Sync LeetCode submission Runtime - 1116 ms (48.86%), Memory - 17.7 MB (92.86%)

Author: jimit105

2802-find-the-punishment-number-of-an-integer/README.md

@@ -0,0 +1,41 @@
+<p>Given a positive integer <code>n</code>, return <em>the <strong>punishment number</strong></em> of <code>n</code>.</p>
+
+<p>The <strong>punishment number</strong> of <code>n</code> is defined as the sum of the squares of all integers <code>i</code> such that:</p>
+
+<ul>
+	<li><code>1 &lt;= i &lt;= n</code></li>
+	<li>The decimal representation of <code>i * i</code> can be partitioned into contiguous substrings such that the sum of the integer values of these substrings equals <code>i</code>.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 10
+<strong>Output:</strong> 182
+<strong>Explanation:</strong> There are exactly 3 integers i in the range [1, 10] that satisfy the conditions in the statement:
+- 1 since 1 * 1 = 1
+- 9 since 9 * 9 = 81 and 81 can be partitioned into 8 and 1 with a sum equal to 8 + 1 == 9.
+- 10 since 10 * 10 = 100 and 100 can be partitioned into 10 and 0 with a sum equal to 10 + 0 == 10.
+Hence, the punishment number of 10 is 1 + 81 + 100 = 182
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 37
+<strong>Output:</strong> 1478
+<strong>Explanation:</strong> There are exactly 4 integers i in the range [1, 37] that satisfy the conditions in the statement:
+- 1 since 1 * 1 = 1. 
+- 9 since 9 * 9 = 81 and 81 can be partitioned into 8 + 1. 
+- 10 since 10 * 10 = 100 and 100 can be partitioned into 10 + 0. 
+- 36 since 36 * 36 = 1296 and 1296 can be partitioned into 1 + 29 + 6.
+Hence, the punishment number of 37 is 1 + 81 + 100 + 1296 = 1478
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 1000</code></li>
+</ul>

2802-find-the-punishment-number-of-an-integer/solution.py

@@ -0,0 +1,37 @@
+# Approach 2: Recursion of Strings
+
+# Time: O(n * 2 ^ (log n))
+# Space: O(log n)
+
+class Solution:
+    def can_partition(self, string_num, target):
+        # Valid partition
+        if not string_num and target == 0:
+            return True
+
+        # Invalid partition
+        if target < 0:
+            return False
+
+        # Recursively check all partitions for a valid partition
+        for index in range(len(string_num)):
+            left = string_num[: index + 1]
+            right = string_num[index + 1 :]
+            left_num = int(left)
+
+            if self.can_partition(right, target - left_num):
+                return True
+
+        return False
+
+    def punishmentNumber(self, n: int) -> int:
+        punishment_sum = 0
+
+        for current_num in range(1, n + 1):
+            squared_num = current_num * current_num
+
+            if self.can_partition(str(squared_num), current_num):
+                punishment_sum += squared_num
+
+        return punishment_sum
+