Java Solution
Intuition\nTo find the maximum sum subsequence where no two elements are adjacent, we can use dynamic programming. Given multiple queries that update the array,
# Intuition\nTo find the maximum sum subsequence where no two elements are adjacent, we can use dynamic programming. Given multiple queries that update the array, we need to efficiently recompute the maximum sum after each update.\n\n# Approach\n1. **Preprocessing**:\n   - Convert all negative numbers in the array to 0, since they don\'t contribute positively to the sum.\n   - Initialize a DP array to store the maximum sum subsequence up to each index.\n\n2. **Dynamic Programming**:\n   - Use a helper function to compute the maximum sum subsequence using the DP approach.\n   - If there are more than one element, initialize the first two elements of the DP array based on the input array.\n\n3. **Processing Queries**:\n   - For each query, update the specified index in the array and recompute the maximum sum subsequence.\n   - Accumulate the results of each query to get the final result.\n\n4. **Modular Arithmetic**:\n   - Ensure the result remains within the bounds of \\(10^9 + 7\\) by using modular arithmetic.\n\n# Complexity\n- **Time Complexity**: \n  - Preprocessing takes \\(O(n)\\).\n  - Each query processing takes \\(O(n)\\) due to recomputation of the DP array.\n  - The overall time complexity is \\(O(q \\times n)\\) where \\(q\\) is the number of queries and \\(n\\) is the length of the array.\n\n- **Space Complexity**: \n  - \\(O(n)\\) for the DP array and input array.\n\n# Code\n```java\nimport java.util.Arrays;\n\nclass Solution {\n    int mod = 1000000000 + 7;\n    \n    public int maximumSumSubsequence(int[] nums, int[][] queries) {\n        long sum = 0;\n        \n        // Preprocess the array: replace negative values with 0\n        for(int i = 0; i < nums.length; i++) {\n            nums[i] = Math.max(nums[i], 0);\n        }\n        \n        int n = nums.length;\n        long[] dp = new long[n];\n        \n        // Compute the initial maximum sum subsequence\n        if(n > 1) {\n            maxSumSubsequence(nums, dp, n, 0);\n        }\n        \n        // Process each query\n        for(int[] q: queries) {\n            nums[q[0]] = Math.max(q[1], 0);\n            sum = (sum + maxSumSubsequence(nums, dp, n, q[0])) % mod;\n        }\n        \n        return (int) sum;\n    }\n\n    public static long maxSumSubsequence(int[] nums, long[] dp, int n, int i) {\n        if (n == 1) return nums[0];\n        \n        // Initialize the first two elements of the DP array\n        if(i <= 1) {\n            dp[0] = nums[0];\n            dp[1] = Math.max(nums[0], nums[1]);\n            i = 2;\n        }\n        \n        // Fill the DP array\n        for (; 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
['Java']