Create Find the Minimum Number of Fibonacci Numbers Who Sum is K.py

Author: anthonyzhub

Find the Minimum Number of Fibonacci Numbers Who Sum is K.py

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+# https://leetcode.com/problems/find-the-minimum-number-of-fibonacci-numbers-whose-sum-is-k/
+
+class Solution:
+    
+    def generateFib(self, k):
+        
+        # OBJECTIVE: Return a list of fibonacci numbers from 1 to k
+        
+        # Create list
+        fib = [1, 1]
+        
+        # Populate list
+        num = 1
+        while num < k:
+            
+            # If the new number is less than or equal to k, update num and add it to list
+            if fib[-1] + fib[-2] <= k:
+                num = fib[-1] + fib[-2]
+                fib.append(num)
+            
+            # If the new number is greater than k, exit loop
+            else:
+                break
+                
+        return fib
+    
+    def findK(self, fib, n, target):
+        
+        # OBJECTIVE: Return index of element that equals or closest to target
+        
+        # If target is 0, exit function
+        if target == 0:
+            return 0
+        
+        # If target is greater than or equal to fib's last element, return it
+        if target >= fib[n - 1]:
+            return n - 1
+        
+        # Create left and right pointers
+        leftPtr = 0
+        rightPtr = n - 1
+        
+        # Do binary search to find target in fib
+        while leftPtr < rightPtr:
+            
+            # Create middle pointer
+            midPtr = (leftPtr + rightPtr) // 2
+            
+            # If middle element is greater than target, move rightPtr
+            if fib[midPtr] > target:
+                rightPtr = midPtr
+            
+            # If middle element is less than or equal to target, move leftPtr
+            else:
+                leftPtr = midPtr + 1
+        
+        # Return left pointer
+        return leftPtr - 1
+    
+    def findMinFibonacciNumbers(self, k: int) -> int:
+        
+        # OBJECTIVE: Return the minimum number of fibonacci numbers whose sum is equal to k
+        
+        # If k equals to 1, return 1.
+        # NOTE: Added this to avoid O(n) space and run time
+        if k == 1:
+            return 1
+        
+        # Generate a list of fibonacci numbers from 1 to k
+        fib = self.generateFib(k)
+        n = len(fib)
+        
+        # If last element equals to k, return 1
+        if fib[-1] == k:
+            return 1
+        
+        # Create a variable to hold minimum number of fibonacci numbers
+        minLen = 0
+        
+        # Get minimum number of fibonacci numbers
+        while k > 0:
+            
+            # Find k (or value closest to k) in fib
+            idx = self.findK(fib, n, k)
+            
+            # Increment minLen
+            minLen += 1
+            
+            # Update k by subtracting idx element from it
+            k -= fib[idx]
+            
+        return minLen