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| 1 | +<h2><a href="https://leetcode.com/problems/maximum-frequency-of-an-element-after-performing-operations-ii/">3347. Maximum Frequency of an Element After Performing Operations II</a></h2><h3>Hard</h3><hr><div><p>You are given an integer array <code>nums</code> and two integers <code>k</code> and <code>numOperations</code>.</p> |
| 2 | + |
| 3 | +<p>You must perform an <strong>operation</strong> <code>numOperations</code> times on <code>nums</code>, where in each operation you:</p> |
| 4 | + |
| 5 | +<ul> |
| 6 | + <li>Select an index <code>i</code> that was <strong>not</strong> selected in any previous operations.</li> |
| 7 | + <li>Add an integer in the range <code>[-k, k]</code> to <code>nums[i]</code>.</li> |
| 8 | +</ul> |
| 9 | + |
| 10 | +<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> |
| 11 | + |
| 12 | +<p> </p> |
| 13 | +<p><strong class="example">Example 1:</strong></p> |
| 14 | + |
| 15 | +<div class="example-block"> |
| 16 | +<p><strong>Input:</strong> <span class="example-io">nums = [1,4,5], k = 1, numOperations = 2</span></p> |
| 17 | + |
| 18 | +<p><strong>Output:</strong> <span class="example-io">2</span></p> |
| 19 | + |
| 20 | +<p><strong>Explanation:</strong></p> |
| 21 | + |
| 22 | +<p>We can achieve a maximum frequency of two by:</p> |
| 23 | + |
| 24 | +<ul> |
| 25 | + <li>Adding 0 to <code>nums[1]</code>, after which <code>nums</code> becomes <code>[1, 4, 5]</code>.</li> |
| 26 | + <li>Adding -1 to <code>nums[2]</code>, after which <code>nums</code> becomes <code>[1, 4, 4]</code>.</li> |
| 27 | +</ul> |
| 28 | +</div> |
| 29 | + |
| 30 | +<p><strong class="example">Example 2:</strong></p> |
| 31 | + |
| 32 | +<div class="example-block"> |
| 33 | +<p><strong>Input:</strong> <span class="example-io">nums = [5,11,20,20], k = 5, numOperations = 1</span></p> |
| 34 | + |
| 35 | +<p><strong>Output:</strong> <span class="example-io">2</span></p> |
| 36 | + |
| 37 | +<p><strong>Explanation:</strong></p> |
| 38 | + |
| 39 | +<p>We can achieve a maximum frequency of two by:</p> |
| 40 | + |
| 41 | +<ul> |
| 42 | + <li>Adding 0 to <code>nums[1]</code>.</li> |
| 43 | +</ul> |
| 44 | +</div> |
| 45 | + |
| 46 | +<p> </p> |
| 47 | +<p><strong>Constraints:</strong></p> |
| 48 | + |
| 49 | +<ul> |
| 50 | + <li><code>1 <= nums.length <= 10<sup>5</sup></code></li> |
| 51 | + <li><code>1 <= nums[i] <= 10<sup>9</sup></code></li> |
| 52 | + <li><code>0 <= k <= 10<sup>9</sup></code></li> |
| 53 | + <li><code>0 <= numOperations <= nums.length</code></li> |
| 54 | +</ul> |
| 55 | +</div> |
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