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| 1 | +# https://leetcode.com/problems/find-the-minimum-number-of-fibonacci-numbers-whose-sum-is-k/ |
| 2 | + |
| 3 | +class Solution: |
| 4 | + |
| 5 | + def generateFib(self, k): |
| 6 | + |
| 7 | + # OBJECTIVE: Return a list of fibonacci numbers from 1 to k |
| 8 | + |
| 9 | + # Create list |
| 10 | + fib = [1, 1] |
| 11 | + |
| 12 | + # Populate list |
| 13 | + num = 1 |
| 14 | + while num < k: |
| 15 | + |
| 16 | + # If the new number is less than or equal to k, update num and add it to list |
| 17 | + if fib[-1] + fib[-2] <= k: |
| 18 | + num = fib[-1] + fib[-2] |
| 19 | + fib.append(num) |
| 20 | + |
| 21 | + # If the new number is greater than k, exit loop |
| 22 | + else: |
| 23 | + break |
| 24 | + |
| 25 | + return fib |
| 26 | + |
| 27 | + def findK(self, fib, n, target): |
| 28 | + |
| 29 | + # OBJECTIVE: Return index of element that equals or closest to target |
| 30 | + |
| 31 | + # If target is 0, exit function |
| 32 | + if target == 0: |
| 33 | + return 0 |
| 34 | + |
| 35 | + # If target is greater than or equal to fib's last element, return it |
| 36 | + if target >= fib[n - 1]: |
| 37 | + return n - 1 |
| 38 | + |
| 39 | + # Create left and right pointers |
| 40 | + leftPtr = 0 |
| 41 | + rightPtr = n - 1 |
| 42 | + |
| 43 | + # Do binary search to find target in fib |
| 44 | + while leftPtr < rightPtr: |
| 45 | + |
| 46 | + # Create middle pointer |
| 47 | + midPtr = (leftPtr + rightPtr) // 2 |
| 48 | + |
| 49 | + # If middle element is greater than target, move rightPtr |
| 50 | + if fib[midPtr] > target: |
| 51 | + rightPtr = midPtr |
| 52 | + |
| 53 | + # If middle element is less than or equal to target, move leftPtr |
| 54 | + else: |
| 55 | + leftPtr = midPtr + 1 |
| 56 | + |
| 57 | + # Return left pointer |
| 58 | + return leftPtr - 1 |
| 59 | + |
| 60 | + def findMinFibonacciNumbers(self, k: int) -> int: |
| 61 | + |
| 62 | + # OBJECTIVE: Return the minimum number of fibonacci numbers whose sum is equal to k |
| 63 | + |
| 64 | + # If k equals to 1, return 1. |
| 65 | + # NOTE: Added this to avoid O(n) space and run time |
| 66 | + if k == 1: |
| 67 | + return 1 |
| 68 | + |
| 69 | + # Generate a list of fibonacci numbers from 1 to k |
| 70 | + fib = self.generateFib(k) |
| 71 | + n = len(fib) |
| 72 | + |
| 73 | + # If last element equals to k, return 1 |
| 74 | + if fib[-1] == k: |
| 75 | + return 1 |
| 76 | + |
| 77 | + # Create a variable to hold minimum number of fibonacci numbers |
| 78 | + minLen = 0 |
| 79 | + |
| 80 | + # Get minimum number of fibonacci numbers |
| 81 | + while k > 0: |
| 82 | + |
| 83 | + # Find k (or value closest to k) in fib |
| 84 | + idx = self.findK(fib, n, k) |
| 85 | + |
| 86 | + # Increment minLen |
| 87 | + minLen += 1 |
| 88 | + |
| 89 | + # Update k by subtracting idx element from it |
| 90 | + k -= fib[idx] |
| 91 | + |
| 92 | + return minLen |
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