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.
Move from brute-force thinking to an efficient approach using math strategy.
Given two integers tomatoSlices and cheeseSlices. The ingredients of different burgers are as follows:
4 tomato slices and 1 cheese slice.2 Tomato slices and 1 cheese slice.Return [total_jumbo, total_small] so that the number of remaining tomatoSlices equal to 0 and the number of remaining cheeseSlices equal to 0. If it is not possible to make the remaining tomatoSlices and cheeseSlices equal to 0 return [].
Example 1:
Input: tomatoSlices = 16, cheeseSlices = 7 Output: [1,6] Explantion: To make one jumbo burger and 6 small burgers we need 4*1 + 2*6 = 16 tomato and 1 + 6 = 7 cheese. There will be no remaining ingredients.
Example 2:
Input: tomatoSlices = 17, cheeseSlices = 4 Output: [] Explantion: There will be no way to use all ingredients to make small and jumbo burgers.
Example 3:
Input: tomatoSlices = 4, cheeseSlices = 17 Output: [] Explantion: Making 1 jumbo burger there will be 16 cheese remaining and making 2 small burgers there will be 15 cheese remaining.
Constraints:
0 <= tomatoSlices, cheeseSlices <= 107Problem summary: Given two integers tomatoSlices and cheeseSlices. The ingredients of different burgers are as follows: Jumbo Burger: 4 tomato slices and 1 cheese slice. Small Burger: 2 Tomato slices and 1 cheese slice. Return [total_jumbo, total_small] so that the number of remaining tomatoSlices equal to 0 and the number of remaining cheeseSlices equal to 0. If it is not possible to make the remaining tomatoSlices and cheeseSlices equal to 0 return [].
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math
16 7
17 4
4 17
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1276: Number of Burgers with No Waste of Ingredients
class Solution {
public List<Integer> numOfBurgers(int tomatoSlices, int cheeseSlices) {
int k = 4 * cheeseSlices - tomatoSlices;
int y = k / 2;
int x = cheeseSlices - y;
return k % 2 != 0 || y < 0 || x < 0 ? List.of() : List.of(x, y);
}
}
// Accepted solution for LeetCode #1276: Number of Burgers with No Waste of Ingredients
func numOfBurgers(tomatoSlices int, cheeseSlices int) []int {
k := 4*cheeseSlices - tomatoSlices
y := k / 2
x := cheeseSlices - y
if k%2 != 0 || x < 0 || y < 0 {
return []int{}
}
return []int{x, y}
}
# Accepted solution for LeetCode #1276: Number of Burgers with No Waste of Ingredients
class Solution:
def numOfBurgers(self, tomatoSlices: int, cheeseSlices: int) -> List[int]:
k = 4 * cheeseSlices - tomatoSlices
y = k // 2
x = cheeseSlices - y
return [] if k % 2 or y < 0 or x < 0 else [x, y]
// Accepted solution for LeetCode #1276: Number of Burgers with No Waste of Ingredients
impl Solution {
pub fn num_of_burgers(tomato_slices: i32, cheese_slices: i32) -> Vec<i32> {
let k = 4 * cheese_slices - tomato_slices;
let y = k / 2;
let x = cheese_slices - y;
if k % 2 != 0 || y < 0 || x < 0 {
Vec::new()
} else {
vec![x, y]
}
}
}
// Accepted solution for LeetCode #1276: Number of Burgers with No Waste of Ingredients
function numOfBurgers(tomatoSlices: number, cheeseSlices: number): number[] {
const k = 4 * cheeseSlices - tomatoSlices;
const y = k >> 1;
const x = cheeseSlices - y;
return k % 2 || y < 0 || x < 0 ? [] : [x, y];
}
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.