feat: word-ladder-challenge challenge

Author: IbatoLionDev

…intervals/proposed-solution/intervals.py …intervals/proposed-solution/intervals.py2-NORMAL/Python/11-merge-overlapping-intervals/proposed-solution/intervals.py renamed to 2-NORMAL/Python/6-merge-overlapping-intervals/proposed-solution/intervals.py 2-NORMAL/Python/11-merge-overlapping-intervals/proposed-solution/intervals.py renamed to 2-NORMAL/Python/6-merge-overlapping-intervals/proposed-solution/intervals.py

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+# Challenge: Given two words (start and end) and a dictionary, implement an algorithm to find the shortest transformation sequence where each word differs by exactly one character from the previous one. This challenge requires applying search techniques like BFS.
+
+from word_ladder import word_ladder
+
+def main():
+    begin_word = "hit"
+    end_word = "cog"
+    word_list = ["hot", "dot", "dog", "lot", "log", "cog"]
+    path = word_ladder(begin_word, end_word, word_list)
+    print("Shortest transformation sequence:", path)
+
+if __name__ == "__main__":
+    main()

…ping-intervals/proposed-solution/main.py …ping-intervals/proposed-solution/main.py2-NORMAL/Python/11-merge-overlapping-intervals/proposed-solution/main.py renamed to 2-NORMAL/Python/6-merge-overlapping-intervals/proposed-solution/main.py 2-NORMAL/Python/11-merge-overlapping-intervals/proposed-solution/main.py renamed to 2-NORMAL/Python/6-merge-overlapping-intervals/proposed-solution/main.py

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+from collections import deque
+
+def word_ladder(begin_word, end_word, word_list):
+    word_set = set(word_list)
+    if end_word not in word_set:
+        return []
+
+    queue = deque([(begin_word, [begin_word])])
+    visited = set([begin_word])
+
+    while queue:
+        current_word, path = queue.popleft()
+        if current_word == end_word:
+            return path
+
+        for i in range(len(current_word)):
+            for c in 'abcdefghijklmnopqrstuvwxyz':
+                next_word = current_word[:i] + c + current_word[i+1:]
+                if next_word in word_set and next_word not in visited:
+                    visited.add(next_word)
+                    queue.append((next_word, path + [next_word]))
+
+    return []