<p>You are given two <strong>0-indexed</strong> integer arrays <code>nums1</code> and <code>nums2</code>, of equal length <code>n</code>.</p>

<p>In one operation, you can swap the values of any two indices of <code>nums1</code>. The <strong>cost</strong> of this operation is the <strong>sum</strong> of the indices.</p>

<p>Find the <strong>minimum</strong> total cost of performing the given operation <strong>any</strong> number of times such that <code>nums1[i] != nums2[i]</code> for all <code>0 &lt;= i &lt;= n - 1</code> after performing all the operations.</p>

<p>Return <em>the <strong>minimum total cost</strong> such that </em><code>nums1</code> and <code>nums2</code><em> satisfy the above condition</em>. In case it is not possible, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [1,2,3,4,5], nums2 = [1,2,3,4,5]
<strong>Output:</strong> 10
<strong>Explanation:</strong> 
One of the ways we can perform the operations is:
- Swap values at indices 0 and 3, incurring cost = 0 + 3 = 3. Now, nums1 = [4,2,3,1,5]
- Swap values at indices 1 and 2, incurring cost = 1 + 2 = 3. Now, nums1 = [4,3,2,1,5].
- Swap values at indices 0 and 4, incurring cost = 0 + 4 = 4. Now, nums1 =[5,3,2,1,4].
We can see that for each index i, nums1[i] != nums2[i]. The cost required here is 10.
Note that there are other ways to swap values, but it can be proven that it is not possible to obtain a cost less than 10.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [2,2,2,1,3], nums2 = [1,2,2,3,3]
<strong>Output:</strong> 10
<strong>Explanation:</strong> 
One of the ways we can perform the operations is:
- Swap values at indices 2 and 3, incurring cost = 2 + 3 = 5. Now, nums1 = [2,2,1,2,3].
- Swap values at indices 1 and 4, incurring cost = 1 + 4 = 5. Now, nums1 = [2,3,1,2,2].
The total cost needed here is 10, which is the minimum possible.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [1,2,2], nums2 = [1,2,2]
<strong>Output:</strong> -1
<strong>Explanation:</strong> 
It can be shown that it is not possible to satisfy the given conditions irrespective of the number of operations we perform.
Hence, we return -1.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == nums1.length == nums2.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums1[i], nums2[i] &lt;= n</code></li>
</ul>

from collections import Counter

class Solution:
    def minimumTotalCost(self, nums1, nums2):
        n = len(nums1)
        
        # Step 1: Identify indices where nums1[i] == nums2[i]
        same_indices = {i for i in range(n) if nums1[i] == nums2[i]}
        
        # If there are no such indices, no swaps are needed, return 0
        if not same_indices:
            return 0
        
        # Step 2: Count the frequency of each number in nums1 at the same indices
        count = Counter(nums1[i] for i in same_indices)
        
        # Step 3: Find the most frequent number (major) in the same_indices
        major = max(count, key=count.get)
        
        # Step 4: Calculate how many more indices need to be added to avoid major being too frequent
        to_add = max(count[major] * 2 - len(same_indices), 0)
        
        # Step 5: Add indices to same_indices to balance the frequency of major
        for i in range(n):
            if to_add and nums1[i] != major != nums2[i] and i not in same_indices:
                to_add -= 1
                same_indices.add(i)
        
        # Step 6: If we still need to add more indices, it's not possible to satisfy the condition
        if to_add:
            return -1
        
        # Step 7: Return the sum of indices in same_indices as the minimum cost
        return sum(same_indices)

# Explanation:
# - We first identify indices where nums1 and nums2 have the same values.
# - If there are no such indices, the arrays are already satisfying the condition, so the cost is 0.
# - We count the frequency of each number in nums1 at these indices to find the most frequent one (major).
# - If the major number appears too frequently, we need to add more indices to balance it.
# - We add indices where neither nums1[i] nor nums2[i] is the major number until the frequency condition is met.
# - If we can't balance the frequency, return -1. Otherwise, return the sum of the indices as the cost.
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