Easy java solution using dynamic programming
Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem can be solved using dynamic programming, similar to the classic problem of 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem can be solved using dynamic programming, similar to the classic problem of finding the maximum sum of a subsequence with non-adjacent elements. We need to update the nums array according to each query and then compute the maximum sum of a subsequence with non-adjacent elements.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Iterate over each query, update the nums array at the specified position.\n2. After each update, calculate the maximum sum of a subsequence with non-adjacent elements using dynamic programming.\n3. Sum up the results of each query, ensuring the result stays within the modulo 10^9 + 7.\n\n# Complexity\n- Time complexity: The time complexity for each query is O(n), where n is the length of nums. Thus for m queries, the overall time complexity is O(n * m)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1) for the additional space used.\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n    public int maximumSumSubsequence(int[] nums, int[][] q) {\n        long ans = 0;\n        int MOD = 1000000007;\n        for(int i=0; i<q.length; i++) {\n            nums[q[i][0]] = q[i][1];\n            ans += maxSubSeq(nums);\n            ans %= MOD;\n        }\n        \n        return (int) ans;\n    }\n    \n    private long maxSubSeq(int[] nums) {\n        int n = nums.length;\n        if (n == 0) return 0;\n        if (n == 1) return Math.max(0, nums[0]);\n        \n        long prev2 = Math.max(0, nums[0]);\n        long prev1 = Math.max(prev2, nums[1]);\n        \n        for (int i = 2; i < n; i++) {\n            long curr = Math.max(prev1, prev2 + nums[i]);\n            prev2 = prev1;\n            prev1 = curr;\n        }\n        \n        return prev1;\n    }\n}\n```
['Dynamic Programming', 'Java']