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.
We are playing the Guessing Game. The game will work as follows:
1 and n.x, you will pay x dollars. If you run out of money, you lose the game.Given a particular n, return the minimum amount of money you need to guarantee a win regardless of what number I pick.
Example 1:
Input: n = 10 Output: 16 Explanation: The winning strategy is as follows: - The range is [1,10]. Guess 7. - If this is my number, your total is $0. Otherwise, you pay $7. - If my number is higher, the range is [8,10]. Guess 9. - If this is my number, your total is $7. Otherwise, you pay $9. - If my number is higher, it must be 10. Guess 10. Your total is $7 + $9 = $16. - If my number is lower, it must be 8. Guess 8. Your total is $7 + $9 = $16. - If my number is lower, the range is [1,6]. Guess 3. - If this is my number, your total is $7. Otherwise, you pay $3. - If my number is higher, the range is [4,6]. Guess 5. - If this is my number, your total is $7 + $3 = $10. Otherwise, you pay $5. - If my number is higher, it must be 6. Guess 6. Your total is $7 + $3 + $5 = $15. - If my number is lower, it must be 4. Guess 4. Your total is $7 + $3 + $5 = $15. - If my number is lower, the range is [1,2]. Guess 1. - If this is my number, your total is $7 + $3 = $10. Otherwise, you pay $1. - If my number is higher, it must be 2. Guess 2. Your total is $7 + $3 + $1 = $11. The worst case in all these scenarios is that you pay $16. Hence, you only need $16 to guarantee a win.
Example 2:
Input: n = 1 Output: 0 Explanation: There is only one possible number, so you can guess 1 and not have to pay anything.
Example 3:
Input: n = 2 Output: 1 Explanation: There are two possible numbers, 1 and 2. - Guess 1. - If this is my number, your total is $0. Otherwise, you pay $1. - If my number is higher, it must be 2. Guess 2. Your total is $1. The worst case is that you pay $1.
Constraints:
1 <= n <= 200Problem summary: We are playing the Guessing Game. The game will work as follows: I pick a number between 1 and n. You guess a number. If you guess the right number, you win the game. If you guess the wrong number, then I will tell you whether the number I picked is higher or lower, and you will continue guessing. Every time you guess a wrong number x, you will pay x dollars. If you run out of money, you lose the game. Given a particular n, return the minimum amount of money you need to guarantee a win regardless of what number I pick.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math · Dynamic Programming
10
1
2
flip-game-ii)guess-number-higher-or-lower)can-i-win)find-k-closest-elements)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #375: Guess Number Higher or Lower II
class Solution {
public int getMoneyAmount(int n) {
int[][] f = new int[n + 1][n + 1];
for (int i = n - 1; i > 0; --i) {
for (int j = i + 1; j <= n; ++j) {
f[i][j] = j + f[i][j - 1];
for (int k = i; k < j; ++k) {
f[i][j] = Math.min(f[i][j], Math.max(f[i][k - 1], f[k + 1][j]) + k);
}
}
}
return f[1][n];
}
}
// Accepted solution for LeetCode #375: Guess Number Higher or Lower II
func getMoneyAmount(n int) int {
f := make([][]int, n+1)
for i := range f {
f[i] = make([]int, n+1)
}
for i := n - 1; i > 0; i-- {
for j := i + 1; j <= n; j++ {
f[i][j] = j + f[i][j-1]
for k := i; k < j; k++ {
f[i][j] = min(f[i][j], k+max(f[i][k-1], f[k+1][j]))
}
}
}
return f[1][n]
}
# Accepted solution for LeetCode #375: Guess Number Higher or Lower II
class Solution:
def getMoneyAmount(self, n: int) -> int:
f = [[0] * (n + 1) for _ in range(n + 1)]
for i in range(n - 1, 0, -1):
for j in range(i + 1, n + 1):
f[i][j] = j + f[i][j - 1]
for k in range(i, j):
f[i][j] = min(f[i][j], max(f[i][k - 1], f[k + 1][j]) + k)
return f[1][n]
// Accepted solution for LeetCode #375: Guess Number Higher or Lower II
struct Solution;
use std::collections::HashMap;
impl Solution {
fn get_money_amount(n: i32) -> i32 {
let mut memo: HashMap<(i32, i32), i32> = HashMap::new();
Self::dp(1, n, &mut memo)
}
fn dp(left: i32, right: i32, memo: &mut HashMap<(i32, i32), i32>) -> i32 {
if left == right {
0
} else {
if let Some(&res) = memo.get(&(left, right)) {
return res;
}
let mut res = std::i32::MAX;
for i in left..right {
let a = if i != left {
Self::dp(left, i - 1, memo)
} else {
0
};
let b = if i != right {
Self::dp(i + 1, right, memo)
} else {
0
};
res = res.min(i + a.max(b));
}
memo.insert((left, right), res);
res
}
}
}
#[test]
fn test() {
let n = 10;
let res = 16;
assert_eq!(Solution::get_money_amount(n), res);
}
// Accepted solution for LeetCode #375: Guess Number Higher or Lower II
function getMoneyAmount(n: number): number {
const f: number[][] = Array.from({ length: n + 1 }, () => Array(n + 1).fill(0));
for (let i = n - 1; i; --i) {
for (let j = i + 1; j <= n; ++j) {
f[i][j] = j + f[i][j - 1];
for (let k = i; k < j; ++k) {
f[i][j] = Math.min(f[i][j], k + Math.max(f[i][k - 1], f[k + 1][j]));
}
}
}
return f[1][n];
}
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.