Given an array A of size N. The value of an array is denoted by the bit-wise XOR of all elements it contains. Find the bit-wise XOR of the values of all subarrays of A.
Question
Given an array A of size N. The value of an array is denoted by the bit-wise XOR of all elements it contains. Find the bit-wise XOR of the values of all subarrays of A.
Solution
To find the bit-wise XOR of the values of all subarrays of array A, we can follow these steps:
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Initialize a variable, let's call it "result", to 0. This variable will store the final XOR value.
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Iterate through each element in the array A.
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For each element, iterate through all possible subarrays starting from that element.
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For each subarray, calculate the XOR value by performing a bit-wise XOR operation on all the elements in that subarray.
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Update the "result" variable by performing a bit-wise XOR operation between the current value of "result" and the XOR value of the subarray.
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Repeat steps 3 to 5 for all elements in the array.
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After iterating through all elements and subarrays, the final value of "result" will be the bit-wise XOR of the values of all subarrays of array A.
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Return the final value of "result" as the answer.
Note: The time complexity of this approach is O(N^3), where N is the size of the array A.
Similar Questions
You are given an array 𝐴A of size 𝑁N.You can rearrange the elements of 𝐴A as you want. After that, construct an array 𝐵B of size 𝑁N as:𝐵1=𝐴1B 1 =A 1 ;𝐵𝑖=𝐵𝑖−1⊕𝐴𝑖B i =B i−1 ⊕A i for all 2≤𝑖≤𝑁2≤i≤N, where ⊕⊕ denotes the bitwise XOR operator.Find a rearrangement of 𝐴A such that 𝐵𝑖≠0B i =0 for all 1≤𝑖≤𝑁1≤i≤N.Input FormatThe first line of input will contain a single integer 𝑇T, denoting the number of test cases.Each test case consists of multiple lines of input.The first line of each test case contains 𝑁N, the size of the array 𝐴A.The second line of each test case contains 𝑁N space-separated integers : 𝐴1,𝐴2,…,𝐴𝑁A 1 ,A 2 ,…,A N .Output FormatFor each test case, output on a new line, 𝑁N space-separated integers denoting the rearrangement of 𝐴A satisfying the conditions.In case no such rearrangement exists, print −1−1 instead.Constraints1≤𝑇≤2⋅1041≤T≤2⋅10 4 1≤𝑁≤2⋅1051≤N≤2⋅10 5 1≤𝐴𝑖≤1091≤A i ≤10 9 The sum of 𝑁N over all test cases does not exceed 2⋅1052⋅10 5 .Sample 1:InputOutput431 2 351 2 3 4 531 2 211-11 2 4 3 51 2 21Explanation:Test case 11 : It can be shown that it's impossible to rearrange 𝐴A in a valid way.Test case 22 : A possible rearrangement of array 𝐴A is [1,2,4,3,5][1,2,4,3,5] and the corresponding array 𝐵B is [1,3,7,4,1][1,3,7,4,1]. Here 𝐵𝑖≠0B i =0 for all 1≤𝑖≤𝑁1≤i≤N.
You are given an array of integers. Find the sum of XOR of all pairs formed by the elements of the array.Input FormatThe first line of input contains T - the number of test cases. It's followed by 2T lines, the first line contains N - the size of the array. The second line contains the elements of the array.Output FormatFor each test case, print the sum of XOR of all pairs formed by the elements of the array, separated by a new line.Constraints20 points1 <= T <= 1001 <= N <= 10000 <= A[i] <= 10580 points1 <= T <= 1001 <= N <= 1050 <= A[i] <= 105ExampleInput335 12 854 10 54 11 8615 35 25 10 15 12Output52560680ExplanationTest-Case 1(5 ^ 5) = 0(5 ^ 12) = 9(5 ^ 8) = 13(12 ^ 5) = 9(12 ^ 12) = 0(12 ^ 8) = 4(8 ^ 5) = 13(8 ^ 12) = 4(8 ^ 8) = 0The sum of all the above xor products = 52
You are given an array nums, where each number in the array appears either once or twice.Return the bitwise XOR of all the numbers that appear twice in the array, or 0 if no number appears twice. Example 1:Input: nums = [1,2,1,3]Output: 1Explanation:The only number that appears twice in nums is 1.Example 2:Input: nums = [1,2,3]Output: 0Explanation:No number appears twice in nums.Example 3:Input: nums = [1,2,2,1]Output: 3Explanation:Numbers 1 and 2 appeared twice. 1 XOR 2 == 3. Constraints:1 <= nums.length <= 501 <= nums[i] <= 50Each number in nums appears either once or twice.
Given N, print the XOR of all numbers between (1-N).
Given an array Arr of N positive integers. Your task is to find the elements whose value is equal to that of its index value ( Consider 1-based indexing ).
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