add quesno 386

Author: agarwalhimanshugaya

dsa-solutions/lc-solutions/0300-0399/0386-Lexicographical-Numbers.md

@@ -0,0 +1,107 @@
+---
+id: lexicographical-numbers
+title: Lexicographical-Numbers
+sidebar_label: 0389-Lexicographical-Numbers
+tags:
+  - Leet code
+description: "Solution to leetocde 386"
+---
+
+### Problem Description
+
+Given an integer `n`, return all the numbers in the range `[1, n]` sorted in lexicographical order.
+
+You must write an algorithm that runs in $O(n)$ time and uses $O(1)$ extra space. 
+
+Example 1:
+
+```
+Input: n = 13
+Output: [1,10,11,12,13,2,3,4,5,6,7,8,9]
+
+```
+
+Example 2:
+
+```
+Input: n = 2
+Output: [1,2]
+
+```
+
+### Constraints:
+
+- `1 <= n <= 5 * 10^4`
+
+### Algorithm
+
+The basic idea is to find the next number to add.
+Take 45 for example: if the current number is 45, the next one will be `450 (450 == 45 * 10)(if 450 <= n)`, or `46 (46 == 45 + 1) (if 46 <= n)` or `5 (5 == 45 / 10 + 1)`(5 is less than 45 so it is for sure less than n).
+We should also consider `n = 600`, and the current number = 499, the next number is `5` because there are all "9"s after "4" in `"499"` so we should divide `499 by 10` until the last digit is not `"9"`.
+It is like a tree, and we are easy to get a sibling, a left most child and the parent of any node.
+
+### Code Implementation
+
+**Python:**
+
+```python
+def lexicalOrder(self, n):
+    top = 1
+    while top * 10 <= n:
+        top *= 10
+    def mycmp(a, b, top=top):
+        while a < top: a *= 10
+        while b < top: b *= 10
+        return -1 if a < b else b < a
+    return sorted(xrange(1, n+1), mycmp)
+    
+```
+
+**C++:**
+
+```c++
+class Solution {
+public:
+    vector<int> lexicalOrder(int n) {
+        vector<int> res(n);
+        int cur = 1;
+        for (int i = 0; i < n; i++) {
+            res[i] = cur;
+            if (cur * 10 <= n) {
+                cur *= 10;
+            } else {
+                if (cur >= n) 
+                    cur /= 10;
+                cur += 1;
+                while (cur % 10 == 0)
+                    cur /= 10;
+            }
+        }
+        return res;
+    }
+};
+```
+
+**Java:**
+
+```java
+public List<Integer> lexicalOrder(int n) {
+        List<Integer> list = new ArrayList<>(n);
+        int curr = 1;
+        for (int i = 1; i <= n; i++) {
+            list.add(curr);
+            if (curr * 10 <= n) {
+                curr *= 10;
+            } else if (curr % 10 != 9 && curr + 1 <= n) {
+                curr++;
+            } else {
+                while ((curr / 10) % 10 == 9) {
+                    curr /= 10;
+                }
+                curr = curr / 10 + 1;
+            }
+        }
+        return list;
+    }
+```
+