diff --git a/Others/2417.Closest-Fair-Integer/2417.Closest-Fair-Integer.cpp b/Others/2417.Closest-Fair-Integer/2417.Closest-Fair-Integer.cpp
new file mode 100644
index 000000000..b5732b550
--- /dev/null
+++ b/Others/2417.Closest-Fair-Integer/2417.Closest-Fair-Integer.cpp
@@ -0,0 +1,48 @@
+class Solution {
+public:
+    int closestFair(int n) 
+    {                
+        int count = 0;
+        int ret = helper(n, count);
+        if (ret!=-1) return ret;            
+        
+        while (n>=0)
+        {
+            int ret = helper(n+1, count);
+            if (ret!=-1) return ret;
+            
+            ret = helper(n+2, count);
+            if (ret!=-1) return ret;
+            
+            n = n/10;
+            count++;                
+        }
+        return -1;
+    }
+    
+    int helper(int n, int count)
+    {
+        int m = n;
+        int odd = 0, even = 0;
+        while (m>0)
+        {
+            if (m%2==0)
+                even++;
+            else
+                odd++;  
+            m/=10;
+        }
+        
+        int sum = count, diff = odd-even;
+        if ((sum + diff) % 2 != 0) return -1;
+        int a = (sum + diff) / 2;   // # of 0
+        int b = (sum - diff) / 2;   // # of 1           
+        if (a<0 || b<0) return -1;
+            
+        for (int i=0; i<a; i++)
+            n = n*10 + 0;
+        for (int i=0; i<b; i++)
+            n = n*10 + 1;
+        return n;
+    }
+};
diff --git a/Others/2417.Closest-Fair-Integer/Readme.md b/Others/2417.Closest-Fair-Integer/Readme.md
new file mode 100644
index 000000000..a4322a885
--- /dev/null
+++ b/Others/2417.Closest-Fair-Integer/Readme.md
@@ -0,0 +1,13 @@
+### 2417.Closest-Fair-Integer
+
+为了寻找最小的答案，我们尽量保持高位不变。假设“仅”令前k-1位不变，如何构造一个比n稍微大一点的数呢？
+
+显然将第k位增加1，后面剩余的位数置零（假设有count个），这就是最小的方案了。此时我们基于这个新数num，用一个函数`helper(num,count)`来寻找最接近的答案：如果此时前k位的digit的奇偶个数之差是diff，那么这个diff可以由后面count位来抵消，怎么抵消呢，显然应该从count拿出一部分置为1，一部分置为0，其中将所有的1放在末尾即可。于是，简单的“和差问题”即可求出最后的count位里需要填充多少个0和多少1能够满足条件。注意，也有可能无解。
+
+如果无解，我们需要尝试将第k位为增加2，继续调用之前的`helper(num,count)`。
+
+如果依然无解，那么不需要再增加第k位的数字了，因为增加3对于前k位的奇偶数字数目不会带来变化，等同于第一种情况。于是我们应该减少k，即“仅”保持前k-2位不变，再重复之前的步骤。
+
+最终本题一定有解。
+
+注意，我们首先要检查原数n本身是否是符合条件的。
diff --git a/Readme.md b/Readme.md
index 740a76c7c..dd6494bf0 100644
--- a/Readme.md
+++ b/Readme.md
@@ -1368,6 +1368,7 @@
 [357.Count-Numbers-with-Unique-Digits](https://github.com/wisdompeak/LeetCode/tree/master/Others/357.Count-Numbers-with-Unique-Digits) (M)     
 [2376.Count-Special-Integers](https://github.com/wisdompeak/LeetCode/tree/master/Others/2376.Count-Special-Integers) (M+)     
 [1012.Numbers-With-Repeated-Digits](https://github.com/wisdompeak/LeetCode/tree/master/Math/1012.Numbers-With-Repeated-Digits) (H-)    
+[2417.Closest-Fair-Integer](https://github.com/wisdompeak/LeetCode/tree/master/Others/2417.Closest-Fair-Integer) (H-)      
 
 #### [LeetCode Cup](https://github.com/wisdompeak/LeetCode/tree/master/LCCUP)
 [LCP23.魔术排列](https://github.com/wisdompeak/LeetCode/tree/master/LCCUP/2020Fall/LCP23.%E9%AD%94%E6%9C%AF%E6%8E%92%E5%88%97)