Overflow in intermediate arithmetic
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Build confidence with an intuition-first walkthrough focused on math fundamentals.
Given two integers, num and t. A number x is achievable if it can become equal to num after applying the following operation at most t times:
x by 1, and simultaneously increase or decrease num by 1.Return the maximum possible value of x.
Example 1:
Input: num = 4, t = 1
Output: 6
Explanation:
Apply the following operation once to make the maximum achievable number equal to num:
num by 1.Example 2:
Input: num = 3, t = 2
Output: 7
Explanation:
Apply the following operation twice to make the maximum achievable number equal to num:
num by 1.Constraints:
1 <= num, t <= 50Problem summary: Given two integers, num and t. A number x is achievable if it can become equal to num after applying the following operation at most t times: Increase or decrease x by 1, and simultaneously increase or decrease num by 1. Return the maximum possible value of x.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math
4 1
3 2
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2769: Find the Maximum Achievable Number
class Solution {
public int theMaximumAchievableX(int num, int t) {
return num + t * 2;
}
}
// Accepted solution for LeetCode #2769: Find the Maximum Achievable Number
func theMaximumAchievableX(num int, t int) int {
return num + t*2
}
# Accepted solution for LeetCode #2769: Find the Maximum Achievable Number
class Solution:
def theMaximumAchievableX(self, num: int, t: int) -> int:
return num + t * 2
// Accepted solution for LeetCode #2769: Find the Maximum Achievable Number
// 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 #2769: Find the Maximum Achievable Number
// class Solution {
// public int theMaximumAchievableX(int num, int t) {
// return num + t * 2;
// }
// }
// Accepted solution for LeetCode #2769: Find the Maximum Achievable Number
function theMaximumAchievableX(num: number, t: number): number {
return num + t * 2;
}
Use this to step through a reusable interview workflow for this problem.
Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.
Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.
Review these before coding to avoid predictable interview regressions.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.