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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;
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);
}
}
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