Odd Even Jump Problem

Problem

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 smallest possible value. If there are multiple such indices j, you can only jump to the smallest such index j.
During even-numbered jumps (i.e., jumps 2, 4, 6, …), 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.

Examples

Example 1:

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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:

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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:

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Input:
arr = [5,1,3,4,2]
Output:
 3
Explanation: We can reach the end from starting indices 1, 2, and 4.

Solution

Method 1 – Dynamic Programming with Monotonic Stack

Intuition

To determine from which indices we can reach the end using odd/even jumps, we need to efficiently find the next higher or lower index for each position. We use a monotonic stack and sorting to precompute these jumps, then use dynamic programming to check reachability.

Approach

For each index, precompute the next higher and next lower index using sorting and a monotonic stack.
Use two DP arrays: odd[i] and even[i] to represent if index i can reach the end with an odd or even jump.
Start from the end and fill DP arrays backwards.
The answer is the count of indices where odd[i] is True.

Code

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class Solution {
public:
    int oddEvenJumps(vector<int>& arr) {
        int n = arr.size();
        vector<int> next_higher(n, -1), next_lower(n, -1);
        vector<pair<int,int>> idx;
        for (int i = 0; i < n; ++i) idx.push_back({arr[i], i});
        sort(idx.begin(), idx.end());
        stack<int> st;
        for (auto& [val, i] : idx) {
            while (!st.empty() && st.top() < i) {
                next_higher[st.top()] = i;
                st.pop();
            }
            st.push(i);
        }
        idx.clear();
        for (int i = 0; i < n; ++i) idx.push_back({-arr[i], i});
        sort(idx.begin(), idx.end());
        while (!st.empty()) st.pop();
        for (auto& [val, i] : idx) {
            while (!st.empty() && st.top() < i) {
                next_lower[st.top()] = i;
                st.pop();
            }
            st.push(i);
        }
        vector<bool> odd(n, false), even(n, false);
        odd[n-1] = even[n-1] = true;
        for (int i = n-2; i >= 0; --i) {
            if (next_higher[i] != -1) odd[i] = even[next_higher[i]];
            if (next_lower[i] != -1) even[i] = odd[next_lower[i]];
        }
        return count(odd.begin(), odd.end(), true);
    }
};

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func oddEvenJumps(arr []int) int {
    n := len(arr)
    next_higher := make([]int, n)
    next_lower := make([]int, n)
    for i := range next_higher { next_higher[i], next_lower[i] = -1, -1 }
    idx := make([]int, n)
    for i := range arr { idx[i] = i }
    sort.Slice(idx, func(i, j int) bool { return arr[idx[i]] < arr[idx[j]] || (arr[idx[i]] == arr[idx[j]] && idx[i] < idx[j]) })
    st := []int{}
    for _, i := range idx {
        for len(st) > 0 && st[len(st)-1] < i {
            next_higher[st[len(st)-1]] = i
            st = st[:len(st)-1]
        }
        st = append(st, i)
    }
    idx2 := make([]int, n)
    for i := range arr { idx2[i] = i }
    sort.Slice(idx2, func(i, j int) bool { return arr[idx2[i]] > arr[idx2[j]] || (arr[idx2[i]] == arr[idx2[j]] && idx2[i] < idx2[j]) })
    st = []int{}
    for _, i := range idx2 {
        for len(st) > 0 && st[len(st)-1] < i {
            next_lower[st[len(st)-1]] = i
            st = st[:len(st)-1]
        }
        st = append(st, i)
    }
    odd := make([]bool, n)
    even := make([]bool, n)
    odd[n-1], even[n-1] = true, true
    for i := n-2; i >= 0; i-- {
        if next_higher[i] != -1 { odd[i] = even[next_higher[i]] }
        if next_lower[i] != -1 { even[i] = odd[next_lower[i]] }
    }
    ans := 0
    for _, v := range odd {
        if v { ans++ }
    }
    return ans
}

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import java.util.*;
class Solution {
    public int oddEvenJumps(int[] arr) {
        int n = arr.length;
        int[] next_higher = new int[n], next_lower = new int[n];
        Arrays.fill(next_higher, -1); Arrays.fill(next_lower, -1);
        Integer[] idx = new Integer[n];
        for (int i = 0; i < n; i++) idx[i] = i;
        Arrays.sort(idx, Comparator.comparingInt(i -> arr[i]));
        Stack<Integer> st = new Stack<>();
        for (int i : idx) {
            while (!st.isEmpty() && st.peek() < i) {
                next_higher[st.pop()] = i;
            }
            st.push(i);
        }
        Arrays.sort(idx, (i, j) -> arr[j] - arr[i]);
        st.clear();
        for (int i : idx) {
            while (!st.isEmpty() && st.peek() < i) {
                next_lower[st.pop()] = i;
            }
            st.push(i);
        }
        boolean[] odd = new boolean[n], even = new boolean[n];
        odd[n-1] = even[n-1] = true;
        for (int i = n-2; i >= 0; i--) {
            if (next_higher[i] != -1) odd[i] = even[next_higher[i]];
            if (next_lower[i] != -1) even[i] = odd[next_lower[i]];
        }
        int ans = 0;
        for (boolean v : odd) if (v) ans++;
        return ans;
    }
}

