LeetCode-in-Java

403. Frog Jump

Hard

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.

Solution

import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;

public class Solution {
    // global hashmap to store visited index -> set of jump lengths from that index
    private HashMap<Integer, HashSet<Integer>> visited = new HashMap<>();

    public boolean canCross(int[] stones) {
        // a mathematical check before going in the recursion
        for (int i = 3; i < stones.length; i++) {
            if (stones[i] > stones[i - 1] * 2) {
                return false;
            }
        }
        // map of values -> index to make sure we get the next index quickly
        HashMap<Integer, Integer> rocks = new HashMap<>();
        for (int i = 0; i < stones.length; i++) {
            rocks.put(stones[i], i);
        }
        return jump(stones, 0, 1, 0, rocks);
    }

    private boolean jump(
            int[] stones, int index, int jumpLength, int expectedVal, Map<Integer, Integer> rocks) {
        // overshoot and going backwards not allowed
        if (index >= stones.length || jumpLength <= 0) {
            return false;
        }
        // reached the last index
        if (index == stones.length - 1) {
            return expectedVal == stones[index];
        }
        // check if this index -> jumpLength pair was seen before, otherwise record it
        HashSet<Integer> rememberJumps = visited.getOrDefault(index, new HashSet<>());
        if (stones[index] > expectedVal || rememberJumps.contains(jumpLength)) {
            return false;
        }
        rememberJumps.add(jumpLength);
        visited.put(index, rememberJumps);
        // check for jumpLength-1, jumpLength, jumpLength+1 for a new expected value
        return jump(
                        stones,
                        rocks.getOrDefault(stones[index] + jumpLength, stones.length),
                        jumpLength + 1,
                        stones[index] + jumpLength,
                        rocks)
                || jump(
                        stones,
                        rocks.getOrDefault(stones[index] + jumpLength, stones.length),
                        jumpLength,
                        stones[index] + jumpLength,
                        rocks)
                || jump(
                        stones,
                        rocks.getOrDefault(stones[index] + jumpLength, stones.length),
                        jumpLength - 1,
                        stones[index] + jumpLength,
                        rocks);
    }
}

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