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 a positive integer x.
You are initially at position 0 in the array and you can visit other positions according to the following rules:
i, then you can move to any position j such that i < j.i that you visit, you get a score of nums[i].i to a position j and the parities of nums[i] and nums[j] differ, then you lose a score of x.Return the maximum total score you can get.
Note that initially you have nums[0] points.
Example 1:
Input: nums = [2,3,6,1,9,2], x = 5 Output: 13 Explanation: We can visit the following positions in the array: 0 -> 2 -> 3 -> 4. The corresponding values are 2, 6, 1 and 9. Since the integers 6 and 1 have different parities, the move 2 -> 3 will make you lose a score of x = 5. The total score will be: 2 + 6 + 1 + 9 - 5 = 13.
Example 2:
Input: nums = [2,4,6,8], x = 3 Output: 20 Explanation: All the integers in the array have the same parities, so we can visit all of them without losing any score. The total score is: 2 + 4 + 6 + 8 = 20.
Constraints:
2 <= nums.length <= 1051 <= nums[i], x <= 106Problem summary: You are given a 0-indexed integer array nums and a positive integer x. You are initially at position 0 in the array and you can visit other positions according to the following rules: If you are currently in position i, then you can move to any position j such that i < j. For each position i that you visit, you get a score of nums[i]. If you move from a position i to a position j and the parities of nums[i] and nums[j] differ, then you lose a score of x. Return the maximum total score you can get. Note that initially you have nums[0] points.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Dynamic Programming
[2,3,6,1,9,2] 5
[2,4,6,8] 3
jump-game-ii)stone-game)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2786: Visit Array Positions to Maximize Score
class Solution {
public long maxScore(int[] nums, int x) {
long[] f = new long[2];
Arrays.fill(f, -(1L << 60));
f[nums[0] & 1] = nums[0];
for (int i = 1; i < nums.length; ++i) {
int v = nums[i];
f[v & 1] = Math.max(f[v & 1], f[v & 1 ^ 1] - x) + v;
}
return Math.max(f[0], f[1]);
}
}
// Accepted solution for LeetCode #2786: Visit Array Positions to Maximize Score
func maxScore(nums []int, x int) int64 {
const inf int = 1 << 40
f := [2]int{-inf, -inf}
f[nums[0]&1] = nums[0]
for _, v := range nums[1:] {
f[v&1] = max(f[v&1], f[v&1^1]-x) + v
}
return int64(max(f[0], f[1]))
}
# Accepted solution for LeetCode #2786: Visit Array Positions to Maximize Score
class Solution:
def maxScore(self, nums: List[int], x: int) -> int:
f = [-inf] * 2
f[nums[0] & 1] = nums[0]
for v in nums[1:]:
f[v & 1] = max(f[v & 1], f[v & 1 ^ 1] - x) + v
return max(f)
// Accepted solution for LeetCode #2786: Visit Array Positions to Maximize Score
// 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 #2786: Visit Array Positions to Maximize Score
// class Solution {
// public long maxScore(int[] nums, int x) {
// long[] f = new long[2];
// Arrays.fill(f, -(1L << 60));
// f[nums[0] & 1] = nums[0];
// for (int i = 1; i < nums.length; ++i) {
// int v = nums[i];
// f[v & 1] = Math.max(f[v & 1], f[v & 1 ^ 1] - x) + v;
// }
// return Math.max(f[0], f[1]);
// }
// }
// Accepted solution for LeetCode #2786: Visit Array Positions to Maximize Score
function maxScore(nums: number[], x: number): number {
const f: number[] = Array(2).fill(-Infinity);
f[nums[0] & 1] = nums[0];
for (let i = 1; i < nums.length; ++i) {
const v = nums[i];
f[v & 1] = Math.max(f[v & 1], f[(v & 1) ^ 1] - x) + v;
}
return Math.max(...f);
}
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.