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class Solution {
    fun oddEvenJumps(arr: IntArray): Int {
        val n = arr.size
        val next_higher = IntArray(n) { -1 }
        val next_lower = IntArray(n) { -1 }
        val idx = arr.indices.sortedWith(compareBy({ arr[it] }, { it }))
        val st = ArrayDeque<Int>()
        for (i in idx) {
            while (st.isNotEmpty() && st.last() < i) {
                next_higher[st.removeLast()] = i
            }
            st.addLast(i)
        }
        val idx2 = arr.indices.sortedWith(compareBy({ -arr[it] }, { it }))
        st.clear()
        for (i in idx2) {
            while (st.isNotEmpty() && st.last() < i) {
                next_lower[st.removeLast()] = i
            }
            st.addLast(i)
        }
        val odd = BooleanArray(n)
        val even = BooleanArray(n)
        odd[n-1] = true; even[n-1] = true
        for (i in n-2 downTo 0) {
            if (next_higher[i] != -1) odd[i] = even[next_higher[i]]
            if (next_lower[i] != -1) even[i] = odd[next_lower[i]]
        }
        return odd.count { it }
    }
}

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def oddEvenJumps(arr: list[int]) -> int:
    n = len(arr)
    next_higher = [-1]*n
    next_lower = [-1]*n
    idx = sorted(range(n), key=lambda i: (arr[i], i))
    st = []
    for i in idx:
        while st and st[-1] < i:
            next_higher[st.pop()] = i
        st.append(i)
    idx2 = sorted(range(n), key=lambda i: (-arr[i], i))
    st = []
    for i in idx2:
        while st and st[-1] < i:
            next_lower[st.pop()] = i
        st.append(i)
    odd = [False]*n
    even = [False]*n
    odd[-1] = even[-1] = True
    for i in range(n-2, -1, -1):
        if next_higher[i] != -1:
            odd[i] = even[next_higher[i]]
        if next_lower[i] != -1:
            even[i] = odd[next_lower[i]]
    return sum(odd)

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impl Solution {
    pub fn odd_even_jumps(arr: Vec<i32>) -> i32 {
        let n = arr.len();
        let mut next_higher = vec![-1; n];
        let mut next_lower = vec![-1; n];
        let mut idx: Vec<_> = (0..n).collect();
        idx.sort_by_key(|&i| (arr[i], i));
        let mut st = Vec::new();
        for &i in &idx {
            while let Some(&top) = st.last() {
                if top < i {
                    next_higher[top] = i as i32;
                    st.pop();
                } else { break; }
            }
            st.push(i);
        }
        idx.sort_by_key(|&i| (-arr[i], i));
        st.clear();
        for &i in &idx {
            while let Some(&top) = st.last() {
                if top < i {
                    next_lower[top] = i as i32;
                    st.pop();
                } else { break; }
            }
            st.push(i);
        }
        let mut odd = vec![false; n];
        let mut even = vec![false; n];
        odd[n-1] = true; even[n-1] = true;
        for i in (0..n-1).rev() {
            if next_higher[i] != -1 {
                odd[i] = even[next_higher[i] as usize];
            }
            if next_lower[i] != -1 {
                even[i] = odd[next_lower[i] as usize];
            }
        }
        odd.iter().filter(|&&x| x).count() as i32
    }
}

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class Solution {
    oddEvenJumps(arr: number[]): number {
        const n = arr.length;
        const next_higher = Array(n).fill(-1);
        const next_lower = Array(n).fill(-1);
        const idx = Array.from({length: n}, (_, i) => i).sort((i, j) => arr[i] - arr[j] || i - j);
        const st: number[] = [];
        for (const i of idx) {
            while (st.length && st[st.length-1] < i) {
                next_higher[st.pop()!] = i;
            }
            st.push(i);
        }
        const idx2 = Array.from({length: n}, (_, i) => i).sort((i, j) => arr[j] - arr[i] || i - j);
        st.length = 0;
        for (const i of idx2) {
            while (st.length && st[st.length-1] < i) {
                next_lower[st.pop()!] = i;
            }
            st.push(i);
        }
        const odd = Array(n).fill(false);
        const even = Array(n).fill(false);
        odd[n-1] = even[n-1] = true;
        for (let i = n-2; i >= 0; i--) {
            if (next_higher[i] !== -1) odd[i] = even[next_higher[i]];
            if (next_lower[i] !== -1) even[i] = odd[next_lower[i]];
        }
        return odd.filter(x => x).length;
    }
}

Complexity

⏰ Time complexity: O(n log n), due to sorting and stack operations.
🧺 Space complexity: O(n), for DP and jump arrays.

Odd Even Jump Problem#

Problem#

Examples#

Solution#

Method 1 – Dynamic Programming with Monotonic Stack#

Intuition#

Approach#

Code#

Complexity#