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.
The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0 and 1. That is,
F(0) = 0, F(1) = 1 F(n) = F(n - 1) + F(n - 2), for n > 1.
Given n, calculate F(n).
Example 1:
Input: n = 2 Output: 1 Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.
Example 2:
Input: n = 3 Output: 2 Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.
Example 3:
Input: n = 4 Output: 3 Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.
Constraints:
0 <= n <= 30Problem summary: The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0 and 1. That is, F(0) = 0, F(1) = 1 F(n) = F(n - 1) + F(n - 2), for n > 1. Given n, calculate F(n).
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math · Dynamic Programming
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climbing-stairs)split-array-into-fibonacci-sequence)length-of-longest-fibonacci-subsequence)n-th-tribonacci-number)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #509: Fibonacci Number
class Solution {
public int fib(int n) {
int a = 0, b = 1;
while (n-- > 0) {
int c = a + b;
a = b;
b = c;
}
return a;
}
}
// Accepted solution for LeetCode #509: Fibonacci Number
func fib(n int) int {
a, b := 0, 1
for i := 0; i < n; i++ {
a, b = b, a+b
}
return a
}
# Accepted solution for LeetCode #509: Fibonacci Number
class Solution:
def fib(self, n: int) -> int:
a, b = 0, 1
for _ in range(n):
a, b = b, a + b
return a
// Accepted solution for LeetCode #509: Fibonacci Number
impl Solution {
pub fn fib(n: i32) -> i32 {
let mut a = 0;
let mut b = 1;
for _ in 0..n {
let t = b;
b = a + b;
a = t;
}
a
}
}
// Accepted solution for LeetCode #509: Fibonacci Number
function fib(n: number): number {
let [a, b] = [0, 1];
while (n--) {
[a, b] = [b, a + b];
}
return a;
}
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.