LeetCode in Kotlin
1306. Jump Game III
Medium
Given an array of non-negative integers arr, you are initially positioned at start index of the array. When you are at index i, you can jump to i + arr[i] or i - arr[i], check if you can reach to any index with value 0.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [4,2,3,0,3,1,2], start = 5
Output: true
Explanation:
All possible ways to reach at index 3 with value 0 are:
index 5 -> index 4 -> index 1 -> index 3
index 5 -> index 6 -> index 4 -> index 1 -> index 3
Example 2:
Input: arr = [4,2,3,0,3,1,2], start = 0
Output: true
Explanation:
One possible way to reach at index 3 with value 0 is:
index 0 -> index 4 -> index 1 -> index 3
Example 3:
Input: arr = [3,0,2,1,2], start = 2
Output: false
Explanation: There is no way to reach at index 1 with value 0.
Constraints:
1 <= arr.length <= 5 * 1040 <= arr[i] < arr.length0 <= start < arr.length
Solution
class Solution {
private lateinit var dp: BooleanArray
private var found = false
fun canReach(arr: IntArray, start: Int): Boolean {
if (arr[start] == 0) {
return true
}
dp = BooleanArray(arr.size)
dp[start] = true
recurse(arr, start)
return found
}
private fun recurse(arr: IntArray, index: Int) {
if (found) {
return
}
if (index - arr[index] >= 0 && !dp[index - arr[index]]) {
if (arr[index - arr[index]] == 0) {
found = true
return
}
dp[index - arr[index]] = true
recurse(arr, index - arr[index])
}
if (index + arr[index] < arr.size && !dp[index + arr[index]]) {
if (arr[index + arr[index]] == 0) {
found = true
return
}
dp[index + arr[index]] = true
recurse(arr, index + arr[index])
}
}
}