Add 1625

Author: trung-hn

Readme.md

@@ -395,6 +395,7 @@ https://leetcode.com/jummyegg/
 | 1615 | Maximal Network Rank | Medium | O(N) | O(1) | Graph | | |
 | 1618 | Maximum Font to Fit a Sentence in a Screen | Medium | O(logN) | O(1) | Premium, String, Binary Search | | 🔒 |
 | 1624 | Largest Substring Between Two Equal Characters | Easy | O(N) | O(1) | String | | |
+| 1625 | Lexicographically Smallest String After Applying Operations | Medium | O(N*N) | O(N) | BFS, DFS | | |
 | 1629 | Slowest Key | Easy | O(N) | O(1) | String | | |
 | 1631 | Path With Minimum Effort | Medium | O(N\*M\*logH) | O(M\*N) | Binary Search, DFS, BFS, Union Find, Graph | | |
 | 1634 | Add Two Polynomials Represented as Linked Lists | Medium | O(N) | O(1) | Premium, Linked List | | 🔒 |

src/1625.lexicographically-smallest-string-after-applying-operations.py

@@ -0,0 +1,65 @@
+#
+# @lc app=leetcode id=1625 lang=python3
+#
+# [1625] Lexicographically Smallest String After Applying Operations
+#
+
+# @lc code=start
+# TAGS: BFS, DFS
+
+
+class Solution:
+    """
+    We Can bruteforce this problem because we notice that 10a = 0a
+    """
+    # 136 ms, 92.93%. Time: O(N^2). Space: O(N)
+
+    def findLexSmallestString(self, s: str, a: int, b: int) -> str:
+                # Try all shifting positions and compare
+        def get_smaller(ans, s):
+            for _ in range(len(s)):
+                s = s[b:] + s[:b]
+                ans = min(ans, s)
+            return ans
+
+            # Add number to just odd or even digits.
+        def add_a(temp, odd=True):
+            rv = [str((int(c) + a) % 10) if i %
+                  2 == odd else c for i, c in enumerate(temp)]
+            return "".join(rv)
+
+        ans = "9" * len(s)
+        odd = s
+        # Shift Odd digits
+        for _ in range(10):
+            odd = add_a(odd, True)
+            even = odd
+
+            # Shift Even digits if applicable
+            for _ in range(10) if b % 2 else range(1):
+                ans = get_smaller(ans, even)
+                even = add_a(even, False)
+        return ans
+
+    # 1440 ms, 68.48%. Same complexity but slow because of overheads
+
+    def findLexSmallestString1(self, s: str, a: int, b: int) -> str:
+        def add(s, a):
+            return "".join(str((int(c) + a) % 10) if i % 2 else c for i, c in enumerate(s))
+
+        def rotate(s, b):
+            return s[b:] + s[:b]
+
+        visited = set()
+        q = [s]
+        for s in q:
+            if s in visited:
+                continue
+            visited.add(s)
+            q.append(add(s, a))
+            q.append(rotate(s, b))
+
+        return min(visited)
+
+
+# @lc code=end