Merge pull request #3656 from Ishitamukherjee2004/patch-1

Nth fibonacci Number from gfg is added

Author: ajay-dhangar

dsa-solutions/gfg-solutions/Easy problems/Nth-Fibonacci-Number.md

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+---
+id: nth-fibonacci-number
+title: Nth Fibonacci Number
+sidebar_label: Nth Fibonacci Number
+tags:
+  - Easy
+  - Dynamic Programming
+  - Math
+description: "This tutorial covers the solution to the Nth Fibonacci Number problem from the GeeksforGeeks."
+---
+## Problem Description
+
+Given a positive integer `n`, find the nth Fibonacci number. Since the answer can be very large, return the answer modulo `1000000007`.
+
+Note: For this question, take the first Fibonacci number to be 1.
+
+## Examples
+
+**Example 1:**
+
+```
+Input: n = 1
+Output: 1
+```
+
+**Example 2:**
+
+```
+Input: n = 5
+Output: 5
+```
+
+## Your Task
+
+You don't need to read input or print anything. Your task is to complete the function `nthFibonacci()` which takes the integer `n` as input and returns the nth Fibonacci number modulo `1000000007`.
+
+Expected Time Complexity: $O(n)$
+
+Expected Auxiliary Space: $O(n)$ for dynamic programming or $O(1)$ for iterative approach.
+
+## Constraints
+
+* `1 ≤ n ≤ 10^7`
+
+## Problem Explanation
+
+The Fibonacci sequence is a series of numbers where each number is the sum of the two preceding ones, starting from 1 and 1. The nth Fibonacci number can be computed using the formula: `F(n) = F(n-1) + F(n-2)`. For large values of `n`, the result should be returned modulo `1000000007`.
+
+## Code Implementation
+
+<Tabs>
+  <TabItem value="Python" label="Python" default>
+  <SolutionAuthor name="@arunimad6yuq"/>
+
+  ```py
+  MOD = 1000000007
+
+  class Solution:
+      def nthFibonacci(self, n: int) -> int:
+          if n == 1 or n == 2:
+              return 1
+          
+          a, b = 1, 1
+          for i in range(3, n + 1):
+              a, b = b, (a + b) % MOD
+          
+          return b
+
+  # Example usage
+  if __name__ == "__main__":
+      solution = Solution()
+      print(solution.nthFibonacci(1))  # Expected output: 1
+      print(solution.nthFibonacci(5))  # Expected output: 5
+  ```
+
+  </TabItem>
+  <TabItem value="C++" label="C++">
+  <SolutionAuthor name="@YourUsername"/>
+
+  ```cpp
+  //{ Driver Code Starts
+  #include <bits/stdc++.h>
+  using namespace std;
+
+  // } Driver Code Ends
+  class Solution {
+  public:
+      // Function to find nth Fibonacci number
+      int nthFibonacci(int n) {
+          const int MOD = 1000000007;
+          if (n == 1 || n == 2) return 1;
+
+          int a = 1, b = 1, c;
+          for (int i = 3; i <= n; ++i) {
+              c = (a + b) % MOD;
+              a = b;
+              b = c;
+          }
+          return b;
+      }
+  };
+
+  //{ Driver Code Starts.
+  int main() {
+      int t;
+      cin >> t;
+      while (t--) {
+          int n;
+          cin >> n;
+          Solution obj;
+          cout << obj.nthFibonacci(n) << endl;
+      }
+      return 0;
+  }
+  // } Driver Code Ends
+  ```
+
+  </TabItem>
+
+  <TabItem value="Javascript" label="Javascript" default>
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+
+  ```javascript
+class Solution {
+  nthFibonacci(n) {
+    if (n === 1 || n === 2) return 1;
+    let a = 1, b = 1;
+    for (let i = 3; i <= n; i++) {
+      [a, b] = [b, (a + b) % 1000000007];
+    }
+    return b;
+  }
+}ut: 5
+
+  ```
+
+  </TabItem>
+
+  <TabItem value="Typescript" label="Typescript" default>
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+
+  ```typescript
+class Solution {
+  nthFibonacci(n: number): number {
+    if (n === 1 || n === 2) return 1;
+    let a: number = 1, b: number = 1;
+    for (let i: number = 3; i <= n; i++) {
+      [a, b] = [b, (a + b) % 1000000007];
+    }
+    return b;
+  }
+}
+
+  ```
+
+  </TabItem>
+
+  <TabItem value="Java" label="Java" default>
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+
+  ```java
+public class Solution {
+  public int nthFibonacci(int n) {
+    if (n == 1 || n == 2) return 1;
+    int a = 1, b = 1;
+    for (int i = 3; i <= n; i++) {
+      int sum = (a + b) % 1000000007;
+      a = b;
+      b = sum;
+    }
+    return b;
+  }
+}
+
+  ```
+
+  </TabItem>
+</Tabs>
+
+
+## Time Complexity
+
+* The iterative approach has a time complexity of $O(n)$.
+
+## Space Complexity
+
+* The space complexity is $O(1)$ since we are using only a fixed amount of extra space.