Oct20/21: daily challenge

Smallest ranges II, use sort to understand.

daily/Oct20.cc

@@ -0,0 +1,34 @@
+#include <iostream>
+using namespace std;
+
+class Solution {
+ public:
+ /**
+  * @brief LCQ1: sequence of strings appeared on screen
+  * Time: O(N^2), Space: O(1)
+  *
+  * @param target
+  * @return vector<string>
+  */
+  vector<string> stringSequence(string target) {
+    int len = target.size();
+    vector<string> ans;
+
+    for (int i = 0; i < len; i++) {
+      char ch = 'a';
+
+      while (target[i] - ch >= 0) {
+        string temp = target.substr(0, i);
+        temp += ch;
+        ans.push_back(temp);
+
+        ch++;
+      }
+    }
+
+    return ans;
+  }
+
+
+
+};

daily/Oct21.cc

@@ -0,0 +1,26 @@
+#include <iostream>
+using namespace std;
+
+class Solution {
+ public:
+  /**
+   * @brief LC910: Smallest Range II
+   * Time: O(NlogN), Sort is O(NlogN), Space: O(1)
+   *
+   * @param nums
+   * @param k
+   * @return int
+   */
+  int smallestRangeII(vector<int>& nums, int k) {
+    sort(nums.begin(), nums.end());
+    int ans = nums.back() - nums[0];
+
+    for (int i = 1; i < nums.size(); i++) {
+      int mx = max(nums[i - 1] + k, nums.back() - k);
+      int mn = min(nums[0] + k, nums[i] - k);
+      ans = min(ans, mx - mn);
+    }
+
+    return ans;
+  }
+};