Maximum Frequency of an Element After Performing Operations I
<p>You are given an integer array <code>nums</code> and two integers <code>k</code> and <code>numOperations</code>.</p>

<p>You must perform an <strong>operation</strong> <code>numOperations</code> times on <code>nums</code>, where in each operation you:</p>

<ul>
	<li>Select an index <code>i</code> that was <strong>not</strong> selected in any previous operations.</li>
	<li>Add an integer in the range <code>[-k, k]</code> to <code>nums[i]</code>.</li>
</ul>

<p>Return the <strong>maximum</strong> possible <span data-keyword="frequency-array">frequency</span> of any element in <code>nums</code> after performing the <strong>operations</strong>.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1,4,5], k = 1, numOperations = 2</span></p>

<p><strong>Output:</strong> <span class="example-io">2</span></p>

<p><strong>Explanation:</strong></p>

<p>We can achieve a maximum frequency of two by:</p>

<ul>
	<li>Adding 0 to <code>nums[1]</code>. <code>nums</code> becomes <code>[1, 4, 5]</code>.</li>
	<li>Adding -1 to <code>nums[2]</code>. <code>nums</code> becomes <code>[1, 4, 4]</code>.</li>
</ul>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [5,11,20,20], k = 5, numOperations = 1</span></p>

<p><strong>Output:</strong> <span class="example-io">2</span></p>

<p><strong>Explanation:</strong></p>

<p>We can achieve a maximum frequency of two by:</p>

<ul>
	<li>Adding 0 to <code>nums[1]</code>.</li>
</ul>
</div>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= k &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= numOperations &lt;= nums.length</code></li>
</ul>
Array; Binary Search; Prefix Sum; Sorting; Sliding Window
class Solution:
    def maxFrequency(self, nums: List[int], k: int, numOperations: int) -> int:
        cnt = defaultdict(int)
        d = defaultdict(int)
        for x in nums:
            cnt[x] += 1
            d[x] += 0
            d[x - k] += 1
            d[x + k + 1] -= 1
        ans = s = 0
        for x, t in sorted(d.items()):
            s += t
            ans = max(ans, min(s, cnt[x] + numOperations))
        return ans
