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This approach involves using bit manipulation to find out which bits are different between the XOR of the array and k. For each differing bit, calculate the minimum number of flips needed across all elements to switch that bit in the resultant XOR.
Time Complexity: O(n), where n is the length of the array, since we only traverse it twice.
Space Complexity: O(1), as no additional data structures are used.
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JavaScript uses similar logic by looping over the array and calculating the XOR difference with k to count necessary bit flips.
In this approach, the focus is on counting the set bits of two numbers (XOR sum and k) - i.e., how many bits are 1 in their binary representation. We aim to equalize the set bits count by flipping specific bits in nums.
Time Complexity: O(n + log(max(nums[i])))
Space Complexity: O(1).
1def countSetBits(n):
2 count = 0
3 while n:
4 count += n & 1
5 n >>= 1
6 return count
7
8
9def minOperations(nums, k):
10 xor_sum = 0
11 for num in nums:
12 xor_sum ^= num
13 set_bits_xor = countSetBits(xor_sum)
14 set_bits_k = countSetBits(k)
15 return abs(set_bits_xor - set_bits_k)
16
17
18# Example usage
19nums = [2, 1, 3, 4]
20k = 1
21print(minOperations(nums, k))
22Contains a function to count set bits which is used to determine the difference needed in the number of operations.