<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:
    def minimizeMax(self, nums: List[int], p: int) -> int:
        def count_pair(diff):
            count = 0
            i = 0
            while i < n - 1:
                if nums[i + 1] - nums[i] <= diff:
                    count += 1
                    i += 2
                else:
                    i += 1
            return count
        nums.sort()
        n = len(nums)
        l, r = 0, max(nums) - min(nums)
        while l < r:
            m = (l + r) // 2
            count = count_pair(m)
            if count >= p:
                r = m
            else:
                l = m + 1
        return l
            

# 10, 1, 2, 7, 1, 3
# [1, 1], [2, 3]

# 1, 1, 2, 3, 7, 10
# dp[i] = min max diff at ith
# dp[0] = 0
# dp[1] = nums[0] - nums[1]
# dp[2] = max(nums[2] - nums[1])
717