Merge pull request Dijkstra-Edu#33 from mrid88/feature/climbing-stairs

Solution Dijkstra-Edu#70 - Mridul/Edited - 16/03/2025

Author: JRS296

Problems/#70 - Climbing Stairs - Easy/Explanation.md

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+# Explanation of the `climbStairs` Code
+
+This Java class provides a solution to the classic "Climbing Stairs" problem using dynamic programming. The goal is to determine the number of distinct ways to climb a staircase with `n` steps, where you can take either 1 step or 2 steps at a time.
+
+---
+
+## Class: `Solution`
+
+### Method: `climbStairs(int n)`
+
+#### **Purpose**
+This method calculates the number of ways to climb a staircase with `n` steps.
+
+#### **Parameters**
+- `n`: An integer representing the total number of steps in the staircase.
+
+#### **Logic**
+1. **Base Cases**:
+   - If `n == 1`, return `1` since there is only one way to climb a single step.
+   - If `n == 2`, return `2` since there are two ways to climb two steps: either `1+1` or `2`.
+
+2. **Dynamic Programming Array**:
+   - Create an integer array `a` of size `n`.
+   - Initialize the first two elements:
+     - `a[0] = 1` (1 way to climb 1 step).
+     - `a[1] = 2` (2 ways to climb 2 steps).
+
+3. **Iterative Calculation**:
+   - Use a loop from `i = 2` to `n - 1`:
+     - For each step `i`, calculate the number of ways to reach it as the sum of the ways to reach the previous two steps:
+       
+
+\[
+       a[i] = a[i - 1] + a[i - 2]
+       \]
+
+
+   - This is based on the observation that to reach step `i`, you can either:
+     - Take one step from `i - 1` (contributing `a[i - 1]` ways).
+     - Take two steps from `i - 2` (contributing `a[i - 2]` ways).
+
+4. **Return Result**:
+   - Return `a[n - 1]`, which contains the total number of ways to climb `n` steps.
+
+---
+
+### Example Execution
+#### Input:
+```java
+int n = 5;
+
+Time Complexity: O(n)
+Space Complexity: O(n)

Problems/#70 - Climbing Stairs - Easy/Solution.java

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+class Solution {
+    public int climbStairs(int n) {
+        if(n==1) return 1;
+        
+        if(n==2) return 2;
+
+        int[] a =  new int[n];
+        a[0]=1;
+        a[1]=2;
+
+        for(int i=2;i<n;i++){
+            a[i]=a[i-1]+a[i-2];
+        }
+        return a[n-1];
+    }
+}