<p>A <strong>swap</strong> is defined as taking two <strong>distinct</strong> positions in an array and swapping the values in them.</p>

<p>A <strong>circular</strong> array is defined as an array where we consider the <strong>first</strong> element and the <strong>last</strong> element to be <strong>adjacent</strong>.</p>

<p>Given a <strong>binary</strong> <strong>circular</strong> array <code>nums</code>, return <em>the minimum number of swaps required to group all </em><code>1</code><em>&#39;s present in the array together at <strong>any location</strong></em>.</p>

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

<pre>
<strong>Input:</strong> nums = [0,1,0,1,1,0,0]
<strong>Output:</strong> 1
<strong>Explanation:</strong> Here are a few of the ways to group all the 1&#39;s together:
[0,<u>0</u>,<u>1</u>,1,1,0,0] using 1 swap.
[0,1,<u>1</u>,1,<u>0</u>,0,0] using 1 swap.
[1,1,0,0,0,0,1] using 2 swaps (using the circular property of the array).
There is no way to group all 1&#39;s together with 0 swaps.
Thus, the minimum number of swaps required is 1.
</pre>

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

<pre>
<strong>Input:</strong> nums = [0,1,1,1,0,0,1,1,0]
<strong>Output:</strong> 2
<strong>Explanation:</strong> Here are a few of the ways to group all the 1&#39;s together:
[1,1,1,0,0,0,0,1,1] using 2 swaps (using the circular property of the array).
[1,1,1,1,1,0,0,0,0] using 2 swaps.
There is no way to group all 1&#39;s together with 0 or 1 swaps.
Thus, the minimum number of swaps required is 2.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,1,0,0,1]
<strong>Output:</strong> 0
<strong>Explanation:</strong> All the 1&#39;s are already grouped together due to the circular property of the array.
Thus, the minimum number of swaps required is 0.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>nums[i]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")

static const bool Booster = [](){
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    return true;
}();

int helper(vector<int>& nums) {
    int n = nums.size();
    int ones = 0;
    for(auto& i : nums) ones += i;
    if(ones == 0 || ones == n) return 0;
    
    int cnt = 0; // 1st window
    for(int i = 0; i < ones; i++) cnt += nums[i];
    int res = cnt;
    for(int i = 0; i < n; i++) {
        cnt -= nums[i];
        cnt += nums[(i + ones) % n];
        res = max(res, cnt);
    }
    
    return ones - res;
}

void parse_input_and_solve(std::ofstream& out, std::string& s) {
    s.erase(s.begin());
    s.pop_back();
    string w;
    istringstream iss(s);
    vector<int> nums;
    while(getline(iss, w, ',')) {
        nums.push_back(stoi(w));
    }

    out << helper(nums) << endl;
}

bool Solve = [](){
    std::ofstream out("user.out");
    for (std::string s; std::getline(std::cin, s);) {
        parse_input_and_solve(out, s);
    }
    out.flush();
    exit(0);
    return true;
}();

class Solution {
public:
    int minSwaps(vector<int>& nums) {
        int numOne = accumulate(nums.begin(), nums.end(), 0);
        int size = nums.size();
        int currentNumOne = accumulate(nums.begin(), nums.begin()+numOne, 0);
        int maxOne = currentNumOne;
        for (int i = numOne; i < size+numOne; i++) {
            currentNumOne += nums[i % size] - nums[(i - numOne) % size];
            maxOne = max(maxOne, currentNumOne); 
        }
        return numOne - maxOne;
    }
};
20160