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.
Move from brute-force thinking to an efficient approach using array strategy.
You are given a 0-indexed integer array nums and an integer k.
You are initially standing at index 0. In one move, you can jump at most k steps forward without going outside the boundaries of the array. That is, you can jump from index i to any index in the range [i + 1, min(n - 1, i + k)] inclusive.
You want to reach the last index of the array (index n - 1). Your score is the sum of all nums[j] for each index j you visited in the array.
Return the maximum score you can get.
Example 1:
Input: nums = [1,-1,-2,4,-7,3], k = 2 Output: 7 Explanation: You can choose your jumps forming the subsequence [1,-1,4,3] (underlined above). The sum is 7.
Example 2:
Input: nums = [10,-5,-2,4,0,3], k = 3 Output: 17 Explanation: You can choose your jumps forming the subsequence [10,4,3] (underlined above). The sum is 17.
Example 3:
Input: nums = [1,-5,-20,4,-1,3,-6,-3], k = 2 Output: 0
Constraints:
1 <= nums.length, k <= 105-104 <= nums[i] <= 104Problem summary: You are given a 0-indexed integer array nums and an integer k. You are initially standing at index 0. In one move, you can jump at most k steps forward without going outside the boundaries of the array. That is, you can jump from index i to any index in the range [i + 1, min(n - 1, i + k)] inclusive. You want to reach the last index of the array (index n - 1). Your score is the sum of all nums[j] for each index j you visited in the array. Return the maximum score you can get.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Dynamic Programming · Monotonic Queue
[1,-1,-2,4,-7,3] 2
[10,-5,-2,4,0,3] 3
[1,-5,-20,4,-1,3,-6,-3] 2
sliding-window-maximum)jump-game-vii)jump-game-viii)maximize-value-of-function-in-a-ball-passing-game)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1696: Jump Game VI
class Solution {
public int maxResult(int[] nums, int k) {
int n = nums.length;
int[] f = new int[n];
Deque<Integer> q = new ArrayDeque<>();
q.offer(0);
for (int i = 0; i < n; ++i) {
if (i - q.peekFirst() > k) {
q.pollFirst();
}
f[i] = nums[i] + f[q.peekFirst()];
while (!q.isEmpty() && f[q.peekLast()] <= f[i]) {
q.pollLast();
}
q.offerLast(i);
}
return f[n - 1];
}
}
// Accepted solution for LeetCode #1696: Jump Game VI
func maxResult(nums []int, k int) int {
n := len(nums)
f := make([]int, n)
q := Deque{}
q.PushBack(0)
for i := 0; i < n; i++ {
if i-q.Front() > k {
q.PopFront()
}
f[i] = nums[i] + f[q.Front()]
for !q.Empty() && f[i] >= f[q.Back()] {
q.PopBack()
}
q.PushBack(i)
}
return f[n-1]
}
type Deque struct{ l, r []int }
func (q Deque) Empty() bool {
return len(q.l) == 0 && len(q.r) == 0
}
func (q Deque) Size() int {
return len(q.l) + len(q.r)
}
func (q *Deque) PushFront(v int) {
q.l = append(q.l, v)
}
func (q *Deque) PushBack(v int) {
q.r = append(q.r, v)
}
func (q *Deque) PopFront() (v int) {
if len(q.l) > 0 {
q.l, v = q.l[:len(q.l)-1], q.l[len(q.l)-1]
} else {
v, q.r = q.r[0], q.r[1:]
}
return
}
func (q *Deque) PopBack() (v int) {
if len(q.r) > 0 {
q.r, v = q.r[:len(q.r)-1], q.r[len(q.r)-1]
} else {
v, q.l = q.l[0], q.l[1:]
}
return
}
func (q Deque) Front() int {
if len(q.l) > 0 {
return q.l[len(q.l)-1]
}
return q.r[0]
}
func (q Deque) Back() int {
if len(q.r) > 0 {
return q.r[len(q.r)-1]
}
return q.l[0]
}
func (q Deque) Get(i int) int {
if i < len(q.l) {
return q.l[len(q.l)-1-i]
}
return q.r[i-len(q.l)]
}
# Accepted solution for LeetCode #1696: Jump Game VI
class Solution:
