Add C++ solution of problem #1400

Author: Cuda-Chen

solution.cpp

@@ -0,0 +1,20 @@
+// math (referenced from LeetCode Hint in this problem)
+// Count the # characters with odd counts then check
+// whether it is less than or equal to k.
+// Edge case: if s.length < k then just return false.
+
+class Solution {
+public:
+    bool canConstruct(string s, int k) {
+        if(s.size() < k)
+            return false;
+
+        vector<int> cnt(26);
+        for(char c : s)
+            cnt[c - 'a']++;
+        int odd_cnt = 0;
+        for(int i = 0; i < 26; i++)
+            odd_cnt += (cnt[i] % 2);
+        return odd_cnt <= k;
+    }
+};