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.
Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
There are buckets buckets of liquid, where exactly one of the buckets is poisonous. To figure out which one is poisonous, you feed some number of (poor) pigs the liquid to see whether they will die or not. Unfortunately, you only have minutesToTest minutes to determine which bucket is poisonous.
You can feed the pigs according to these steps:
minutesToDie minutes. You may not feed any other pigs during this time.minutesToDie minutes have passed, any pigs that have been fed the poisonous bucket will die, and all others will survive.Given buckets, minutesToDie, and minutesToTest, return the minimum number of pigs needed to figure out which bucket is poisonous within the allotted time.
Example 1:
Input: buckets = 4, minutesToDie = 15, minutesToTest = 15 Output: 2 Explanation: We can determine the poisonous bucket as follows: At time 0, feed the first pig buckets 1 and 2, and feed the second pig buckets 2 and 3. At time 15, there are 4 possible outcomes: - If only the first pig dies, then bucket 1 must be poisonous. - If only the second pig dies, then bucket 3 must be poisonous. - If both pigs die, then bucket 2 must be poisonous. - If neither pig dies, then bucket 4 must be poisonous.
Example 2:
Input: buckets = 4, minutesToDie = 15, minutesToTest = 30 Output: 2 Explanation: We can determine the poisonous bucket as follows: At time 0, feed the first pig bucket 1, and feed the second pig bucket 2. At time 15, there are 2 possible outcomes: - If either pig dies, then the poisonous bucket is the one it was fed. - If neither pig dies, then feed the first pig bucket 3, and feed the second pig bucket 4. At time 30, one of the two pigs must die, and the poisonous bucket is the one it was fed.
Constraints:
1 <= buckets <= 10001 <= minutesToDie <= minutesToTest <= 100Problem summary: There are buckets buckets of liquid, where exactly one of the buckets is poisonous. To figure out which one is poisonous, you feed some number of (poor) pigs the liquid to see whether they will die or not. Unfortunately, you only have minutesToTest minutes to determine which bucket is poisonous. You can feed the pigs according to these steps: Choose some live pigs to feed. For each pig, choose which buckets to feed it. The pig will consume all the chosen buckets simultaneously and will take no time. Each pig can feed from any number of buckets, and each bucket can be fed from by any number of pigs. Wait for minutesToDie minutes. You may not feed any other pigs during this time. After minutesToDie minutes have passed, any pigs that have been fed the poisonous bucket will die, and all others will survive. Repeat this process until you run out of time. Given buckets, minutesToDie, and
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math · Dynamic Programming
4 15 15
4 15 30
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #458: Poor Pigs
class Solution {
public int poorPigs(int buckets, int minutesToDie, int minutesToTest) {
int base = minutesToTest / minutesToDie + 1;
int res = 0;
for (int p = 1; p < buckets; p *= base) {
++res;
}
return res;
}
}
// Accepted solution for LeetCode #458: Poor Pigs
func poorPigs(buckets int, minutesToDie int, minutesToTest int) int {
base := minutesToTest/minutesToDie + 1
res := 0
for p := 1; p < buckets; p *= base {
res++
}
return res
}
# Accepted solution for LeetCode #458: Poor Pigs
class Solution:
def poorPigs(self, buckets: int, minutesToDie: int, minutesToTest: int) -> int:
base = minutesToTest // minutesToDie + 1
res, p = 0, 1
while p < buckets:
p *= base
res += 1
return res
// Accepted solution for LeetCode #458: Poor Pigs
struct Solution;
impl Solution {
fn poor_pigs(buckets: i32, minutes_to_die: i32, minutes_to_test: i32) -> i32 {
let mut pigs = 0;
let t = minutes_to_test / minutes_to_die + 1;
while t.pow(pigs) < buckets {
pigs += 1;
}
pigs as i32
}
}
#[test]
fn test() {
let buckets = 1000;
let minutes_to_test = 60;
let minutes_to_die = 15;
let res = 5;
assert_eq!(
Solution::poor_pigs(buckets, minutes_to_die, minutes_to_test),
res
);
}
// Accepted solution for LeetCode #458: Poor Pigs
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #458: Poor Pigs
// class Solution {
// public int poorPigs(int buckets, int minutesToDie, int minutesToTest) {
// int base = minutesToTest / minutesToDie + 1;
// int res = 0;
// for (int p = 1; p < buckets; p *= base) {
// ++res;
// }
// return res;
// }
// }
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
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: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.