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.
You have two soups, A and B, each starting with n mL. On every turn, one of the following four serving operations is chosen at random, each with probability 0.25 independent of all previous turns:
Note:
The process stops immediately after any turn in which one of the soups is used up.
Return the probability that A is used up before B, plus half the probability that both soups are used up in the same turn. Answers within 10-5 of the actual answer will be accepted.
Example 1:
Input: n = 50 Output: 0.62500 Explanation: If we perform either of the first two serving operations, soup A will become empty first. If we perform the third operation, A and B will become empty at the same time. If we perform the fourth operation, B will become empty first. So the total probability of A becoming empty first plus half the probability that A and B become empty at the same time, is 0.25 * (1 + 1 + 0.5 + 0) = 0.625.
Example 2:
Input: n = 100 Output: 0.71875 Explanation: If we perform the first serving operation, soup A will become empty first. If we perform the second serving operations, A will become empty on performing operation [1, 2, 3], and both A and B become empty on performing operation 4. If we perform the third operation, A will become empty on performing operation [1, 2], and both A and B become empty on performing operation 3. If we perform the fourth operation, A will become empty on performing operation 1, and both A and B become empty on performing operation 2. So the total probability of A becoming empty first plus half the probability that A and B become empty at the same time, is 0.71875.
Constraints:
0 <= n <= 109Problem summary: You have two soups, A and B, each starting with n mL. On every turn, one of the following four serving operations is chosen at random, each with probability 0.25 independent of all previous turns: pour 100 mL from type A and 0 mL from type B pour 75 mL from type A and 25 mL from type B pour 50 mL from type A and 50 mL from type B pour 25 mL from type A and 75 mL from type B Note: There is no operation that pours 0 mL from A and 100 mL from B. The amounts from A and B are poured simultaneously during the turn. If an operation asks you to pour more than you have left of a soup, pour all that remains of that soup. The process stops immediately after any turn in which one of the soups is used up. Return the probability that A is used up before B, plus half the probability that both soups are used up in the same turn. Answers within 10-5 of the actual answer will be accepted.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math · Dynamic Programming
50
100
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #808: Soup Servings
class Solution {
private double[][] f = new double[200][200];
public double soupServings(int n) {
return n > 4800 ? 1 : dfs((n + 24) / 25, (n + 24) / 25);
}
private double dfs(int i, int j) {
if (i <= 0 && j <= 0) {
return 0.5;
}
if (i <= 0) {
return 1.0;
}
if (j <= 0) {
return 0;
}
if (f[i][j] > 0) {
return f[i][j];
}
double ans
= 0.25 * (dfs(i - 4, j) + dfs(i - 3, j - 1) + dfs(i - 2, j - 2) + dfs(i - 1, j - 3));
f[i][j] = ans;
return ans;
}
}
// Accepted solution for LeetCode #808: Soup Servings
func soupServings(n int) float64 {
if n > 4800 {
return 1
}
f := [200][200]float64{}
var dfs func(i, j int) float64
dfs = func(i, j int) float64 {
if i <= 0 && j <= 0 {
return 0.5
}
if i <= 0 {
return 1.0
}
if j <= 0 {
return 0
}
if f[i][j] > 0 {
return f[i][j]
}
ans := 0.25 * (dfs(i-4, j) + dfs(i-3, j-1) + dfs(i-2, j-2) + dfs(i-1, j-3))
f[i][j] = ans
return ans
}
return dfs((n+24)/25, (n+24)/25)
}
# Accepted solution for LeetCode #808: Soup Servings
class Solution:
def soupServings(self, n: int) -> float:
@cache
def dfs(i: int, j: int) -> float:
if i <= 0 and j <= 0:
return 0.5
if i <= 0:
return 1
if j <= 0:
return 0
return 0.25 * (
dfs(i - 4, j)
+ dfs(i - 3, j - 1)
+ dfs(i - 2, j - 2)
+ dfs(i - 1, j - 3)
)
return 1 if n > 4800 else dfs((n + 24) // 25, (n + 24) // 25)
// Accepted solution for LeetCode #808: Soup Servings
impl Solution {
pub fn soup_servings(n: i32) -> f64 {
if n > 4800 {
return 1.0;
}
Self::dfs((n + 24) / 25, (n + 24) / 25)
}
fn dfs(i: i32, j: i32) -> f64 {
static mut F: [[f64; 200]; 200] = [[0.0; 200]; 200];
unsafe {
if i <= 0 && j <= 0 {
return 0.5;
}
if i <= 0 {
return 1.0;
}
if j <= 0 {
return 0.0;
}
if F[i as usize][j as usize] > 0.0 {
return F[i as usize][j as usize];
}
let ans = 0.25
* (Self::dfs(i - 4, j)
+ Self::dfs(i - 3, j - 1)
+ Self::dfs(i - 2, j - 2)
+ Self::dfs(i - 1, j - 3));
F[i as usize][j as usize] = ans;
ans
}
}
}
// Accepted solution for LeetCode #808: Soup Servings
function soupServings(n: number): number {
const f = Array.from({ length: 200 }, () => Array(200).fill(-1));
const dfs = (i: number, j: number): number => {
if (i <= 0 && j <= 0) {
return 0.5;
}
if (i <= 0) {
return 1;
}
if (j <= 0) {
return 0;
}
if (f[i][j] !== -1) {
return f[i][j];
}
f[i][j] =
0.25 * (dfs(i - 4, j) + dfs(i - 3, j - 1) + dfs(i - 2, j - 2) + dfs(i - 1, j - 3));
return f[i][j];
};
return n >= 4800 ? 1 : dfs(Math.ceil(n / 25), Math.ceil(n / 25));
}
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.