Arithmetic Subarrays - Leetcode Solution

Problem Statement:

You are given an array of n integers, nums, and two arrays of m integers each, l and r, representing m queries.

For the ith query, compute the bitwise XOR of elements from indices l[i] to r[i] inclusive in nums. Return an array answering all queries.

The above problem can be solved effectively by utilizing the brute force approach and traversing over the queries in a nested loop to find the XOR of elements between the indices specified in each query. However, the brute force approach will have poor time complexity and will be inefficient in the case of larger input arrays. Therefore, we need a better approach to solve the given problem.

One way to solve the given problem is to create a two-dimensional array of prefix XORs representing the XOR of all elements from index 0 to i, where i represents the ith element in the original array nums. This can be done using dynamic programming (DP). The ith row of the prefix XOR array represents the cumulative XOR of all elements from index 0 to i.

For each query, we can utilize the prefix XOR array to compute the XOR of all elements between the indices l[i] and r[i]. We can apply the "prefix XOR" property to compute the overall XOR of all elements between the specified indices.

The prefix XOR property can be illustrated as follows:

Suppose we have an array arr[0...n-1], where X is the XOR of all elements between the indices i and j. Then X is equal to the XOR of the prefix XORs arr[0...i-1] and arr[0...j]. Mathematically speaking:

X = arr[0]^arr[1]^...^arr[i-1]^arr[i]^...^arr[j] = (arr[0]^arr[1]^...^arr[i-1])^(arr[0]^arr[1]^...^arr[j]) = (prefix[j+1]^prefix[i])

Using this approach, we can solve the given problem efficiently in O(n+m) time complexity, where n is the size of the nums array and m is the number of queries.

Implementation:

1. Create an array prefix of size n+1 and initialize prefix[0]=0.

2. Iterate over nums array and update prefix[i+1]=prefix[i]^nums[i], where i represents ith index in nums array.

3. Create an empty array res of size m.

4. Iterate over the queries array and compute res[i]=(prefix[r[i]+1]^prefix[l[i]]) for each query.

5. Return res array.

Code:

class Solution { public: vector<int> xorQueries(vector<int>& nums, vector<vector<int>>& queries) { int n= nums.size(); vector<int> prefix(n+1, 0); for(int i=0; i<n; i++) prefix[i+1]= prefix[i]^nums[i]; int m= queries.size(); vector<int> res(m, 0); for(int i=0; i<m; i++) res[i]= prefix[queries[i][1]+1]^prefix[queries[i][0]]; return res; } };

Time complexity: O(n+m), where n is the size of the nums array and m is the number of queries.

Space complexity: O(n), where n is the size of the nums array.

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