<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> and an integer <code>p</code>. Find <code>p</code> pairs of indices of <code>nums</code> such that the <strong>maximum</strong> difference amongst all the pairs is <strong>minimized</strong>. Also, ensure no index appears more than once amongst the <code>p</code> pairs.</p>

<p>Note that for a pair of elements at the index <code>i</code> and <code>j</code>, the difference of this pair is <code>|nums[i] - nums[j]|</code>, where <code>|x|</code> represents the <strong>absolute</strong> <strong>value</strong> of <code>x</code>.</p>

<p>Return <em>the <strong>minimum</strong> <strong>maximum</strong> difference among all </em><code>p</code> <em>pairs.</em> We define the maximum of an empty set to be zero.</p>

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

<pre>
<strong>Input:</strong> nums = [10,1,2,7,1,3], p = 2
<strong>Output:</strong> 1
<strong>Explanation:</strong> The first pair is formed from the indices 1 and 4, and the second pair is formed from the indices 2 and 5. 
The maximum difference is max(|nums[1] - nums[4]|, |nums[2] - nums[5]|) = max(0, 1) = 1. Therefore, we return 1.
</pre>

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

<pre>
<strong>Input:</strong> nums = [4,2,1,2], p = 1
<strong>Output:</strong> 0
<strong>Explanation:</strong> Let the indices 1 and 3 form a pair. The difference of that pair is |2 - 2| = 0, which is the minimum we can attain.
</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>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>0 &lt;= p &lt;= (nums.length)/2</code></li>
</ul>

class Solution {
public:
    int minimizeMax(vector<int>& nums, int p) {
                ios::sync_with_stdio(false);
        cin.tie(nullptr); 
        cout.tie(nullptr);
        if (p == 0) {
            return 0;
        }

        sort(nums.begin(), nums.end());

        int l = 0, r = nums[nums.size() - 1];

        while (l < r) {
            int mid = l + (r - l) / 2;

            int validPairs = 0, i = 0;

            while (i < nums.size() - 1) {
                if (nums[i + 1] - nums[i] <= mid) {
                    validPairs++;
                    i++;
                }
                i++;
            }

            if (validPairs < p) {
                l = mid + 1;
            } else {
                r = mid;
            }
        }

        return l;
    }
};


// 4 5 8 12
// 1 5 7 8 9 16

// 1 2 2 2 3 3 4
// 1 0 0 1 0 1
85800