In this guide, you will get 975. Odd Even Jump LeetCode Solution with the best time and space complexity. The solution to Odd Even Jump problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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 j such that arr[i] = arr[j] and arr[j] is the largest possible value. If there are multiple such indices j, you can only jump to the smallest such index j.
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.
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.
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 * 104
0 <= arr[i] < 105
Complexity Analysis
Time Complexity: O(n\log n)
Space Complexity: O(n)
975. Odd Even Jump LeetCode Solution in C++
class Solution {
public:
int oddEvenJumps(vector<int>& arr) {
const int n = arr.size();
map<int, int> map; // {num: min index}
// inc[i] := true if can reach arr[n - 1] from i using increasing jumps
vector<bool> inc(n);
// dec[i] := true if can reach arr[n - 1] from i using decreasing jumps
vector<bool> dec(n);
map[arr[n - 1]] = n - 1;
inc.back() = true;
dec.back() = true;
for (int i = n - 2; i >= 0; --i) {
const auto lo = map.lower_bound(arr[i]); // the minimum value >= arr[i]
const auto hi = map.upper_bound(arr[i]); // the minimum value > arr[i]
if (lo != map.cend())
inc[i] = dec[lo->second];
if (hi != map.cbegin())
dec[i] = inc[prev(hi)->second];
map[arr[i]] = i;
}
return ranges::count(inc, true);
}
}; /* code provided by PROGIEZ */
975. Odd Even Jump LeetCode Solution in Java
class Solution {
public int oddEvenJumps(int[] arr) {
final int n = arr.length;
TreeMap<Integer, Integer> map = new TreeMap<>(); // {num: min index}
// inc[i] := true if can reach arr[n - 1] from i using increasing jumps
int[] inc = new int[n];
// dec[i] := true if can reach arr[n - 1] from i using increasing jumps
int[] dec = new int[n];
map.put(arr[n - 1], n - 1);
inc[n - 1] = 1;
dec[n - 1] = 1;
for (int i = n - 2; i >= 0; --i) {
// the minimum value >= arr[i]
Map.Entry<Integer, Integer> lo = map.ceilingEntry(arr[i]);
// the maximum value <= arr[i]
Map.Entry<Integer, Integer> hi = map.floorEntry(arr[i]);
if (lo != null)
inc[i] = dec[(int) lo.getValue()];
if (hi != null)
dec[i] = inc[(int) hi.getValue()];
map.put(arr[i], i);
}
return Arrays.stream(inc).sum();
}
} // code provided by PROGIEZ
