diff --git a/content/posts/leetcode.md b/content/posts/leetcode.md
index 9369462..3b9fa61 100644
--- a/content/posts/leetcode.md
+++ b/content/posts/leetcode.md
@@ -6441,3 +6441,113 @@ func checkRecord(n int) int {
     return (before[0] + before[1] + before[2] + before[3] + before[4] + before[5]) % 1000000007
 }
 ```
+
+## day91 2024-05-27
+
+### 1608. Special Array With X Elements Greater Than or Equal X
+
+You are given an array nums of non-negative integers. nums is considered special if there exists a number x such that there are exactly x numbers in nums that are greater than or equal to x.
+
+Notice that x does not have to be an element in nums.
+
+Return x if the array is special, otherwise, return -1. It can be proven that if nums is special, the value for x is unique.
+
+![0527iTr6Az7fGaKl](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/0527iTr6Az7fGaKl.png)
+
+### 题解
+
+最简单的思路就是找到数组中的最大值, 再从最大值开始向下遍历各个整数, 并遍历数组查看是否满足条件. 但显然这样做时间复杂度很高. 如果先将数组排序, 则任一下标处的数后面的数必然大于等于这个数本身, 此时就可以用到数组末尾的距离来表示大于等于这个数的个数. 此时可以从后到前依次遍历并将当前下标到末尾的距离设为x, 并判断x是否大于前面的数, 如果大于则x满足条件. 对于这样在有序数组中依次遍历查找满足条件的x的值的问题都可以用二分法来加速解决. 设初始start为0, end为数组末尾下标, mid为二者和的一半. 因为题目中要求大于等于x的数的个数为x, 可以假设mid到数组末尾的距离为x, 判断这个数前面的数是否小于x并且当前mid下标对应的数大于等于x即可. 如果都满足则x为所求, 否则根据条件更新start或end的值.
+
+![0527aTgFw8IMG_D9B340F7B960-1](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/0527aTgFw8IMG_D9B340F7B960-1.jpeg)
+
+### 代码
+
+```go
+func specialArray(nums []int) int {
+    sort.Ints(nums)
+    length := len(nums)
+    start := 0
+    end := length - 1
+    mid := 0
+    x := 0
+    last := 0
+    for start <= end{
+        mid = (start + end) / 2
+        x = length - mid
+        if mid == 0{
+            last = 0
+        }else{
+            last = nums[mid-1]
+        }
+        if x <= nums[mid] && x > last{
+            return x
+        }else if x > nums[mid]{
+            start = mid + 1
+        }else if x <= last{
+            end = mid - 1
+        }
+    }
+    return -1
+}
+```
+
+## day92 2024-05-28
+
+### 1208. Get Equal Substrings Within Budget
+
+You are given two strings s and t of the same length and an integer maxCost.
+
+You want to change s to t. Changing the ith character of s to ith character of t costs |s[i] - t[i]| (i.e., the absolute difference between the ASCII values of the characters).
+
+Return the maximum length of a substring of s that can be changed to be the same as the corresponding substring of t with a cost less than or equal to maxCost. If there is no substring from s that can be changed to its corresponding substring from t, return 0.
+
+![0528RccbVPDYGNL4](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/0528RccbVPDYGNL4.png)
+
+### 题解
+
+本题先计算出两个字符串各个位上的差值, 再使用滑动窗口找到窗口内和小于maxCost的最长窗口即可.
+
+### 代码
+
+```go
+func equalSubstring(s string, t string, maxCost int) int {
+    differs := []int{}
+    length := len(s)
+    differ := 0
+    for i,_ := range s{
+        differ = abs(s[i], t[i])
+        differs = append(differs, differ)
+    }
+    fmt.Println(differs)
+
+    begin := 0
+    end := 0
+    cost := 0
+    result := 0
+    for end < length{
+        // fmt.Println(end)
+        cost += differs[end]
+        if cost <= maxCost{
+            result = max(result, end-begin+1)
+        }else{
+            for begin<=end{
+                cost -= differs[begin]
+                begin++
+                if cost <= maxCost{
+                    break
+                }
+            }
+        }
+        end++
+    }
+    return result
+}
+
+func abs(i byte, j byte)int{
+    if i > j{
+        return int(i - j)
+    }else{
+        return int(j - i)
+    }
+}
+```