<ul><li>The problem is to create a third list, nums3, by taking the XOR of one number from nums1 with one number from nums2.</li><li>The naive approach calculates the XOR for every pair of numbers from nums1 and nums2, which is computationally expensive for large inputs.</li><li>The optimized solution improves efficiency by taking advantage of the mathematical properties of XOR and avoids explicitly iterating over all pairs.</li><li>The approach has a time complexity of O(n+m), where n and m are the lengths of nums1 and nums2, respectively.</li></ul>

Leetcode Daily: Bitwise XOR of All Pairings

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