From 4eebd265df5c9b1a2d1d24631407d3c47eab0e7a Mon Sep 17 00:00:00 2001
From: rimgosu <rimgosu@users.noreply.github.com>
Date: Wed, 23 Sep 2026 01:07:15 +0000
Subject: [PATCH] 1137

---
 .../v3/1137. N-th Tribonacci Number.py"       | 25 +++++++++++++++++++
 1 file changed, 25 insertions(+)
 create mode 100644 "leetcode3/\353\263\200\354\247\200\355\230\221/v3/1137. N-th Tribonacci Number.py"

diff --git "a/leetcode3/\353\263\200\354\247\200\355\230\221/v3/1137. N-th Tribonacci Number.py" "b/leetcode3/\353\263\200\354\247\200\355\230\221/v3/1137. N-th Tribonacci Number.py"
new file mode 100644
index 00000000..d8331882
--- /dev/null
+++ "b/leetcode3/\353\263\200\354\247\200\355\230\221/v3/1137. N-th Tribonacci Number.py"	
@@ -0,0 +1,25 @@
+'''
+1. 아이디어 :
+점화식 T(n) = T(n-1) + T(n-2) + T(n-3) 을 그대로 DP로 옮긴다.
+dp[0]=0, dp[1]=1, dp[2]=1 을 초기값으로 두고 n까지 채운 뒤 dp[n]을 반환.
+
+2. 시간복잡도 :
+o(n) - n까지 한 번만 순회
+
+3. 자료구조/알고리즘 :
+DP (1차원 배열), 점화식
+
+'''
+
+class Solution:
+    def tribonacci(self, n: int) -> int:
+        dp = [0] * 40
+
+        dp[0] = 0
+        dp[1] = 1
+        dp[2] = 1
+
+        for i in range(2,n+1):
+            dp[i] = dp[i-1] + dp[i-2] + dp[i-3]
+
+        return dp[n]