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.
You are given a positive integer num consisting only of digits 6 and 9.
Return the maximum number you can get by changing at most one digit (6 becomes 9, and 9 becomes 6).
Example 1:
Input: num = 9669 Output: 9969 Explanation: Changing the first digit results in 6669. Changing the second digit results in 9969. Changing the third digit results in 9699. Changing the fourth digit results in 9666. The maximum number is 9969.
Example 2:
Input: num = 9996 Output: 9999 Explanation: Changing the last digit 6 to 9 results in the maximum number.
Example 3:
Input: num = 9999 Output: 9999 Explanation: It is better not to apply any change.
Constraints:
1 <= num <= 104num consists of only 6 and 9 digits.Problem summary: You are given a positive integer num consisting only of digits 6 and 9. Return the maximum number you can get by changing at most one digit (6 becomes 9, and 9 becomes 6).
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math · Greedy
9669
9996
9999
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1323: Maximum 69 Number
class Solution {
public int maximum69Number(int num) {
return Integer.valueOf(String.valueOf(num).replaceFirst("6", "9"));
}
}
// Accepted solution for LeetCode #1323: Maximum 69 Number
func maximum69Number(num int) int {
s := strings.Replace(strconv.Itoa(num), "6", "9", 1)
ans, _ := strconv.Atoi(s)
return ans
}
# Accepted solution for LeetCode #1323: Maximum 69 Number
class Solution:
def maximum69Number(self, num: int) -> int:
return int(str(num).replace("6", "9", 1))
// Accepted solution for LeetCode #1323: Maximum 69 Number
impl Solution {
pub fn maximum69_number(num: i32) -> i32 {
num.to_string().replacen('6', "9", 1).parse().unwrap()
}
}
// Accepted solution for LeetCode #1323: Maximum 69 Number
function maximum69Number(num: number): number {
return Number((num + '').replace('6', '9'));
}
Use this to step through a reusable interview workflow for this problem.
Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.
Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.
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.
Wrong move: Locally optimal choices may fail globally.
Usually fails on: Counterexamples appear on crafted input orderings.
Fix: Verify with exchange argument or monotonic objective before committing.