Algorithm
Problem Name: 975. Odd Even Jump
Problem Link: https://leetcode.com/problems/odd-even-jump/
You are given an integer array arr. From some starting index, you can make a series of jumps. The (1st, 3rd, 5th, ...) jumps in the series are called odd-numbered jumps, and the (2nd, 4th, 6th, ...) jumps in the series are called even-numbered jumps. Note that the jumps are numbered, not the indices.
You may jump forward from index i to index j (with i < j) in the following way:
- During odd-numbered jumps (i.e., jumps 1, 3, 5, ...), you jump to the index
jsuch thatarr[i] <= arr[j]andarr[j]is the smallest possible value. If there are multiple such indicesj, you can only jump to the smallest such indexj. - During even-numbered jumps (i.e., jumps 2, 4, 6, ...), you jump to the index
jsuch thatarr[i] >= arr[j]andarr[j]is the largest possible value. If there are multiple such indicesj, you can only jump to the smallest such indexj. - It may be the case that for some index
i, there are no legal jumps.
A starting index is good if, starting from that index, you can reach the end of the array (index arr.length - 1) by jumping some number of times (possibly 0 or more than once).
Return the number of good starting indices.
Example 1:
Input: arr = [10,13,12,14,15] Output: 2 Explanation: From starting index i = 0, we can make our 1st jump to i = 2 (since arr[2] is the smallest among arr[1], arr[2], arr[3], arr[4] that is greater or equal to arr[0]), then we cannot jump any more. From starting index i = 1 and i = 2, we can make our 1st jump to i = 3, then we cannot jump any more. From starting index i = 3, we can make our 1st jump to i = 4, so we have reached the end. From starting index i = 4, we have reached the end already. In total, there are 2 different starting indices i = 3 and i = 4, where we can reach the end with some number of jumps.
Example 2:
Input: arr = [2,3,1,1,4] Output: 3 Explanation: From starting index i = 0, we make jumps to i = 1, i = 2, i = 3: During our 1st jump (odd-numbered), we first jump to i = 1 because arr[1] is the smallest value in [arr[1], arr[2], arr[3], arr[4]] that is greater than or equal to arr[0]. During our 2nd jump (even-numbered), we jump from i = 1 to i = 2 because arr[2] is the largest value in [arr[2], arr[3], arr[4]] that is less than or equal to arr[1]. arr[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3 During our 3rd jump (odd-numbered), we jump from i = 2 to i = 3 because arr[3] is the smallest value in [arr[3], arr[4]] that is greater than or equal to arr[2]. We can't jump from i = 3 to i = 4, so the starting index i = 0 is not good. In a similar manner, we can deduce that: From starting index i = 1, we jump to i = 4, so we reach the end. From starting index i = 2, we jump to i = 3, and then we can't jump anymore. From starting index i = 3, we jump to i = 4, so we reach the end. From starting index i = 4, we are already at the end. In total, there are 3 different starting indices i = 1, i = 3, and i = 4, where we can reach the end with some number of jumps.
Example 3:
Input: arr = [5,1,3,4,2] Output: 3 Explanation: We can reach the end from starting indices 1, 2, and 4.
Constraints:
1 <= arr.length <= 2 * 1040 <= arr[i] < 105
Code Examples
#1 Code Example with C Programming
Code -
C Programming
int smallest_among_greaters(int k, int *map, int sz) {
for (int i = k; i < sz; i ++) {
if (map[i] != 0) return map[i];
}
return 0;
}
int greatest_among_smallers(int k, int *map, int sz) {
for (int i = k; i >= 0; i --) {
if (map[i] != 0) return map[i];
}
return 0;
}
int oddEvenJumps(int* A, int ASize){
// if jump can to to a high or low number on current index
bool hi[20000] = { false }, lo[20000] = { false };
// index of a number is on
int map[100000] = { 0 };
#define sz (sizeof(map) / sizeof(map[0]))
int i, h, l, ans;
hi[ASize - 1] = lo[ASize - 1] = true;
map[A[ASize - 1]] = ASize; // index + 1
ans = 1;
for (i = ASize - 2; i >= 0; i --) {
h = smallest_among_greaters(A[i], map, sz);
l = greatest_among_smallers(A[i], map, sz);
if (h && lo[h - 1]) { hi[i] = true; ans ++; }
if (l && hi[l - 1]) lo[i] = true;
map[A[i]] = i + 1;
}
return ans;
}
Input
