LeetCode 1323. Maximum 69 Number

Author: raul-sauco

README.md

@@ -286,6 +286,7 @@ Solutions to LeetCode problems. The first column links to the problem in LeetCod
 | [1268. Search Suggestions System][lc1268]                                      | 🟠 Medium  | [![python](res/py.png)][lc1268py]                                                   |
 | [1293. Shortest Path in a Grid with Obstacles Elimination][lc1293]             | 🔴 Hard    | [![python](res/py.png)][lc1293py]                                                   |
 | [1296. Divide Array in Sets of K Consecutive Numbers][lc1296]                  | 🟠 Medium  | [![python](res/py.png)][lc1296py]                                                   |
+| [1323. Maximum 69 Number][lc1323]                                              | 🟠 Medium  | [![python](res/py.png)][lc1323py]                                                   |
 | [1328. Break a Palindrome][lc1328]                                             | 🟠 Medium  | [![python](res/py.png)][lc1328py]                                                   |
 | [1329. Sort the Matrix Diagonally][lc1329]                                     | 🟠 Medium  | [![python](res/py.png)][lc1329py]                                                   |
 | [1332. Remove Palindromic Subsequences][lc1332]                                | 🟢 Easy    | [![python](res/py.png)][lc1332py]                                                   |
@@ -867,6 +868,8 @@ Solutions to LeetCode problems. The first column links to the problem in LeetCod
 [lc1293py]: leetcode/shortest-path-in-a-grid-with-obstacles-elimination.py
 [lc1296]: https://leetcode.com/problems/divide-array-in-sets-of-k-consecutive-numbers/
 [lc1296py]: leetcode/divide-array-in-sets-of-k-consecutive-numbers.py
+[lc1323]: https://leetcode.com/problems/maximum-69-number/
+[lc1323py]: leetcode/maximum-69-number.py
 [lc1328]: https://leetcode.com/problems/break-a-palindrome/
 [lc1328py]: leetcode/break-a-palindrome.py
 [lc1329]: https://leetcode.com/problems/sort-the-matrix-diagonally/

leetcode/maximum-69-number.py

@@ -0,0 +1,84 @@
+# 1323. Maximum 69 Number
+# 🟢 Easy
+#
+# https://leetcode.com/problems/maximum-69-number/
+#
+# Tags: Math - Greedy
+
+import timeit
+
+
+# We can obtain the largest number converting the 6 with the highest
+# order into a 9, if none are found, then it is optimal to not modify
+# the input.
+#
+# Time complexity: O(1) - We iterate 3 times over the input, converting
+# to a list of strings, finding the first 6 and casting back to int.
+# Since the input has a maximum of 4 digits, time is bounded.
+# Space complexity: O(1) - Where n is the number of digits in the input,
+# we create a list of digits. Since the input has a maximum of 4 digits,
+# space is bounded.
+#
+# Runtime: 38 ms, faster than 84.45%
+# Memory Usage: 13.8 MB, less than 54.69%
+class Parsing:
+    def maximum69Number(self, num: int) -> int:
+        return int(str(num).replace("6", "9", 1))
+
+
+# Use divmod inside a loop to find the index of the furthest left 6 in
+# the input, then add 3 at that position to update that value to a 9.
+#
+# Time complexity: O(1) - The loop will execute a maximum of 4 times
+# because num <= 10^4.
+# Space complexity: O(1) - Constant space.
+#
+# Runtime: 67 ms, faster than 10.70%
+# Memory Usage: 14 MB, less than 9.69%
+class Math:
+    def maximum69Number(self, num: int) -> int:
+        # The digit that we are visiting left to right.
+        i = 0
+        # Make a copy of num to mutate it.
+        quotient = num
+        # Store the furthest left index at which we find a number 6.
+        highest_six = -1
+        while quotient > 0:
+            quotient, remainder = divmod(quotient, 10)
+            if remainder == 6:
+                highest_six = i
+            i += 1
+        # If we found any 6, update the highest 6 found to become a 9
+        # by adding 3 at that position in the input number.
+        return (num + 3 * (10**highest_six)) if highest_six != -1 else num
+
+
+def test():
+    executors = [
+        Parsing,
+        Math,
+    ]
+    tests = [
+        [9669, 9969],
+        [9996, 9999],
+        [9999, 9999],
+    ]
+    for executor in executors:
+        start = timeit.default_timer()
+        for _ in range(1):
+            for col, t in enumerate(tests):
+                sol = executor()
+                result = sol.maximum69Number(t[0])
+                exp = t[1]
+                assert result == exp, (
+                    f"\033[93m» {result} <> {exp}\033[91m for"
+                    + f" test {col} using \033[1m{executor.__name__}"
+                )
+        stop = timeit.default_timer()
+        used = str(round(stop - start, 5))
+        cols = "{0:20}{1:10}{2:10}"
+        res = cols.format(executor.__name__, used, "seconds")
+        print(f"\033[92m» {res}\033[0m")
+
+
+test()