climbStairs

Dynamic Programming/Climbing stairs/README.md

@@ -4,8 +4,8 @@ This program provides a solution to the "Climbing Stairs" problem, a classic exa
 Problem Statement:
 A person is at the bottom of a staircase with n steps and wants to reach the top. They can take either 1 or 2 steps at a time. The task is to determine the number of distinct ways the person can reach the top of the staircase.
 
-Approach:
-Dynamic Programming:
+Approach:s
+Dynamic Programming
 
 This problem has overlapping subproblems, making it a suitable candidate for dynamic programming.
 Define dp[i] as the number of ways to reach the i-th step.

Dynamic Programming/Climbing stairs/program.c

@@ -1,18 +1,26 @@
-class Solution {
-    public int climbStairs(int n) 
+#include <stdio.h>
+
+int climbStairs(int n) 
+{
+    if (n == 0 || n == 1) 
     {
-        if (n == 0 || n == 1) 
-        {
-            return 1;
-        }
+        return 1;
+    }
+
+    int dp[n + 1];
+    dp[0] = dp[1] = 1;
 
-        int[] dp = new int[n+1];
-        dp[0] = dp[1] = 1;
-
-        for (int i = 2; i <= n; i++) 
-        {
-            dp[i] = dp[i-1] + dp[i-2];
-        }
-        return dp[n];
+    for (int i = 2; i <= n; i++) 
+    {
+        dp[i] = dp[i - 1] + dp[i - 2];
     }
-}
+
+    return dp[n];
+}
+
+int main() 
+{
+    int n = 5; // Example input
+    printf("Ways to climb %d stairs: %d\n", n, climbStairs(n));
+    return 0;
+}