery: Miss the memo, explode the runtime (and maybe your soul).
• Time/space: There are at most O(N^2) (position, jumpSize) combos. Manageable—but don’t feed the input steroids.

🧠 Interviewer Brain-Tattoo

“If you don’t memoize, this frog and your interview performance both drown. Track your state—or get ready for wet socks.”