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 three positive integers num1, num2, and num3.
The key of num1, num2, and num3 is defined as a four-digit number such that:
ith digit (1 <= i <= 4) of the key is generated by taking the smallest digit among the ith digits of num1, num2, and num3.Return the key of the three numbers without leading zeros (if any).
Example 1:
Input: num1 = 1, num2 = 10, num3 = 1000
Output: 0
Explanation:
On padding, num1 becomes "0001", num2 becomes "0010", and num3 remains "1000".
1st digit of the key is min(0, 0, 1).2nd digit of the key is min(0, 0, 0).3rd digit of the key is min(0, 1, 0).4th digit of the key is min(1, 0, 0).Hence, the key is "0000", i.e. 0.
Example 2:
Input: num1 = 987, num2 = 879, num3 = 798
Output: 777
Example 3:
Input: num1 = 1, num2 = 2, num3 = 3
Output: 1
Constraints:
1 <= num1, num2, num3 <= 9999Problem summary: You are given three positive integers num1, num2, and num3. The key of num1, num2, and num3 is defined as a four-digit number such that: Initially, if any number has less than four digits, it is padded with leading zeros. The ith digit (1 <= i <= 4) of the key is generated by taking the smallest digit among the ith digits of num1, num2, and num3. Return the key of the three numbers without leading zeros (if any).
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math
1 10 1000
987 879 798
1 2 3
largest-number)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3270: Find the Key of the Numbers
class Solution {
public int generateKey(int num1, int num2, int num3) {
int ans = 0, k = 1;
for (int i = 0; i < 4; ++i) {
int x = Math.min(Math.min(num1 / k % 10, num2 / k % 10), num3 / k % 10);
ans += x * k;
k *= 10;
}
return ans;
}
}
// Accepted solution for LeetCode #3270: Find the Key of the Numbers
func generateKey(num1 int, num2 int, num3 int) (ans int) {
k := 1
for i := 0; i < 4; i++ {
x := min(min(num1/k%10, num2/k%10), num3/k%10)
ans += x * k
k *= 10
}
return
}
# Accepted solution for LeetCode #3270: Find the Key of the Numbers
class Solution:
def generateKey(self, num1: int, num2: int, num3: int) -> int:
ans, k = 0, 1
for _ in range(4):
x = min(num1 // k % 10, num2 // k % 10, num3 // k % 10)
ans += x * k
k *= 10
return ans
// Accepted solution for LeetCode #3270: Find the Key of the Numbers
/**
* [3270] Find the Key of the Numbers
*/
pub struct Solution {}
// submission codes start here
impl Solution {
pub fn generate_key(num1: i32, num2: i32, num3: i32) -> i32 {
let mut array = [num1, num2, num3];
let mut result = 0;
for i in 0..4 {
result += array.iter().map(|x| *x % 10).min().unwrap() * 10_i32.pow(i);
for i in 0..3 {
array[i] /= 10;
}
}
result
}
}
// submission codes end
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn test_3270() {
assert_eq!(0, Solution::generate_key(1, 10, 1000));
assert_eq!(777, Solution::generate_key(987, 879, 798));
assert_eq!(1, Solution::generate_key(1, 2, 3));
}
}
// Accepted solution for LeetCode #3270: Find the Key of the Numbers
function generateKey(num1: number, num2: number, num3: number): number {
let [ans, k] = [0, 1];
for (let i = 0; i < 4; ++i) {
const x = Math.min(((num1 / k) | 0) % 10, ((num2 / k) | 0) % 10, ((num3 / k) | 0) % 10);
ans += x * k;
k *= 10;
}
return ans;
}
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.