update: May 6 and May 10

daily/May10.cpp

@@ -0,0 +1,46 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+class Solution {
+ public:
+ /**
+  * @brief Use the 2-loop
+  * Time: O(N^2)
+  * 
+  * @param batteryPercentages 
+  * @return int 
+  */
+  int countTestedDevices(vector<int>& batteryPercentages) {
+    int res = 0;
+    int i = 0;
+    int len = batteryPercentages.size();
+    while (i < len) {
+      if (batteryPercentages[i] > 0) {
+        res++;
+        for (int j = i; j < len; j++) {
+          batteryPercentages[j] = max(0, batteryPercentages[j] - 1);
+        }
+      }
+
+      i++;
+    }
+
+    return res;
+  }
+
+  /**
+   * @brief This is optimized version
+   * Time: O(N)
+   * 
+   * @param batteryPercentages 
+   * @return int 
+   */
+  int countTestedDevicesOP(vector<int>& batteryPercentages) {
+    int res = 0;
+    for (int x : batteryPercentages) {
+      res += res < x;
+    }
+
+    return res;
+  }
+};

daily/May6.cpp

@@ -0,0 +1,26 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+class Solution {
+ public:
+  int cherryPickup(vector<vector<int>>& grid) {
+    int n = grid.size();
+    vector<vector<int>> f(n + 1, vector<int>(n + 1, INT_MIN));
+    f[1][1] = grid[0][0];
+    for (int t = 1; t < n * 2 - 1; t++) {
+      for (int j = min(t, n - 1); j >= max(t - n + 1, 0); j--) {
+        for (int k = min(t, n - 1); k >= j; k--) {
+          if (grid[t - j][j] < 0 || grid[t - k][k] < 0) {
+            f[j + 1][k + 1] = INT_MIN;
+          } else {
+            f[j + 1][k + 1] =
+                max({f[j + 1][k + 1], f[j + 1][k], f[j][k + 1], f[j][k]}) +
+                grid[t - j][j] + (k != j ? grid[t - k][k] : 0);
+          }
+        }
+      }
+    }
+
+    return max(f[n][n], 0);
+  }
+};