Count Triplets That Can Form Two Arrays of Equal XOR | Leetcode 1442 | codestorywithMIK
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This is the 16th Video of our Playlist "Bit Manipulation : Popular Interview Problems" by codestorywithMIK
In this video we will try to solve a good Bit Manipulation Problem : Count Triplets That Can Form Two Arrays of Equal XOR | Multiple Approaches | Leetcode 1442 | codestorywithMIK
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We will do live coding after explanation and see if we are able to pass all the test cases.
Also, please note that my Github solution link below contains both C++ as well as JAVA code.
Problem Name : Count Triplets That Can Form Two Arrays of Equal XOR | Multiple Approaches | Leetcode 1442 | codestorywithMIK
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Summary :
Initialization:
We create a prefixXor array of length arr.length + 1 to store the cumulative XOR values, with the first element initialized to 0.
Computing the prefix XOR:
The for-loop computes the cumulative XOR up to each index i and stores it in the prefixXor array. This is equivalent to the C++ operation with the prefix XOR computation.
Counting the triplets:
Nested loops iterate over all pairs (i, k) where i less than k. If prefixXor[k] is equal to prefixXor[i], it means the XOR of the subarray arr[i] to arr[k-1] is 0. We then count the number of valid j values (i.e., j between i and k), which is k - i - 1.
✨ Timelines✨
00:00 - Introduction
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