LeetCode #2770 — MEDIUM

Maximum Number of Jumps to Reach the Last Index

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

Solve on LeetCode

The Problem

Problem Statement

You are given a 0-indexed array nums of n integers and an integer target.

You are initially positioned at index 0. In one step, you can jump from index i to any index j such that:

0 <= i < j < n
-target <= nums[j] - nums[i] <= target

Return the maximum number of jumps you can make to reach index n - 1.

If there is no way to reach index n - 1, return -1.

Example 1:

Input: nums = [1,3,6,4,1,2], target = 2
Output: 3
Explanation: To go from index 0 to index n - 1 with the maximum number of jumps, you can perform the following jumping sequence:
- Jump from index 0 to index 1. 
- Jump from index 1 to index 3.
- Jump from index 3 to index 5.
It can be proven that there is no other jumping sequence that goes from 0 to n - 1 with more than 3 jumps. Hence, the answer is 3.

Example 2:

Input: nums = [1,3,6,4,1,2], target = 3
Output: 5
Explanation: To go from index 0 to index n - 1 with the maximum number of jumps, you can perform the following jumping sequence:
- Jump from index 0 to index 1.
- Jump from index 1 to index 2.
- Jump from index 2 to index 3.
- Jump from index 3 to index 4.
- Jump from index 4 to index 5.
It can be proven that there is no other jumping sequence that goes from 0 to n - 1 with more than 5 jumps. Hence, the answer is 5.

Example 3:

Input: nums = [1,3,6,4,1,2], target = 0
Output: -1
Explanation: It can be proven that there is no jumping sequence that goes from 0 to n - 1. Hence, the answer is -1.

Constraints:

2 <= nums.length == n <= 1000
-10⁹ <= nums[i] <= 10⁹
0 <= target <= 2 * 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed array nums of n integers and an integer target. You are initially positioned at index 0. In one step, you can jump from index i to any index j such that: 0 <= i < j < n -target <= nums[j] - nums[i] <= target Return the maximum number of jumps you can make to reach index n - 1. If there is no way to reach index n - 1, return -1.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming

Example 1

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

Example 2

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

Example 3

[1,3,6,4,1,2]
0

Core Insight

What unlocks the optimal approach

Use a dynamic programming approach.
Define a dynamic programming array dp of size n, where dp[i] represents the maximum number of jumps from index 0 to index i.
For each j iterate over all i < j. Set dp[j] = max(dp[j], dp[i] + 1) if -target <= nums[j] - nums[i] <= target.

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 #2770: Maximum Number of Jumps to Reach the Last Index
class Solution {
    private Integer[] f;
    private int[] nums;
    private int n;
    private int target;

    public int maximumJumps(int[] nums, int target) {
        n = nums.length;
        this.target = target;
        this.nums = nums;
        f = new Integer[n];
        int ans = dfs(0);
        return ans < 0 ? -1 : ans;
    }

    private int dfs(int i) {
        if (i == n - 1) {
            return 0;
        }
        if (f[i] != null) {
            return f[i];
        }
        int ans = -(1 << 30);
        for (int j = i + 1; j < n; ++j) {
            if (Math.abs(nums[i] - nums[j]) <= target) {
                ans = Math.max(ans, 1 + dfs(j));
            }
        }
        return f[i] = ans;
    }
}

// Accepted solution for LeetCode #2770: Maximum Number of Jumps to Reach the Last Index
func maximumJumps(nums []int, target int) int {
	n := len(nums)
	f := make([]int, n)
	for i := range f {
		f[i] = -1
	}
	var dfs func(int) int
	dfs = func(i int) int {
		if i == n-1 {
			return 0
		}
		if f[i] != -1 {
			return f[i]
		}
		f[i] = -(1 << 30)
		for j := i + 1; j < n; j++ {
			if abs(nums[i]-nums[j]) <= target {
				f[i] = max(f[i], 1+dfs(j))
			}
		}
		return f[i]
	}
	ans := dfs(0)
	if ans < 0 {
		return -1
	}
	return ans
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #2770: Maximum Number of Jumps to Reach the Last Index
class Solution:
    def maximumJumps(self, nums: List[int], target: int) -> int:
        @cache
        def dfs(i: int) -> int:
            if i == n - 1:
                return 0
            ans = -inf
            for j in range(i + 1, n):
                if abs(nums[i] - nums[j]) <= target:
                    ans = max(ans, 1 + dfs(j))
            return ans

        n = len(nums)
        ans = dfs(0)
        return -1 if ans < 0 else ans

// Accepted solution for LeetCode #2770: Maximum Number of Jumps to Reach the Last Index
// 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 #2770: Maximum Number of Jumps to Reach the Last Index
// class Solution {
//     private Integer[] f;
//     private int[] nums;
//     private int n;
//     private int target;
// 
//     public int maximumJumps(int[] nums, int target) {
//         n = nums.length;
//         this.target = target;
//         this.nums = nums;
//         f = new Integer[n];
//         int ans = dfs(0);
//         return ans < 0 ? -1 : ans;
//     }
// 
//     private int dfs(int i) {
//         if (i == n - 1) {
//             return 0;
//         }
//         if (f[i] != null) {
//             return f[i];
//         }
//         int ans = -(1 << 30);
//         for (int j = i + 1; j < n; ++j) {
//             if (Math.abs(nums[i] - nums[j]) <= target) {
//                 ans = Math.max(ans, 1 + dfs(j));
//             }
//         }
//         return f[i] = ans;
//     }
// }

// Accepted solution for LeetCode #2770: Maximum Number of Jumps to Reach the Last Index
function maximumJumps(nums: number[], target: number): number {
    const n = nums.length;
    const f: number[] = Array(n).fill(-1);
    const dfs = (i: number): number => {
        if (i === n - 1) {
            return 0;
        }
        if (f[i] !== -1) {
            return f[i];
        }
        f[i] = -(1 << 30);
        for (let j = i + 1; j < n; ++j) {
            if (Math.abs(nums[i] - nums[j]) <= target) {
                f[i] = Math.max(f[i], 1 + dfs(j));
            }
        }
        return f[i];
    };
    const ans = dfs(0);
    return ans < 0 ? -1 : 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.