LeetCode #975 — HARD

Odd Even Jump

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an integer array arr. From some starting index, you can make a series of jumps. The (1^st, 3^rd, 5^th, ...) jumps in the series are called odd-numbered jumps, and the (2^nd, 4^th, 6^th, ...) 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.

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 * 10⁴
0 <= arr[i] < 10⁵

Step 01

Brute Force Baseline

Problem summary: 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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Stack · Segment Tree

Example 1

[10,13,12,14,15]

Example 2

[2,3,1,1,4]

Example 3

[5,1,3,4,2]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #975: Odd Even Jump
class Solution {
    private int n;
    private Integer[][] f;
    private int[][] g;

    public int oddEvenJumps(int[] arr) {
        TreeMap<Integer, Integer> tm = new TreeMap<>();
        n = arr.length;
        f = new Integer[n][2];
        g = new int[n][2];
        for (int i = n - 1; i >= 0; --i) {
            var hi = tm.ceilingEntry(arr[i]);
            g[i][1] = hi == null ? -1 : hi.getValue();
            var lo = tm.floorEntry(arr[i]);
            g[i][0] = lo == null ? -1 : lo.getValue();
            tm.put(arr[i], i);
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans += dfs(i, 1);
        }
        return ans;
    }

    private int dfs(int i, int k) {
        if (i == n - 1) {
            return 1;
        }
        if (g[i][k] == -1) {
            return 0;
        }
        if (f[i][k] != null) {
            return f[i][k];
        }
        return f[i][k] = dfs(g[i][k], k ^ 1);
    }
}

// Accepted solution for LeetCode #975: Odd Even Jump
func oddEvenJumps(arr []int) (ans int) {
	n := len(arr)
	rbt := redblacktree.NewWithIntComparator()
	f := make([][2]int, n)
	g := make([][2]int, n)
	for i := n - 1; i >= 0; i-- {
		if v, ok := rbt.Ceiling(arr[i]); ok {
			g[i][1] = v.Value.(int)
		} else {
			g[i][1] = -1
		}
		if v, ok := rbt.Floor(arr[i]); ok {
			g[i][0] = v.Value.(int)
		} else {
			g[i][0] = -1
		}
		rbt.Put(arr[i], i)
	}
	var dfs func(int, int) int
	dfs = func(i, k int) int {
		if i == n-1 {
			return 1
		}
		if g[i][k] == -1 {
			return 0
		}
		if f[i][k] != 0 {
			return f[i][k]
		}
		f[i][k] = dfs(g[i][k], k^1)
		return f[i][k]
	}
	for i := 0; i < n; i++ {
		if dfs(i, 1) == 1 {
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #975: Odd Even Jump
class Solution:
    def oddEvenJumps(self, arr: List[int]) -> int:
        @cache
        def dfs(i: int, k: int) -> bool:
            if i == n - 1:
                return True
            if g[i][k] == -1:
                return False
            return dfs(g[i][k], k ^ 1)

        n = len(arr)
        g = [[0] * 2 for _ in range(n)]
        sd = SortedDict()
        for i in range(n - 1, -1, -1):
            j = sd.bisect_left(arr[i])
            g[i][1] = sd.values()[j] if j < len(sd) else -1
            j = sd.bisect_right(arr[i]) - 1
            g[i][0] = sd.values()[j] if j >= 0 else -1
            sd[arr[i]] = i
        return sum(dfs(i, 1) for i in range(n))

// Accepted solution for LeetCode #975: Odd Even Jump
use std::collections::BTreeMap;

impl Solution {
    pub fn odd_even_jumps(arr: Vec<i32>) -> i32 {
        let n: usize = arr.len();
        let mut f: Vec<Vec<Option<i32>>> = vec![vec![None; 2]; n];
        let mut g: Vec<Vec<i32>> = vec![vec![-1; 2]; n];
        let mut tm: BTreeMap<i32, usize> = BTreeMap::new();

        for i in (0..n).rev() {
            if let Some((_, &v)) = tm.range(arr[i]..).next() {
                g[i][1] = v as i32;
            }
            if let Some((_, &v)) = tm.range(..=arr[i]).next_back() {
                g[i][0] = v as i32;
            }
            tm.insert(arr[i], i);
        }

        fn dfs(
            i: usize,
            k: usize,
            n: usize,
            f: &mut Vec<Vec<Option<i32>>>,
            g: &Vec<Vec<i32>>,
        ) -> i32 {
            if i == n - 1 {
                return 1;
            }
            if g[i][k] == -1 {
                return 0;
            }
            if let Some(v) = f[i][k] {
                return v;
            }
            let res = dfs(g[i][k] as usize, k ^ 1, n, f, g);
            f[i][k] = Some(res);
            res
        }

        let mut ans: i32 = 0;
        for i in 0..n {
            ans += dfs(i, 1, n, &mut f, &g);
        }
        ans
    }
}

// Accepted solution for LeetCode #975: Odd Even Jump
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #975: Odd Even Jump
// class Solution {
//     private int n;
//     private Integer[][] f;
//     private int[][] g;
// 
//     public int oddEvenJumps(int[] arr) {
//         TreeMap<Integer, Integer> tm = new TreeMap<>();
//         n = arr.length;
//         f = new Integer[n][2];
//         g = new int[n][2];
//         for (int i = n - 1; i >= 0; --i) {
//             var hi = tm.ceilingEntry(arr[i]);
//             g[i][1] = hi == null ? -1 : hi.getValue();
//             var lo = tm.floorEntry(arr[i]);
//             g[i][0] = lo == null ? -1 : lo.getValue();
//             tm.put(arr[i], i);
//         }
//         int ans = 0;
//         for (int i = 0; i < n; ++i) {
//             ans += dfs(i, 1);
//         }
//         return ans;
//     }
// 
//     private int dfs(int i, int k) {
//         if (i == n - 1) {
//             return 1;
//         }
//         if (g[i][k] == -1) {
//             return 0;
//         }
//         if (f[i][k] != null) {
//             return f[i][k];
//         }
//         return f[i][k] = dfs(g[i][k], k ^ 1);
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.