arr = [10,13,12,14,15]
Output
2
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int oddEvenJumps(int[] A) {
int n = A.length;
if (n < = 1) {
return n;
}
boolean[] odd = new boolean[n];
boolean[] even = new boolean[n];
odd[n - 1] = even[n - 1] = true;
TreeMap < Integer, Integer> map = new TreeMap<>();
map.put(A[n - 1], n - 1);
int count = 0;
for (int i = n - 2; i >= 0; i--) {
int val = A[i];
if (map.containsKey(val)) {
odd[i] = even[map.get(val)];
even[i] = odd[map.get(val)];
}
else {
Integer lower = map.lowerKey(val);
Integer higher = map.higherKey(val);
if (lower != null) {
even[i] = odd[map.get(lower)];
}
if (higher != null) {
odd[i] = even[map.get(higher)];
}
}
map.put(val, i);
}
for (boolean b : odd) {
count += b ? 1 : 0;
}
return count;
}
}
Input
arr = [10,13,12,14,15]
Output
2
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const oddEvenJumps = function (A) {
// Creates an array with ONLY the indices of the sorted array
let sorted = A.map((el, idx) => idx).sort((a, b) => A[a] - A[b] || a - b)
// Create an array of '-1's of the same array length for odd and even jumps
let oddJumps = new Array(A.length).fill(-1)
let evenJumps = new Array(A.length).fill(-1)
// Create an empty stack
let stack = []
// Loop the the sorted array of the indices
for (let i of sorted) {
// Loops as long the stack is full OR if the index is greater than the the last index of the stack
while (stack.length && i > stack[stack.length - 1]) {
// Pops the index from the stack and place and add the 'i' index in sortedJumps
oddJumps[stack.pop()] = i
}
// Pushes the index onto the stack
stack.push(i)
}
// Empty the stack
stack = []
// Reverses the sorted index array
let reverseSorted = sorted.sort((a, b) => A[b] - A[a] || a - b)
// Does the exact thing but for even jumps
for (let i of reverseSorted) {
while (stack.length && i > stack[stack.length - 1]) {
evenJumps[stack.pop()] = i
}
stack.push(i)
}
// Starts the count at 0
let count = 1
// Creates a boolean array of false elements for even and odd ends
let oddEnd = new Array(A.length).fill(false)
let evenEnd = new Array(A.length).fill(false)
// Switches the end of each array to true
oddEnd[A.length - 1] = true
evenEnd[A.length - 1] = true
// Loops through the array, starting from the 2nd from the right (since we do not need to worry about the last index)
for (let i = A.length - 2; i >= 0; --i) {
// If even jumps does
if (evenJumps[i] !== -1 && oddEnd[evenJumps[i]]) evenEnd[i] = true
if (oddJumps[i] !== -1 && evenEnd[oddJumps[i]]) {
oddEnd[i] = true
count++
}
}
return count
}
Input
arr = [2,3,1,1,4]
Output
3
#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
def oddEvenJumps(self, A: List[int]) -> int:
n = len(A)
next_higher, next_lower = [0] * n, [0] * n
stack = []
for a, i in sorted([a, i] for i, a in enumerate(A)):
while stack and stack[-1] < i:
next_higher[stack.pop()] = i
stack.append(i)
stack = []
for a, i in sorted([-a, i] for i, a in enumerate(A)):
while stack and stack[-1] < i:
next_lower[stack.pop()] = i
stack.append(i)
higher, lower = [0] * n, [0] * n
higher[-1] = lower[-1] = 1
for i in range(n - 1)[::-1]:
higher[i] = lower[next_higher[i]]
lower[i] = higher[next_lower[i]]
return sum(higher)
Input
arr = [2,3,1,1,4]
Output
3
#5 Code Example with C# Programming
Code -
C# Programming
using System.Collections.Generic;
namespace LeetCode
{
public class _0975_OddEvenJump
{
public int OddEvenJumps(int[] A)
{
var length = A.Length;
var oddPosibility = new bool[length];
var evenPosibility = new bool[length];
oddPosibility[length - 1] = true;
evenPosibility[length - 1] = true;
var map = new SortedList < int, int>
{
{ A[length - 1], length - 1 }
};
var result = 1;
for (int i = length - 2; i >= 0; i--)
{
var existed = map.TryGetValue(A[i], out int index);
map[A[i]] = i;
if (existed)
{
oddPosibility[i] = evenPosibility[index];
evenPosibility[i] = oddPosibility[index];
}
else
{
index = map.IndexOfKey(A[i]);
if (index != map.Count - 1)
oddPosibility[i] = evenPosibility[map.Values[index + 1]];
if (index != 0)
evenPosibility[i] = oddPosibility[map.Values[index - 1]];
}
if (oddPosibility[i])
result++;
}
return result;
}
}
}
Input
arr = [5,1,3,4,2]
Output
3