<a target="_blank" href="https://leetcode.com/problems/minimum-number-of-operations-to-make-array-xor-equal-to-k/description/">https://leetcode.com/problems/minimum-number-of-operations-to-make-array-xor-equal-to-k/description/</a>
<ul>
<li>Firstly, we find XOR of original array
</li>
<li>Find binary of k
</li>
<li>Find different bits count b/w binary of k and binary of XOR(nums)
</li>
</ul>
<pre><code class="lang-python">class Solution:
 def minOperations(self, nums: List[int], k: int) -&gt; int:
 oxor = 0
 for e in nums:
 oxor ^= e

 binK = bin(k)[2:]
 binO = bin(oxor)[2:]
 maxLength = max(len(binO), len(binK))

 b1 = binK.zfill(maxLength)
 b2 = binO.zfill(maxLength)

 ans = 0
 for i in range(maxLength):
 if b1[i] != b2[i]:
 ans += 1

 return ans
</code></pre>

[https://leetcode.com/problems/minimum-number-of-operations-to-make-array-xor-equal-to-k/description/](https://leetcode.com/problems/minimum-number-of-operations-to-make-array-xor-equal-to-k/description/)

* Firstly, we find XOR of original array
    
* Find binary of k
    
* Find different bits count b/w binary of k and binary of XOR(nums)
    

```python
class Solution:
    def minOperations(self, nums: List[int], k: int) -> int:
        oxor = 0
        for e in nums:
            oxor ^= e

        binK = bin(k)[2:]
        binO = bin(oxor)[2:]
        maxLength = max(len(binO), len(binK))

        b1 = binK.zfill(maxLength)
        b2 = binO.zfill(maxLength)

        ans = 0
        for i in range(maxLength):
            if b1[i] != b2[i]:
                ans += 1
        
        return ans
```

Minimum Number of Operations to Make Array XOR Equal to K