Solution as on 09-07-2022 10:00 am

Author: knockcat

1696. Jump Game VI.cpp

@@ -0,0 +1,60 @@
+// 1696.✅ Jump Game VI
+
+/*
+You are given a 0-indexed integer array nums and an integer k.
+
+You are initially standing at index 0. In one move, you can jump at most k steps forward without going outside the boundaries of the array. That is, you can jump from index i to any index in the range [i + 1, min(n - 1, i + k)] inclusive.
+
+You want to reach the last index of the array (index n - 1). Your score is the sum of all nums[j] for each index j you visited in the array.
+
+Return the maximum score you can get.
+
+
+
+Example 1:
+
+Input: nums = [1,-1,-2,4,-7,3], k = 2
+Output: 7
+Explanation: You can choose your jumps forming the subsequence [1,-1,4,3] (underlined above). The sum is 7.
+Example 2:
+
+Input: nums = [10,-5,-2,4,0,3], k = 3
+Output: 17
+Explanation: You can choose your jumps forming the subsequence [10,4,3] (underlined above). The sum is 17.
+Example 3:
+
+Input: nums = [1,-5,-20,4,-1,3,-6,-3], k = 2
+Output: 0
+
+
+Constraints:
+
+1 <= nums.length, k <= 105
+-104 <= nums[i] <= 104
+*/
+
+class Solution
+{
+public:
+    int maxResult(vector<int> &nums, int k)
+    {
+        deque<int> q; // O(k)
+        int n = nums.size();
+        vector<int> dp(n); // O(n)
+
+        dp[n - 1] = nums[n - 1];
+
+        q.push_back(n - 1);
+
+        for (int i = n - 2; i >= 0; --i) // O(k)
+        {
+            if (q.front() - i > k)
+                q.pop_front();
+            dp[i] = nums[i] + dp[q.front()];
+            while (q.size() && dp[q.back()] < dp[i])
+                q.pop_back();
+            q.push_back(i);
+        }
+        return dp[0];
+    }
+};