Java Clean Solution || Using DP+TABULATION APPROACH
Complexity\n- Time complexity:O(nK)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:O(2n)\n Add your space complexity here, e.g. O(n) \n\n# Co
# Complexity\n- Time complexity:$$O(n*K)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$$O(2*n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n\n    int MOD = 1_000_000_007;\n\n    public int maximumSumSubsequence(int[] nums, int[][] queries) {\n        int n = nums.length;\n\n        long[] inc = new long[n];\n        long[] exc = new long[n];\n\n        inc[0] = Math.max(0, nums[0]);\n        exc[0] = 0;\n\n        for (int i = 1; i < n; i++) {\n            inc[i] = Math.max(0, nums[i] + exc[i - 1]);\n            exc[i] = Math.max(inc[i - 1], exc[i - 1]);\n        }\n\n        long max = Math.max(inc[n - 1], exc[n - 1]);\n        long total = 0;\n        for (int[] query : queries) {\n            int pos = query[0];\n            int _val = query[1];\n\n            nums[pos] = _val;\n\n            if (pos == 0) {\n                inc[0] = Math.max(0, nums[0]);\n                exc[0] = 0;\n            }\n\n            for (int i = Math.max(1, pos); i < n; i++) {\n                if (i == pos) {\n                    inc[i] = Math.max(0, nums[i] + (i > 0 ? exc[i - 1] : 0));\n                } else {\n                    inc[i] = Math.max(0, nums[i] + exc[i - 1]);\n                }\n                exc[i] = Math.max(inc[i - 1], exc[i - 1]);\n            }\n\n            max = Math.max(inc[n - 1], exc[n - 1]);\n            total = (total + max) % MOD;\n        }\n        return (int) total;\n    }\n}\n```
['Array', 'Dynamic Programming', 'Recursion', 'Memoization', 'Java']