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+### Solution 1
+
+min heapを使った解法。Arai60の問題に引っ張られて最初にこの解法で書いたがarrayでこれをやるのは結構不自然かも。
+こんなことをやるんだったら配列をソートしてk-1番目を出力した方がいい気がする。
+
+```python
+class Solution:
+    def findKthLargest(self, nums: List[int], k: int) -> int:
+        min_heap = []
+        for num in nums:
+            if len(min_heap) < k:
+                heapq.heappush(min_heap, num)
+            else:
+                if num < min_heap[0]:
+                    continue
+                heapq.heappushpop(min_heap, num)
+        return min_heap[0]
+```
+
+### Solution 2
+
+3-partition quick select
+普通のquick selectは以下と未満の判別をしないので都合が悪い（全部同じ要素、みたいな配列が入ってくるとkをスキップしてしまう）
+
+```python
+class Solution:
+    def findKthLargest(self, nums: List[int], k: int) -> int:
+        if k > len(nums):
+            raise ValueError("invalid k for the list")
+        nums = nums[:]
+        left = 0
+        right = len(nums) - 1
+        while True:
+            pivot = nums[random.randint(left, right)]
+            l = left
+            r = right
+            idx = left
+            while idx <= r:
+                if nums[idx] > pivot:
+                    nums[l], nums[idx] = nums[idx], nums[l]
+                    l += 1
+                    idx += 1
+                elif nums[idx] == pivot:
+                    idx += 1
+                else:
+                    nums[idx], nums[r] = nums[r], nums[idx]
+                    r -= 1
+            if l <= k - 1 <= r:
+                return pivot
+            elif k - 1 < l:
+                right = l - 1
+            elif r < k - 1:
+                left = r + 1
+```