✅ Detailed explanation of implementation in Swift
Intuition\nTo solve the problem of finding the maximum sum of a non-adjacent subsequence for each query efficiently, we can use dynamic programming. The idea is
## Intuition\nTo solve the problem of finding the maximum sum of a non-adjacent subsequence for each query efficiently, we can use **dynamic programming**. The idea is to track the maximum sums that include and exclude the previous elements and update these values as we process each query.\n\n## Approach\n1. **Initialize Variables**:\n   - Copy the input array `nums` to modify it as per the queries.\n   - Set a `modulus` constant to handle large sums and prevent overflow.\n   - Initialize `totalSum` to accumulate the results for each query.\n\n2. **Process Each Query**:\n   - For each query, update the value at the specified position in the array.\n   - Calculate the maximum sum of a non-adjacent subsequence using a helper function `calculateMaxSum`.\n\n3. **Dynamic Programming in `calculateMaxSum`**:\n   - Use two variables, `includePrevious` and `excludePrevious`, to keep track of the maximum sums including and excluding the previous element, respectively.\n   - Iterate through the array and update these variables:\n     - `newInclude` stores the sum when the current element is included.\n     - `excludePrevious` is updated to the maximum of the previous `includePrevious` and `excludePrevious`.\n     - `includePrevious` is updated to `newInclude`.\n   - Return the maximum of `includePrevious` and `excludePrevious` as the result for the current state of the array.\n\n4. **Accumulate Results**:\n   - After processing each query, add the result of the current state of the array to `totalSum`.\n   - Use modulus operation to handle large sums and prevent overflow.\n\n## Complexity\n- **Time complexity**: \n  - Each query requires recalculating the maximum sum of a non-adjacent subsequence which takes $$O(n)$$ time.\n  - With $$q$$ queries, the total time complexity is $$O(n \\times q)$$.\n\n- **Space complexity**:\n  - The space complexity is $$O(1)$$ since we only use a few extra variables to track the sums and perform the calculations.\n\n![Screenshot 2024-05-27 at 00.21.01.png](https://assets.leetcode.com/users/images/57b41cf1-716f-42e6-9663-70fb7bfc60b0_1716758488.1851132.png)\n\n## Code\n\n```swift []\nclass Solution {\n    func maximumSumSubsequence(_ nums: [Int], _ queries: [[Int]]) -> Int {\n        var nums = nums\n        let modulus = 1_000_000_007\n        var totalSum: Int64 = 0\n        \n        // Iterate through each query\n        for query in queries {\n            let position = query[0]\n            let newValue = query[1]\n            \n            // Update the value at the specified position\n            nums[position] = newValue\n            \n            // Calculate the maximum sum of a non-adjacent subsequence\n            let currentSum = calculateMaxSum(nums)\n            \n            // Add the current sum to the total sum\n            totalSum = (totalSum + currentSum) % Int64(modulus)\n        }\n        \n        return Int(totalSum)\n    }\n    \n    // Function to calculate the maximum sum of a non-adjacent subsequence\n    private func calculateMaxSum(_ array: [Int]) -> Int64 {\n        var includePrevious: Int64 = 0 // Max sum including the previous element\n        var excludePrevious: Int64 = 0 // Max sum excluding the previous element\n        \n        for num in array {\n            // Calculate new value including the current element\n            let newInclude = Int64(num) + excludePrevious\n            \n            // Update exclude to be the maximum of previous include and exclude\n            excludePrevious = max(includePrevious, excludePrevious)\n            \n            // Update include to the new calculated value\n            includePrevious = newInclude\n        }\n        \n        // Return the maximum of including or excluding the last element\n        return max(includePrevious, excludePrevious)\n    }\n}\n```
['Dynamic Programming', 'Swift']