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 an integer array nums.
Two players, Alice and Bob, play a game in turns, with Alice playing first.
nums[l..r] such that r - l + 1 < m, where m is the current length of the array.Alice aims to maximize the final element, while Bob aims to minimize it. Assuming both play optimally, return the value of the final remaining element.
Example 1:
Input: nums = [1,5,2]
Output: 2
Explanation:
One valid optimal strategy:
[1], array becomes [5, 2].[5], array becomes [2]. Thus, the answer is 2.Example 2:
Input: nums = [3,7]
Output: 7
Explanation:
Alice removes [3], leaving the array [7]. Since Bob cannot play a turn now, the answer is 7.
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 105Problem summary: You are given an integer array nums. Two players, Alice and Bob, play a game in turns, with Alice playing first. In each turn, the current player chooses any subarray nums[l..r] such that r - l + 1 < m, where m is the current length of the array. The selected subarray is removed, and the remaining elements are concatenated to form the new array. The game continues until only one element remains. Alice aims to maximize the final element, while Bob aims to minimize it. Assuming both play optimally, return the value of the final remaining element.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Math
[1,5,2]
[3,7]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3828: Final Element After Subarray Deletions
class Solution {
public int finalElement(int[] nums) {
return Math.max(nums[0], nums[nums.length - 1]);
}
}
// Accepted solution for LeetCode #3828: Final Element After Subarray Deletions
func finalElement(nums []int) int {
return max(nums[0], nums[len(nums)-1])
}
# Accepted solution for LeetCode #3828: Final Element After Subarray Deletions
class Solution:
def finalElement(self, nums: List[int]) -> int:
return max(nums[0], nums[-1])
// Accepted solution for LeetCode #3828: Final Element After Subarray Deletions
// 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 #3828: Final Element After Subarray Deletions
// class Solution {
// public int finalElement(int[] nums) {
// return Math.max(nums[0], nums[nums.length - 1]);
// }
// }
// Accepted solution for LeetCode #3828: Final Element After Subarray Deletions
function finalElement(nums: number[]): number {
return Math.max(nums.at(0)!, nums.at(-1)!);
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
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: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.