Sync LeetCode submission Runtime - 3 ms (21.39%), Memory - 17.7 MB (64.68%)

Author: jimit105

0440-k-th-smallest-in-lexicographical-order/README.md

@@ -0,0 +1,24 @@
+<p>Given two integers <code>n</code> and <code>k</code>, return <em>the</em> <code>k<sup>th</sup></code> <em>lexicographically smallest integer in the range</em> <code>[1, n]</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 13, k = 2
+<strong>Output:</strong> 10
+<strong>Explanation:</strong> The lexicographical order is [1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9], so the second smallest number is 10.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 1, k = 1
+<strong>Output:</strong> 1
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= k &lt;= n &lt;= 10<sup>9</sup></code></li>
+</ul>

0440-k-th-smallest-in-lexicographical-order/solution.py

@@ -0,0 +1,29 @@
+# Approach: Prefix tree
+
+class Solution(object):
+    def findKthNumber(self, n, k):
+        curr = 1
+        k -= 1
+
+        while k > 0:
+            step = self._count_steps(n, curr, curr + 1)
+            # If the steps are less than or equal to k, we skip this prefix's subtree
+            if step <= k:
+                # Move to the next prefix and decrease k by the number of steps we skip
+                curr += 1
+                k -= step
+            else:
+                # Move to the next level of the tree and decrement k by 1
+                curr *= 10
+                k -= 1
+
+        return curr
+
+    # To count how many numbers exist between prefix1 and prefix2
+    def _count_steps(self, n, prefix1, prefix2):
+        steps = 0
+        while prefix1 <= n:
+            steps += min(n + 1, prefix2) - prefix1
+            prefix1 *= 10
+            prefix2 *= 10
+        return steps