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 an array of integers nums. You are also given an integer original which is the first number that needs to be searched for in nums.
You then do the following steps:
original is found in nums, multiply it by two (i.e., set original = 2 * original).Return the final value of original.
Example 1:
Input: nums = [5,3,6,1,12], original = 3 Output: 24 Explanation: - 3 is found in nums. 3 is multiplied by 2 to obtain 6. - 6 is found in nums. 6 is multiplied by 2 to obtain 12. - 12 is found in nums. 12 is multiplied by 2 to obtain 24. - 24 is not found in nums. Thus, 24 is returned.
Example 2:
Input: nums = [2,7,9], original = 4 Output: 4 Explanation: - 4 is not found in nums. Thus, 4 is returned.
Constraints:
1 <= nums.length <= 10001 <= nums[i], original <= 1000Problem summary: You are given an array of integers nums. You are also given an integer original which is the first number that needs to be searched for in nums. You then do the following steps: If original is found in nums, multiply it by two (i.e., set original = 2 * original). Otherwise, stop the process. Repeat this process with the new number as long as you keep finding the number. Return the final value of original.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Hash Map
[5,3,6,1,12] 3
[2,7,9] 4
largest-number-at-least-twice-of-others)check-if-n-and-its-double-exist)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2154: Keep Multiplying Found Values by Two
class Solution {
public int findFinalValue(int[] nums, int original) {
Set<Integer> s = new HashSet<>();
for (int num : nums) {
s.add(num);
}
while (s.contains(original)) {
original <<= 1;
}
return original;
}
}
// Accepted solution for LeetCode #2154: Keep Multiplying Found Values by Two
func findFinalValue(nums []int, original int) int {
s := map[int]bool{}
for _, x := range nums {
s[x] = true
}
for s[original] {
original <<= 1
}
return original
}
# Accepted solution for LeetCode #2154: Keep Multiplying Found Values by Two
class Solution:
def findFinalValue(self, nums: List[int], original: int) -> int:
s = set(nums)
while original in s:
original <<= 1
return original
// Accepted solution for LeetCode #2154: Keep Multiplying Found Values by Two
impl Solution {
pub fn find_final_value(nums: Vec<i32>, original: i32) -> i32 {
use std::collections::HashSet;
let s: HashSet<i32> = nums.into_iter().collect();
let mut original = original;
while s.contains(&original) {
original <<= 1;
}
original
}
}
// Accepted solution for LeetCode #2154: Keep Multiplying Found Values by Two
function findFinalValue(nums: number[], original: number): number {
const s: Set<number> = new Set([...nums]);
while (s.has(original)) {
original <<= 1;
}
return original;
}
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: Zero-count keys stay in map and break distinct/count constraints.
Usually fails on: Window/map size checks are consistently off by one.
Fix: Delete keys when count reaches zero.