Minimum Bit Flips To Convert Number - Leetcode Solution

Problem Statement:

You are given two integer arrays nums1 and nums2 of equal length consisting of binary digits (0 and 1).

You can perform the following operation any number of times:

Swap any two elements of nums1. Swap any two elements of nums2. For example, if nums1 = [1,2,3,4] and we swapped the second and third element, the array would become nums1 = [1,3,2,4]. Return the minimum number of times you can perform the operation so that nums1 and nums2 are equal. Equality is defined as follows:

nums1[i] == nums2[i] for every 0 <= i < nums1.length.

Solution:

We can solve this problem using a greedy approach. We can iterate through the two input arrays and check if the corresponding elements of the two arrays are equal.

If the corresponding elements are not equal, we need to perform a swap operation on either nums1 or nums2 to make them equal.

We can keep a counter variable to keep track of the number of swap operations we need to perform. If it is not possible to make nums1 and nums2 equal, we return -1.

The algorithm will be as follows:

1. Initialize a counter variable to 0.
2. Iterate through the two input arrays nums1 and nums2.
3. If the corresponding elements of the two arrays are not equal, perform a swap operation on either nums1 or nums2 to make them equal.
4. Increment the counter variable.
5. If it is not possible to make nums1 and nums2 equal, return -1.
6. Return the counter variable.

Implementation:

Here is the Python implementation of the algorithm:

def minFlips(nums1: list[int], nums2: list[int]) -> int: n = len(nums1) count = 0 for i in range(n): if nums1[i] != nums2[i]: if nums1[i] == 0 and nums2[i] == 1: # swap operation on nums1 if i < n-1 and nums1[i+1] != nums2[i+1]: nums1[i], nums1[i+1] = nums1[i+1], nums1[i] else: nums1[i] = nums2[i] count += 1 elif nums1[i] == 1 and nums2[i] == 0: # swap operation on nums2 if i < n-1 and nums1[i+1] != nums2[i+1]: nums2[i], nums2[i+1] = nums2[i+1], nums2[i] else: nums2[i] = nums1[i] count += 1 else: # nums1[i] and nums2[i] both are 0 count += 1 if nums1 == nums2: return count else: return -1

Time Complexity:

The time complexity of the algorithm is O(n), where n is the length of the input arrays nums1 and nums2.

Space Complexity:

The space complexity of the algorithm is O(1) because we are not using any extra space other than the input arrays.

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