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 an integer k, return the minimum number of Fibonacci numbers whose sum is equal to k. The same Fibonacci number can be used multiple times.
The Fibonacci numbers are defined as:
F1 = 1F2 = 1Fn = Fn-1 + Fn-2 for n > 2.k.
Example 1:
Input: k = 7 Output: 2 Explanation: The Fibonacci numbers are: 1, 1, 2, 3, 5, 8, 13, ... For k = 7 we can use 2 + 5 = 7.
Example 2:
Input: k = 10 Output: 2 Explanation: For k = 10 we can use 2 + 8 = 10.
Example 3:
Input: k = 19 Output: 3 Explanation: For k = 19 we can use 1 + 5 + 13 = 19.
Constraints:
1 <= k <= 109Problem summary: Given an integer k, return the minimum number of Fibonacci numbers whose sum is equal to k. The same Fibonacci number can be used multiple times. The Fibonacci numbers are defined as: F1 = 1 F2 = 1 Fn = Fn-1 + Fn-2 for n > 2. It is guaranteed that for the given constraints we can always find such Fibonacci numbers that sum up to k.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math · Greedy
7
10
19
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1414: Find the Minimum Number of Fibonacci Numbers Whose Sum Is K
class Solution {
public int findMinFibonacciNumbers(int k) {
int a = 1, b = 1;
while (b <= k) {
int c = a + b;
a = b;
b = c;
}
int ans = 0;
while (k > 0) {
if (k >= b) {
k -= b;
++ans;
}
int c = b - a;
b = a;
a = c;
}
return ans;
}
}
// Accepted solution for LeetCode #1414: Find the Minimum Number of Fibonacci Numbers Whose Sum Is K
func findMinFibonacciNumbers(k int) (ans int) {
a, b := 1, 1
for b <= k {
c := a + b
a = b
b = c
}
for k > 0 {
if k >= b {
k -= b
ans++
}
c := b - a
b = a
a = c
}
return
}
# Accepted solution for LeetCode #1414: Find the Minimum Number of Fibonacci Numbers Whose Sum Is K
class Solution:
def findMinFibonacciNumbers(self, k: int) -> int:
a = b = 1
while b <= k:
a, b = b, a + b
ans = 0
while k:
if k >= b:
k -= b
ans += 1
a, b = b - a, a
return ans
// Accepted solution for LeetCode #1414: Find the Minimum Number of Fibonacci Numbers Whose Sum Is K
impl Solution {
pub fn find_min_fibonacci_numbers(mut k: i32) -> i32 {
let mut a = 1;
let mut b = 1;
while b <= k {
let c = a + b;
a = b;
b = c;
}
let mut ans = 0;
while k > 0 {
if k >= b {
k -= b;
ans += 1;
}
let c = b - a;
b = a;
a = c;
}
ans
}
}
// Accepted solution for LeetCode #1414: Find the Minimum Number of Fibonacci Numbers Whose Sum Is K
function findMinFibonacciNumbers(k: number): number {
let [a, b] = [1, 1];
while (b <= k) {
let c = a + b;
a = b;
b = c;
}
let ans = 0;
while (k > 0) {
if (k >= b) {
k -= b;
ans++;
}
let c = b - a;
b = a;
a = c;
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.
Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.
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: Locally optimal choices may fail globally.
Usually fails on: Counterexamples appear on crafted input orderings.
Fix: Verify with exchange argument or monotonic objective before committing.