LeetCode #3660 — MEDIUM

Jump Game IX

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an integer array nums.

From any index i, you can jump to another index j under the following rules:

Jump to index j where j > i is allowed only if nums[j] < nums[i].
Jump to index j where j < i is allowed only if nums[j] > nums[i].

For each index i, find the maximum value in nums that can be reached by following any sequence of valid jumps starting at i.

Return an array ans where ans[i] is the maximum value reachable starting from index i.

Example 1:

Input: nums = [2,1,3]

Output: [2,2,3]

Explanation:

For i = 0: No jump increases the value.
For i = 1: Jump to j = 0 as nums[j] = 2 is greater than nums[i].
For i = 2: Since nums[2] = 3 is the maximum value in nums, no jump increases the value.

Thus, ans = [2, 2, 3].

Example 2:

Input: nums = [2,3,1]

Output: [3,3,3]

Explanation:

For i = 0: Jump forward to j = 2 as nums[j] = 1 is less than nums[i] = 2, then from i = 2 jump to j = 1 as nums[j] = 3 is greater than nums[2].
For i = 1: Since nums[1] = 3 is the maximum value in nums, no jump increases the value.
For i = 2: Jump to j = 1 as nums[j] = 3 is greater than nums[2] = 1.

Thus, ans = [3, 3, 3].

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums. From any index i, you can jump to another index j under the following rules: Jump to index j where j > i is allowed only if nums[j] < nums[i]. Jump to index j where j < i is allowed only if nums[j] > nums[i]. For each index i, find the maximum value in nums that can be reached by following any sequence of valid jumps starting at i. Return an array ans where ans[i] is the maximum value reachable starting from index i.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming

Example 1

[2,1,3]

Example 2

[2,3,1]

Step 02

Core Insight

What unlocks the optimal approach

Think of the array as a directed graph where edges represent valid jumps.
From index <code>i</code>, forward jumps go only to smaller values; backward jumps go only to larger values.
The maximum reachable value from <code>i</code> is the maximum value in the connected component reachable under these jump rules.
You can find connected ranges by looking at prefix maximums and suffix minimums, a cut happens where all values to the left are <= all values to the right.

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

Upper-end input sizes

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 #3660: Jump Game IX
class Solution {
    public int[] maxValue(int[] nums) {
        int n = nums.length;
        int[] ans = new int[n];
        int[] preMax = new int[n];
        preMax[0] = nums[0];
        for (int i = 1; i < n; ++i) {
            preMax[i] = Math.max(preMax[i - 1], nums[i]);
        }
        int sufMin = 1 << 30;
        for (int i = n - 1; i >= 0; --i) {
            ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
            sufMin = Math.min(sufMin, nums[i]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3660: Jump Game IX
func maxValue(nums []int) []int {
	n := len(nums)
	ans := make([]int, n)
	preMax := make([]int, n)
	preMax[0] = nums[0]
	for i := 1; i < n; i++ {
		preMax[i] = max(preMax[i-1], nums[i])
	}
	sufMin := 1 << 30
	for i := n - 1; i >= 0; i-- {
		if preMax[i] > sufMin {
			ans[i] = ans[i+1]
		} else {
			ans[i] = preMax[i]
		}
		sufMin = min(sufMin, nums[i])
	}
	return ans
}

# Accepted solution for LeetCode #3660: Jump Game IX
class Solution:
    def maxValue(self, nums: List[int]) -> List[int]:
        n = len(nums)
        ans = [0] * n
        pre_max = [nums[0]] * n
        for i in range(1, n):
            pre_max[i] = max(pre_max[i - 1], nums[i])
        suf_min = inf
        for i in range(n - 1, -1, -1):
            ans[i] = ans[i + 1] if pre_max[i] > suf_min else pre_max[i]
            suf_min = min(suf_min, nums[i])
        return ans

// Accepted solution for LeetCode #3660: Jump Game IX
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3660: Jump Game IX
// class Solution {
//     public int[] maxValue(int[] nums) {
//         int n = nums.length;
//         int[] ans = new int[n];
//         int[] preMax = new int[n];
//         preMax[0] = nums[0];
//         for (int i = 1; i < n; ++i) {
//             preMax[i] = Math.max(preMax[i - 1], nums[i]);
//         }
//         int sufMin = 1 << 30;
//         for (int i = n - 1; i >= 0; --i) {
//             ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
//             sufMin = Math.min(sufMin, nums[i]);
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3660: Jump Game IX
function maxValue(nums: number[]): number[] {
    const n = nums.length;
    const ans = Array(n).fill(0);
    const preMax = Array(n).fill(nums[0]);
    for (let i = 1; i < n; i++) {
        preMax[i] = Math.max(preMax[i - 1], nums[i]);
    }
    let sufMin = 1 << 30;
    for (let i = n - 1; i >= 0; i--) {
        ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
        sufMin = Math.min(sufMin, nums[i]);
    }
    return ans;
}

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 × m)

Space

O(n × m)

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.