diff --git a/work/1/Medium/Array/1004.max-consecutive-ones-iii.py b/work/1/Medium/Array/1004.max-consecutive-ones-iii.py
new file mode 100644
index 0000000..a0d5fd0
--- /dev/null
+++ b/work/1/Medium/Array/1004.max-consecutive-ones-iii.py
@@ -0,0 +1,57 @@
+"""
+1004. Max Consecutive Ones III
+Difficulty: Medium
+https://leetcode.com/problems/max-consecutive-ones-iii/
+
+──────────────────────────────────────────────────
+
+Given a binary array nums and an integer k, return the maximum number
+of consecutive 1's in the array if you can flip at most k 0's.
+
+
+
+Example 1:
+
+Input: nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2
+Output: 6
+Explanation: [1,1,1,0,0,1,1,1,1,1,1]
+Bolded numbers were flipped from 0 to 1. The longest subarray is
+underlined.
+
+Example 2:
+
+Input: nums = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], k = 3
+Output: 10
+Explanation: [0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]
+Bolded numbers were flipped from 0 to 1. The longest subarray is
+underlined.
+
+
+
+Constraints:
+
+        • 1 <= nums.length <= 10^5
+
+        • nums[i] is either 0 or 1.
+
+        • 0 <= k <= nums.length
+"""
+
+
+class Solution:
+    def longestOnes(self, nums: list[int], k: int) -> int:
+        left = 0
+        cur = 0
+        ans = 0
+
+        for right in range(len(nums)):
+            if nums[right] == 0:
+                cur += 1
+            while cur > k:
+                if nums[left] == 0:
+                    cur -= 1
+                left += 1
+
+            ans = max(ans, right - left + 1)
+
+        return ans