def maxResult(self, nums: List[int], k: int) -> int:
n = len(nums)
f = [0] * n
q = deque([0])
for i in range(n):
if i - q[0] > k:
q.popleft()
f[i] = nums[i] + f[q[0]]
while q and f[q[-1]] <= f[i]:
q.pop()
q.append(i)
return f[-1]
// Accepted solution for LeetCode #1696: Jump Game VI
struct Solution;
use std::collections::VecDeque;
impl Solution {
fn max_result(nums: Vec<i32>, k: i32) -> i32 {
let n = nums.len();
let k = k as usize;
let mut queue: VecDeque<(usize, i32)> = VecDeque::new();
queue.push_back((0, nums[0]));
for i in 1..n {
while let Some((j, _)) = queue.front() {
if j + k < i {
queue.pop_front();
} else {
break;
}
}
let max = queue.front().as_ref().unwrap().1 + nums[i];
while let Some(&(_, sum)) = queue.back() {
if sum < max {
queue.pop_back();
} else {
break;
}
}
queue.push_back((i, max));
}
queue.back().as_ref().unwrap().1
}
}
#[test]
fn test() {
let nums = vec![1, -1, -2, 4, -7, 3];
let k = 2;
let res = 7;
assert_eq!(Solution::max_result(nums, k), res);
let nums = vec![10, -5, -2, 4, 0, 3];
let k = 3;
let res = 17;
assert_eq!(Solution::max_result(nums, k), res);
let nums = vec![1, -5, -20, 4, -1, 3, -6, -3];
let k = 2;
let res = 0;
assert_eq!(Solution::max_result(nums, k), res);
}
// Accepted solution for LeetCode #1696: Jump Game VI
function maxResult(nums: number[], k: number): number {
const n = nums.length;
const f: number[] = Array(n).fill(0);
const q = new Deque<number>();
q.pushBack(0);
for (let i = 0; i < n; ++i) {
if (i - q.frontValue()! > k) {
q.popFront();
}
f[i] = nums[i] + f[q.frontValue()!];
while (!q.isEmpty() && f[i] >= f[q.backValue()!]) {
q.popBack();
}
q.pushBack(i);
}
return f[n - 1];
}
class Node<T> {
value: T;
next: Node<T> | null;
prev: Node<T> | null;
constructor(value: T) {
this.value = value;
this.next = null;
this.prev = null;
}
}
class Deque<T> {
private front: Node<T> | null;
private back: Node<T> | null;
private size: number;
constructor() {
this.front = null;
this.back = null;
this.size = 0;
}
pushFront(val: T): void {
const newNode = new Node(val);
if (this.isEmpty()) {
this.front = newNode;
this.back = newNode;
} else {
newNode.next = this.front;
this.front!.prev = newNode;
this.front = newNode;
}
this.size++;
}
pushBack(val: T): void {
const newNode = new Node(val);
if (this.isEmpty()) {
this.front = newNode;
this.back = newNode;
} else {
newNode.prev = this.back;
this.back!.next = newNode;
this.back = newNode;
}
this.size++;
}
popFront(): T | undefined {
if (this.isEmpty()) {
return undefined;
}
const value = this.front!.value;
this.front = this.front!.next;
if (this.front !== null) {
this.front.prev = null;
} else {
this.back = null;
}
this.size--;
return value;
}
popBack(): T | undefined {
if (this.isEmpty()) {
return undefined;
}
const value = this.back!.value;
this.back = this.back!.prev;
if (this.back !== null) {
this.back.next = null;
} else {
this.front = null;
}
this.size--;
return value;
}
frontValue(): T | undefined {
return this.front?.value;
}
backValue(): T | undefined {
return this.back?.value;
}
getSize(): number {
return this.size;
}
isEmpty(): boolean {
return this.size === 0;
}
}
Use this to step through a reusable interview workflow for this problem.
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.
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.
Review these before coding to avoid predictable interview regressions.
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.
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.