From 8af3a40294831c23d8dbcfdebcf1073be3149c96 Mon Sep 17 00:00:00 2001
From: aritradey-CS <aritra1999.dey@gmail.com>
Date: Thu, 12 Oct 2023 23:56:11 +0530
Subject: [PATCH] Add new code to fix errors in  Frog_Jump.py

---
 CONTRIBUTING.md              |  5 +++++
 Python Projects/Frog_Jump.py | 30 +++++++++++++++++-------------
 2 files changed, 22 insertions(+), 13 deletions(-)

diff --git a/CONTRIBUTING.md b/CONTRIBUTING.md
index 9b4d03d..c5d4f05 100644
--- a/CONTRIBUTING.md
+++ b/CONTRIBUTING.md
@@ -61,3 +61,8 @@
 #### Name: Debaditya Banerji
 - GitHub: https://github.com/devAdityaa/
 ```
+
+```markdown
+#### Name: Aritra Dey
+- GitHub: https://github.com/aritradey-CS/
+```
diff --git a/Python Projects/Frog_Jump.py b/Python Projects/Frog_Jump.py
index a976a42..ae0d148 100644
--- a/Python Projects/Frog_Jump.py	
+++ b/Python Projects/Frog_Jump.py	
@@ -1,23 +1,27 @@
 # Title: 403. Frog Jump
 # Difficulty: Hard
 # Problem: https://leetcode.com/problems/frog-jump/description/
+from typing import List
 
 # using stack (highest efficiency)
 class Solution:
     def canCross(self, stones: List[int]) -> bool:
-        stone_set = set(stones)
-        visited_set = set()
-        stack = [(0, 0)]
-        while stack:
-            stone, jump = stack.pop()
-            for j in [jump-1, jump, jump+1]:
-                s = stone + j
-                if j > 0 and s in stone_set and (s, j) not in visited_set:
-                    if s == stones[-1]:
-                        return True
-                    stack.append((s, j))
-            visited_set.add((stone, jump))
-        return False
+        n = len(stones)
+        dp = {}  # Use a dictionary to store valid jumps for each stone
+
+        for stone in stones:
+            dp[stone] = set()
+
+        dp[0].add(0)  # The first stone can only be reached with a jump of size 0
+
+        for i in range(n):
+            for j in dp[stones[i]]:
+                for step in range(j - 1, j + 2):  # Next jump size could be j-1, j, or j+1
+                    if step > 0 and stones[i] + step in dp:
+                        dp[stones[i] + step].add(step)
+
+        return len(dp[stones[-1]]) > 0
+
 
 # using dp (low efficiency)
 class Solution: