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.
Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
Given an integer array nums and an integer k, return the maximum sum of a non-empty subsequence of that array such that for every two consecutive integers in the subsequence, nums[i] and nums[j], where i < j, the condition j - i <= k is satisfied.
A subsequence of an array is obtained by deleting some number of elements (can be zero) from the array, leaving the remaining elements in their original order.
Example 1:
Input: nums = [10,2,-10,5,20], k = 2 Output: 37 Explanation: The subsequence is [10, 2, 5, 20].
Example 2:
Input: nums = [-1,-2,-3], k = 1 Output: -1 Explanation: The subsequence must be non-empty, so we choose the largest number.
Example 3:
Input: nums = [10,-2,-10,-5,20], k = 2 Output: 23 Explanation: The subsequence is [10, -2, -5, 20].
Constraints:
1 <= k <= nums.length <= 105-104 <= nums[i] <= 104Problem summary: Given an integer array nums and an integer k, return the maximum sum of a non-empty subsequence of that array such that for every two consecutive integers in the subsequence, nums[i] and nums[j], where i < j, the condition j - i <= k is satisfied. A subsequence of an array is obtained by deleting some number of elements (can be zero) from the array, leaving the remaining elements in their original order.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Dynamic Programming · Sliding Window · Monotonic Queue
[10,2,-10,5,20] 2
[-1,-2,-3] 1
[10,-2,-10,-5,20] 2
maximum-element-sum-of-a-complete-subset-of-indices)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1425: Constrained Subsequence Sum
class Solution {
public int constrainedSubsetSum(int[] nums, int k) {
Deque<Integer> q = new ArrayDeque<>();
q.offer(0);
int n = nums.length;
int[] f = new int[n];
int ans = -(1 << 30);
for (int i = 0; i < n; ++i) {
while (i - q.peekFirst() > k) {
q.pollFirst();
}
f[i] = Math.max(0, f[q.peekFirst()]) + nums[i];
ans = Math.max(ans, f[i]);
while (!q.isEmpty() && f[q.peekLast()] <= f[i]) {
q.pollLast();
}
q.offerLast(i);
}
return ans;
}
}
// Accepted solution for LeetCode #1425: Constrained Subsequence Sum
func constrainedSubsetSum(nums []int, k int) int {
q := Deque{}
q.PushFront(0)
n := len(nums)
f := make([]int, n)
ans := nums[0]
for i, x := range nums {
for i-q.Front() > k {
q.PopFront()
}
f[i] = max(0, f[q.Front()]) + x
ans = max(ans, f[i])
for !q.Empty() && f[q.Back()] <= f[i] {
q.PopBack()
}
q.PushBack(i)
}
return ans
}
// template
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 #1425: Constrained Subsequence Sum
class Solution:
def constrainedSubsetSum(self, nums: List[int], k: int) -> int:
q = deque([0])
n = len(nums)
f = [0] * n
ans = -inf
for i, x in enumerate(nums):
while i - q[0] > k:
q.popleft()
f[i] = max(0, f[q[0]]) + x
ans = max(ans, f[i])
while q and f[q[-1]] <= f[i]:
q.pop()
q.append(i)
return ans
// Accepted solution for LeetCode #1425: Constrained Subsequence Sum
/**
* [1425] Constrained Subsequence Sum
*
* Given an integer array nums and an integer k, return the maximum sum of a non-empty subsequence of that array such that for every two consecutive integers in the subsequence, nums[i] and nums[j], where i < j, the condition j - i <= k is satisfied.
* A subsequence of an array is obtained by deleting some number of elements (can be zero) from the array, leaving the remaining elements in their original order.
*
* Example 1:
*
* Input: nums = [10,2,-10,5,20], k = 2
* Output: 37
* Explanation: The subsequence is [10, 2, 5, 20].
*
* Example 2:
*
* Input: nums = [-1,-2,-3], k = 1
* Output: -1
* Explanation: The subsequence must be non-empty, so we choose the largest number.
*
* Example 3:
*
* Input: nums = [10,-2,-10,-5,20], k = 2
* Output: 23
* Explanation: The subsequence is [10, -2, -5, 20].
*
*
* Constraints:
*
* 1 <= k <= nums.length <= 10^5
* -10^4 <= nums[i] <= 10^4
*
*/
pub struct Solution {}
// problem: https://leetcode.com/problems/constrained-subsequence-sum/
// discuss: https://leetcode.com/problems/constrained-subsequence-sum/discuss/?currentPage=1&orderBy=most_votes&query=
// submission codes start here
impl Solution {
// Credit: https://leetcode.com/problems/constrained-subsequence-sum/solutions/3111911/just-a-runnable-solution/
pub fn constrained_subset_sum(nums: Vec<i32>, k: i32) -> i32 {
let mut dp = vec![0; nums.len()];
let mut result = nums[0];
let mut q = std::collections::VecDeque::new();
for i in 0..nums.len() {
dp[i] = nums[i];
if let Some(&j) = q.front() {
dp[i] = std::cmp::max(dp[i], dp[j] + nums[i]);
}
result = std::cmp::max(result, dp[i]);
while let Some(&j) = q.back() {
if dp[j] < dp[i] {
q.pop_back();
} else {
break;
}
}
q.push_back(i);
if let Some(&j) = q.front() {
if i as i32 - j as i32 >= k {
q.pop_front();
}
}
}
result
}
}
// submission codes end
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn test_1425_example_1() {
let nums = vec![10, 2, -10, 5, 20];
let k = 2;
let result = 37;
assert_eq!(Solution::constrained_subset_sum(nums, k), result);
}
#[test]
fn test_1425_example_2() {
let nums = vec![-1, -2, -3];
let k = 1;
let result = -1;
assert_eq!(Solution::constrained_subset_sum(nums, k), result);
}
#[test]
fn test_1425_example_3() {
let nums = vec![10, -2, -10, -5, 20];
let k = 2;
let result = 23;
assert_eq!(Solution::constrained_subset_sum(nums, k), result);
}
}
// Accepted solution for LeetCode #1425: Constrained Subsequence Sum
function constrainedSubsetSum(nums: number[], k: number): number {
const q = new Deque<number>();
const n = nums.length;
q.pushBack(0);
let ans = nums[0];
const f: number[] = Array(n).fill(0);
for (let i = 0; i < n; ++i) {
while (i - q.frontValue()! > k) {
q.popFront();
}
f[i] = Math.max(0, f[q.frontValue()!]!) + nums[i];
ans = Math.max(ans, f[i]);
while (!q.isEmpty() && f[q.backValue()!]! <= f[i]) {
q.popBack();
}
q.pushBack(i);
}
return ans;
}
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.
Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.
Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.
Fix: Shrink in a `while` loop until the invariant is valid again.