Efficient Pair Counting in C#
Intuition\nWhen tackling the problem of counting valid pairs in an array, the initial thought is to leverage dynamic programming to efficiently track the number
# Intuition\nWhen tackling the problem of counting valid pairs in an array, the initial thought is to leverage dynamic programming to efficiently track the number of valid sequences as we progress through the array. Instead of recalculating pair counts repeatedly, we store intermediate results and build upon them, optimizing both time and space.\n\n# Approach\nWe initialize a dynamic programming array dp to keep track of the number of valid pairs that can be formed for each possible value up to 1000 (assuming the elements in A are bounded by this value). As we iterate through the array, we update this dp array by considering the difference between consecutive elements. We update dp in such a way that it reflects the number of valid sequences up to the current element. By updating the array in place, we avoid unnecessary memory usage, making the solution more space-efficient.\n\n1.Initialization: Start by initializing the dp array with all elements set to 1, reflecting the base case where any element can form a valid sequence by itself.\n\n2.Iteration: For each element in A, calculate the difference d with the previous element and update the dp array: Accumulate the counts for valid sequences up to the current value. Adjust the dp array in reverse order to ensure that only valid sequences are considered.\n\n3.Final Computation: Sum all valid counts stored in the dp array to get the final result, taking care to apply modulo to prevent overflow.\n\n# Complexity\n- Time complexity:\nThe time complexity is O(n\xD7m), where n is the length of the array A, and m is the maximum value an element in A can take (1000 in this case). This results from iterating through each element in A and then updating the dp array.\n\n- Space complexity:\nThe space complexity is O(m), as we use a single dp array of size m to store the number of valid sequences.\n\n# Code\n```\npublic class Solution {\n    public int CountOfPairs(int[] A) {\n        int n = A.Length; // Get the length of the array A\n        int m = 1001; // Set m to 1001, the maximum possible value in A\n        int mod = 1000000007; // Modulo constant to prevent overflow\n        int[] dp = new int[m]; // Dynamic programming array to store intermediate results\n        Array.Fill(dp, 1); // Initialize the dp array with 1\n\n        for (int i = 1; i < n; ++i) { // Start iterating from the second element in A\n            int d = Math.Max(0, A[i] - A[i - 1]); // Calculate the difference d between A[i] and A[i-1]\n            for (int j = 1; j < A[i] + 1; ++j) { // Iterate over all possible values j from 1 to A[i]\n                dp[j] = (dp[j] + dp[j - 1]) % mod; // Update dp[j] as the sum of dp[j] and dp[j-1], modulo mod\n            }\n            for (int j = A[i]; j >= d; --j) { // Update dp array in reverse from A[i] down to d\n                dp[j] = dp[j - d]; // Set dp[j] to the value of dp[j-d]\n            }\n            for (int j = d - 1; j >= 0; --j) { // Set values in dp less than d to 0\n                dp[j] = 0;\n            }\n        }\n\n        int res = 0; // Initialize the result variable\n        for (int j = 0; j <= A[n - 1]; ++j) // Iterate over all possible values in dp\n            res = (res + dp[j]) % mod; // Sum up all valid pairs, modulo mod\n\n        return res; // Return the final result\n    }\n}\n```
['C#']