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.
Build confidence with an intuition-first walkthrough focused on array fundamentals.
You are given a 0-indexed integer array nums whose length is a power of 2.
Apply the following algorithm on nums:
n be the length of nums. If n == 1, end the process. Otherwise, create a new 0-indexed integer array newNums of length n / 2.i where 0 <= i < n / 2, assign the value of newNums[i] as min(nums[2 * i], nums[2 * i + 1]).i where 0 <= i < n / 2, assign the value of newNums[i] as max(nums[2 * i], nums[2 * i + 1]).nums with newNums.Return the last number that remains in nums after applying the algorithm.
Example 1:
Input: nums = [1,3,5,2,4,8,2,2] Output: 1 Explanation: The following arrays are the results of applying the algorithm repeatedly. First: nums = [1,5,4,2] Second: nums = [1,4] Third: nums = [1] 1 is the last remaining number, so we return 1.
Example 2:
Input: nums = [3] Output: 3 Explanation: 3 is already the last remaining number, so we return 3.
Constraints:
1 <= nums.length <= 10241 <= nums[i] <= 109nums.length is a power of 2.Problem summary: You are given a 0-indexed integer array nums whose length is a power of 2. Apply the following algorithm on nums: Let n be the length of nums. If n == 1, end the process. Otherwise, create a new 0-indexed integer array newNums of length n / 2. For every even index i where 0 <= i < n / 2, assign the value of newNums[i] as min(nums[2 * i], nums[2 * i + 1]). For every odd index i where 0 <= i < n / 2, assign the value of newNums[i] as max(nums[2 * i], nums[2 * i + 1]). Replace the array nums with newNums. Repeat the entire process starting from step 1. Return the last number that remains in nums after applying the algorithm.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array
[1,3,5,2,4,8,2,2]
[3]
elimination-game)find-triangular-sum-of-an-array)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2293: Min Max Game
class Solution {
public int minMaxGame(int[] nums) {
for (int n = nums.length; n > 1;) {
n >>= 1;
for (int i = 0; i < n; ++i) {
int a = nums[i << 1], b = nums[i << 1 | 1];
nums[i] = i % 2 == 0 ? Math.min(a, b) : Math.max(a, b);
}
}
return nums[0];
}
}
// Accepted solution for LeetCode #2293: Min Max Game
func minMaxGame(nums []int) int {
for n := len(nums); n > 1; {
n >>= 1
for i := 0; i < n; i++ {
a, b := nums[i<<1], nums[i<<1|1]
if i%2 == 0 {
nums[i] = min(a, b)
} else {
nums[i] = max(a, b)
}
}
}
return nums[0]
}
# Accepted solution for LeetCode #2293: Min Max Game
class Solution:
def minMaxGame(self, nums: List[int]) -> int:
n = len(nums)
while n > 1:
n >>= 1
for i in range(n):
a, b = nums[i << 1], nums[i << 1 | 1]
nums[i] = min(a, b) if i % 2 == 0 else max(a, b)
return nums[0]
// Accepted solution for LeetCode #2293: Min Max Game
impl Solution {
pub fn min_max_game(mut nums: Vec<i32>) -> i32 {
let mut n = nums.len();
while n != 1 {
n >>= 1;
for i in 0..n {
nums[i] = (if (i & 1) == 1 { i32::max } else { i32::min })(
nums[i << 1],
nums[(i << 1) | 1],
);
}
}
nums[0]
}
}
// Accepted solution for LeetCode #2293: Min Max Game
function minMaxGame(nums: number[]): number {
for (let n = nums.length; n > 1; ) {
n >>= 1;
for (let i = 0; i < n; ++i) {
const a = nums[i << 1];
const b = nums[(i << 1) | 1];
nums[i] = i % 2 == 0 ? Math.min(a, b) : Math.max(a, b);
}
}
return nums[0];
}
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.