Create Fibonacci number.cpp

Author: sanyathisside

LeetCode/Fibonacci number.cpp

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+Solution 1 - Recursive Approach
+Time Complexity - O(2^N)
+Space Complexity - O(N) given the function call stack size
+
+    int fib(int N) {
+        if(N == 0)  return 0;
+        if(N == 1)  return 1;
+        return fib(N-1) + fib(N-2);
+    }
+
+
+Solution 2 - Dynamic Programming Approach
+Use memoization to store perviously computed fibonacci values.
+Time Complexity - O(N)
+Space Complexity - O(N)
+
+    int fib(int N) {
+        if(N < 2)
+            return N;
+        int memo[N+1];
+        memo[0] = 0;
+        memo[1] = 1;
+        for(int i=2; i<=N; i++)
+            memo[i] = memo[i-1] + memo[i-2];
+        return memo[N];
+    }
+
+
+Solution 3 - Imperative Approach (Bottom Up DP)
+With Imperative approach, we step through the loop and optimize the space by storing only two previous fibonacci values in two variables.
+Time Complexity - O(N)
+Space Complexity - O(1)
+
+    int fib(int N) {
+        if(N < 2) 
+            return N;
+    	int a = 0, b = 1, c = 0;
+        for(int i = 1; i < N; i++)
+        {
+            c = a + b;
+            a = b;
+            b = c;
+        }
+        return c;
+    }
+
+
+Solution 4 - Binet's Nth-term Formula
+Using Binet's Formula for the Nth Fibonacci involves the usage of our golden section number Phi.
+Phi = ( sqrt(5) + 1 ) / 2
+Using approximation equation is good enough here, since we know N >= 0 && N <= 30, we can safely use the following rounded function
+Fib(N) = round( ( Phi^N ) / sqrt(5) )
+Full mathematical explanation of Binet's Formula here
+Time Complexity - O(1)
+Space Complexity - O(1)
+
+    int fib(int N) {
+        double phi = (sqrt(5) + 1) / 2;     
+        return round(pow(phi, N) / sqrt(5));
+    }