🚀 Efficient Approach for Handling Queries with Dynamic Programming 📊
\n\n# Approach\nI have used a dynamic programming approach to solve this problem efficiently. The main idea is to maintain a dp array that keeps track of the ma
\n\n# Approach\nI have used a dynamic programming approach to solve this problem efficiently. The main idea is to maintain a `dp` array that keeps track of the maximum sum possible for each subsequence considering non-adjacent elements.\n\n\n- **Dynamic Programming**: Used to calculate the maximum sum of a subsequence with non-adjacent elements.\n- **Handling Queries**: Each query updates an element, and the `dp` array is recalculated to find the new maximum sum.\n- **Modulo Operation**: The result is returned modulo \\(10^9 + 7\\) to handle large sums.\n\n\n# Complexity\nTime Complexity: O(q * n), where q is the number of queries and n is the length of the array nums.\n\n\nSpace Complexity: O(n) for the dynamic programming array dp.\n\n# Code\n```\nclass Solution {\n    int MOD = 1000000007;\n    public int maximumSumSubsequence(int[] nums, int[][] queries) {\n        int maxSum = 0;\n        for (int[] q : queries) {\n            nums[q[0]] = q[1];\n            maxSum = (maxSum + maxSubSeq(nums)) % MOD;\n        }\n        return maxSum;\n        \n    }\n    int maxSubSeq(int[] nums) {\n         if (nums == null || nums.length == 0) return 0;\n        int n = nums.length;\n        if (n == 1) return Math.max(0, nums[0]);\n        int[] dp = new int[n];\n        dp[0] = Math.max(0, nums[0]);\n        dp[1] = Math.max(dp[0], nums[1]);\n\n        for (int i = 2; i < n; i++) {\n            dp[i] = Math.max(dp[i - 1], nums[i] + dp[i - 2]);\n        }\n        \n        return dp[n - 1];\n        \n    }\n}\n```
['Dynamic Programming', 'Java']