June14: DP state transition w/ fixed size array

Time: O(N), Space: O(1), DP practice

daily/June14.cpp

@@ -0,0 +1,43 @@
+#include <bits/stdc++.h>
+
+#include <limits>
+using namespace std;
+
+class Solution {
+ public:
+  /**
+   * @brief LC: 2786: Visit Array positions to max score
+   * This is a DP problem, with optimized solution
+   * Time: O(N), Space: O(1)
+   * a vector w/ fixed size like 2 is Space complexity of O(1)
+   * But a vector space complexity is usually O(N)
+   *
+   * @param nums
+   * @param x
+   * @return long long
+   */
+  long long maxScore(vector<int>& nums, int x) {
+    // vector<long long> dp = {INT_MIN, INT_MIN};
+    vector<long long> dp(2, numeric_limits<int>::min());
+    dp[nums[0] % 2] = nums[0];
+    for (int i = 1; i < nums.size(); i++) {
+      int state = nums[i] % 2;
+      dp[state] = max(dp[state] + nums[i], dp[abs(state - 1)] + nums[i] - x);
+    }
+    return max(dp[0], dp[1]);
+  }
+
+  // long long maxScoreOP(vector<int>& nums, int x) {
+  // bool isEven = nums[0] % 2 == 0;
+  // long long oddMax = 0, evenMax = 0;
+  // if (isEven) evenMax = nums[0];
+  // else oddMax = nums[0];
+
+  // for (size_t i = 1; i < nums.size(); i++) {
+  //   isEven = nums[i] % 2 == 0;
+  //   if (isEven) evenMax = max(evenMax + nums[i], oddMax + nums[i] - x);
+  //   else oddMax = max(oddMax + nums[i], evenMax + nums[i] - x);
+  // }
+  // return max(oddMax, evenMax);
+  // }
+};