Dynamic Programming Solution (House Robber)
Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem appears to be a combination of two classic Dynamic Programming challenges. 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem appears to be a combination of two classic Dynamic Programming challenges. Firstly, because we need to skip every alternate position, it resembles the House Robber problem. Secondly, to maximize the sum of the subsequence, we can apply a strategy similar to Kadane\'s Algorithm.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe approach involves two main steps. First, update the array based on the given query. Second, find the maximum sum of the array using a dynamic programming technique similar to the House Robber problem, and then add this maximum sum to the total sum.\n\n# Complexity\n- Time complexity: It has nested for-loop with outer loop running $$m$$ times, where $$m$$ is the number of queries and inner loop running $$n - 1$$ times, where $$n$$ is the size of the array nums. Hence, the time complexity is $$O(m*n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: Additional space for only three variables is required. Hence, the space complexity is $$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n    public int maximumSumSubsequence(int[] nums, int[][] queries) {\n        int sum = 0;\n        \n        for (int i = 0; i < queries.length; i++) {\n            nums[queries[i][0]] = queries[i][1];\n            \n            int a = -1;\n            int b = nums[0];\n            \n            for (int j = 1; j < nums.length; j++) {\n                int temp = nums[j];\n\n                if (j != 1 && a > 0) {\n                    temp += a;\n                }\n                \n                temp = Math.max(temp, b);\n\n                a = b;\n                b = temp;\n            }\n            \n            sum = Math.max(sum, sum + b)  % 1000000007;\n        }\n        \n        return sum;\n    }\n}\n```
['Array', 'Dynamic Programming', 'Java']