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Java-based LeetCode algorithm problem solutions, regularly updated
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package g0401_0500.s0403_frog_jump;

// #Hard #Array #Dynamic_Programming #2022_07_15_Time_13_ms_(99.06%)_Space_54.5_MB_(74.78%)

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

/**
 * 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.
**/
public class Solution {
    // global hashmap to store visited index -> set of jump lengths from that index
    private HashMap> 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 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 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 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);
    }
}