LeetCode #1340 — HARD

Jump Game V

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

The Problem

Problem Statement

Given an array of integers arr and an integer d. In one step you can jump from index i to index:

i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.

In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).

You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.

Notice that you can not jump outside of the array at any time.

Example 1:

Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.

Example 2:

Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.

Example 3:

Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.

Constraints:

1 <= arr.length <= 1000
1 <= arr[i] <= 10⁵
1 <= d <= arr.length

Step 01

Brute Force Baseline

Problem summary: Given an array of integers arr and an integer d. In one step you can jump from index i to index: i + x where: i + x < arr.length and 0 < x <= d. i - x where: i - x >= 0 and 0 < x <= d. In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)). You can choose any index of the array and start jumping. Return the maximum number of indices you can visit. Notice that you can not jump outside of the array at any time.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming

Example 1

[6,4,14,6,8,13,9,7,10,6,12]
2

Example 2

[3,3,3,3,3]
3

Example 3

[7,6,5,4,3,2,1]
1

Core Insight

What unlocks the optimal approach

Use dynamic programming. dp[i] is max jumps you can do starting from index i. Answer is max(dp[i]).
dp[i] = 1 + max (dp[j]) where j is all indices you can reach from i.

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 #1340: Jump Game V
class Solution {
    private int n;
    private int d;
    private int[] arr;
    private Integer[] f;

    public int maxJumps(int[] arr, int d) {
        n = arr.length;
        this.d = d;
        this.arr = arr;
        f = new Integer[n];
        int ans = 1;
        for (int i = 0; i < n; ++i) {
            ans = Math.max(ans, dfs(i));
        }
        return ans;
    }

    private int dfs(int i) {
        if (f[i] != null) {
            return f[i];
        }
        int ans = 1;
        for (int j = i - 1; j >= 0; --j) {
            if (i - j > d || arr[j] >= arr[i]) {
                break;
            }
            ans = Math.max(ans, 1 + dfs(j));
        }
        for (int j = i + 1; j < n; ++j) {
            if (j - i > d || arr[j] >= arr[i]) {
                break;
            }
            ans = Math.max(ans, 1 + dfs(j));
        }
        return f[i] = ans;
    }
}

// Accepted solution for LeetCode #1340: Jump Game V
func maxJumps(arr []int, d int) (ans int) {
	n := len(arr)
	f := make([]int, n)
	var dfs func(int) int
	dfs = func(i int) int {
		if f[i] != 0 {
			return f[i]
		}
		ans := 1
		for j := i - 1; j >= 0; j-- {
			if i-j > d || arr[j] >= arr[i] {
				break
			}
			ans = max(ans, 1+dfs(j))
		}
		for j := i + 1; j < n; j++ {
			if j-i > d || arr[j] >= arr[i] {
				break
			}
			ans = max(ans, 1+dfs(j))
		}
		f[i] = ans
		return ans
	}
	for i := 0; i < n; i++ {
		ans = max(ans, dfs(i))
	}
	return
}

# Accepted solution for LeetCode #1340: Jump Game V
class Solution:
    def maxJumps(self, arr: List[int], d: int) -> int:
        @cache
        def dfs(i):
            ans = 1
            for j in range(i - 1, -1, -1):
                if i - j > d or arr[j] >= arr[i]:
                    break
                ans = max(ans, 1 + dfs(j))
            for j in range(i + 1, n):
                if j - i > d or arr[j] >= arr[i]:
                    break
                ans = max(ans, 1 + dfs(j))
            return ans

        n = len(arr)
        return max(dfs(i) for i in range(n))

// Accepted solution for LeetCode #1340: Jump Game V
struct Solution;

use std::collections::HashMap;

impl Solution {
    fn max_jumps(arr: Vec<i32>, d: i32) -> i32 {
        let n = arr.len();
        let d = d as usize;
        let mut memo: HashMap<usize, i32> = HashMap::new();
        let mut res = 0;
        for i in 0..n {
            res = res.max(Self::dp(i, &mut memo, &arr, d, n));
        }
        res
    }

    fn dp(start: usize, memo: &mut HashMap<usize, i32>, arr: &[i32], d: usize, n: usize) -> i32 {
        if let Some(&res) = memo.get(&start) {
            return res;
        }
        let mut max = 0;
        let mut i = 1;
        while i <= d && start >= i && arr[start] > arr[start - i] {
            max = max.max(Self::dp(start - i, memo, arr, d, n));
            i += 1;
        }
        let mut i = 1;
        while i <= d && start + i < n && arr[start] > arr[start + i] {
            max = max.max(Self::dp(start + i, memo, arr, d, n));
            i += 1;
        }
        let res = 1 + max;
        memo.insert(start, res);
        res
    }
}

#[test]
fn test() {
    let arr = vec![6, 4, 14, 6, 8, 13, 9, 7, 10, 6, 12];
    let d = 2;
    let res = 4;
    assert_eq!(Solution::max_jumps(arr, d), res);
    let arr = vec![7, 6, 5, 4, 3, 2, 1];
    let d = 1;
    let res = 7;
    assert_eq!(Solution::max_jumps(arr, d), res);
    let arr = vec![7, 1, 7, 1, 7, 1];
    let d = 2;
    let res = 2;
    assert_eq!(Solution::max_jumps(arr, d), res);
    let arr = vec![66];
    let d = 1;
    let res = 1;
    assert_eq!(Solution::max_jumps(arr, d), res);
}

// Accepted solution for LeetCode #1340: Jump Game V
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1340: Jump Game V
// class Solution {
//     private int n;
//     private int d;
//     private int[] arr;
//     private Integer[] f;
// 
//     public int maxJumps(int[] arr, int d) {
//         n = arr.length;
//         this.d = d;
//         this.arr = arr;
//         f = new Integer[n];
//         int ans = 1;
//         for (int i = 0; i < n; ++i) {
//             ans = Math.max(ans, dfs(i));
//         }
//         return ans;
//     }
// 
//     private int dfs(int i) {
//         if (f[i] != null) {
//             return f[i];
//         }
//         int ans = 1;
//         for (int j = i - 1; j >= 0; --j) {
//             if (i - j > d || arr[j] >= arr[i]) {
//                 break;
//             }
//             ans = Math.max(ans, 1 + dfs(j));
//         }
//         for (int j = i + 1; j < n; ++j) {
//             if (j - i > d || arr[j] >= arr[i]) {
//                 break;
//             }
//             ans = Math.max(ans, 1 + dfs(j));
//         }
//         return f[i] = 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.