From 79082d619cb5277fc41c265b41b867b01456fa2f Mon Sep 17 00:00:00 2001
From: Deven <63876261+devenperez@users.noreply.github.com>
Date: Tue, 16 Jul 2024 21:33:59 -0400
Subject: [PATCH] Time: 101 ms (100%), Space: 16.7 MB (69.71%) - LeetHub

---
 ...shortest-uncommon-substring-in-an-array.py | 29 +++++++++++++++++++
 1 file changed, 29 insertions(+)
 create mode 100644 3076-shortest-uncommon-substring-in-an-array/3076-shortest-uncommon-substring-in-an-array.py

diff --git a/3076-shortest-uncommon-substring-in-an-array/3076-shortest-uncommon-substring-in-an-array.py b/3076-shortest-uncommon-substring-in-an-array/3076-shortest-uncommon-substring-in-an-array.py
new file mode 100644
index 0000000..97c5ea8
--- /dev/null
+++ b/3076-shortest-uncommon-substring-in-an-array/3076-shortest-uncommon-substring-in-an-array.py
@@ -0,0 +1,29 @@
+class Solution:
+    def shortestSubstrings(self, arr: List[str]) -> List[str]:
+        max_len = max([len(x) for x in arr])
+        sol = [""] * len(arr)
+        for l in range(1, max_len + 1):
+            seen = {}
+            for i, x in enumerate(arr):
+                for j in range(len(x) - l + 1):
+                    window = x[j:j + l]
+                    if seen.get(window) == None or seen.get(window) == i:
+                        seen[window] = i
+                    else:
+                        seen[window] = -1
+
+            # Check if we found everything
+            for k in seen.keys():
+                if seen[k] == -1:
+                    continue
+                cur_sol = sol[seen[k]]
+                if len(cur_sol) == 0 or (len(k) <= len(cur_sol) and k < cur_sol):
+                    sol[seen[k]] = k
+
+            for i in range(len(arr)):
+                if len(sol[i]) == 0 and len(arr[i]) > l:
+                    break
+                if i == len(arr) - 1:
+                    return sol
+
+        return sol