From 842ff5294f5e97dfd1e4c09f9d9e01a89ac16e32 Mon Sep 17 00:00:00 2001
From: Hardik Pawar <97388607+Hardvan@users.noreply.github.com>
Date: Wed, 2 Oct 2024 19:25:57 +0530
Subject: [PATCH] Improve comments & readability in ClimbingStairs.java (#5498)

---
 .../dynamicprogramming/ClimbingStairs.java    | 46 ++++++++++++++-----
 1 file changed, 35 insertions(+), 11 deletions(-)

diff --git a/src/main/java/com/thealgorithms/dynamicprogramming/ClimbingStairs.java b/src/main/java/com/thealgorithms/dynamicprogramming/ClimbingStairs.java
index d79ed3c23..0f0f5375b 100644
--- a/src/main/java/com/thealgorithms/dynamicprogramming/ClimbingStairs.java
+++ b/src/main/java/com/thealgorithms/dynamicprogramming/ClimbingStairs.java
@@ -1,31 +1,55 @@
 package com.thealgorithms.dynamicprogramming;
 
-/* A DynamicProgramming solution for Climbing Stairs' problem Returns the
-    distinct ways can you climb to the staircase by either climbing 1 or 2 steps.
-
-    Link : https://medium.com/analytics-vidhya/leetcode-q70-climbing-stairs-easy-444a4aae54e8
-*/
+/*
+ * A dynamic programming solution for the "Climbing Stairs" problem.
+ * Returns the no. of distinct ways to climb to the top
+ * of a staircase when you can climb either 1 or 2 steps at a time.
+ *
+ * For example, if there are 5 steps, the possible ways to climb the
+ * staircase are:
+ * 1. 1-1-1-1-1
+ * 2. 1-1-1-2
+ * 3. 1-2-1-1
+ * 4. 2-1-1-1
+ * 5. 2-2-1
+ * 6. 1-1-2-1
+ * 7. 1-2-2
+ * 8. 2-1-2
+ * Ans: 8 ways
+ */
 public final class ClimbingStairs {
+
     private ClimbingStairs() {
     }
 
+    /**
+     * Calculates the no. of distinct ways to climb a staircase with n steps.
+     *
+     * @param n the no. of steps in the staircase (non-negative integer)
+     * @return the no. of distinct ways to climb to the top
+     *         - Returns 0 if n is 0 (no steps to climb).
+     *         - Returns 1 if n is 1 (only one way to climb).
+     *         - For n > 1, it returns the total no. of ways to climb.
+     */
     public static int numberOfWays(int n) {
 
+        // Base case: if there are no steps or only one step, return n.
         if (n == 1 || n == 0) {
             return n;
         }
-        int prev = 1;
-        int curr = 1;
 
-        int next;
+        int prev = 1; // Ways to reach the step before the current one (step 1)
+        int curr = 1; // Ways to reach the current step (step 2)
+        int next; // Total ways to reach the next step
 
-        for (int i = 2; i <= n; i++) {
+        for (int i = 2; i <= n; i++) { // step 2 to n
             next = curr + prev;
-            prev = curr;
 
+            // Move the pointers to the next step
+            prev = curr;
             curr = next;
         }
 
-        return curr;
+        return curr; // Ways to reach the nth step
     }
 }