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Author: jimit105

3360-minimum-deletions-to-make-string-k-special/README.md

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+<p>You are given a string <code>word</code> and an integer <code>k</code>.</p>
+
+<p>We consider <code>word</code> to be <strong>k-special</strong> if <code>|freq(word[i]) - freq(word[j])| &lt;= k</code> for all indices <code>i</code> and <code>j</code> in the string.</p>
+
+<p>Here, <code>freq(x)</code> denotes the <span data-keyword="frequency-letter">frequency</span> of the character <code>x</code> in <code>word</code>, and <code>|y|</code> denotes the absolute value of <code>y</code>.</p>
+
+<p>Return <em>the <strong>minimum</strong> number of characters you need to delete to make</em> <code>word</code> <strong><em>k-special</em></strong>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
+<p><strong>Input: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">word = &quot;aabcaba&quot;, k = 0</span></p>
+
+<p><strong>Output: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">3</span></p>
+
+<p><strong>Explanation:</strong> We can make <code>word</code> <code>0</code>-special by deleting <code>2</code> occurrences of <code>&quot;a&quot;</code> and <code>1</code> occurrence of <code>&quot;c&quot;</code>. Therefore, <code>word</code> becomes equal to <code>&quot;baba&quot;</code> where <code>freq(&#39;a&#39;) == freq(&#39;b&#39;) == 2</code>.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
+<p><strong>Input: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">word = &quot;dabdcbdcdcd&quot;, k = 2</span></p>
+
+<p><strong>Output: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">2</span></p>
+
+<p><strong>Explanation:</strong> We can make <code>word</code> <code>2</code>-special by deleting <code>1</code> occurrence of <code>&quot;a&quot;</code> and <code>1</code> occurrence of <code>&quot;d&quot;</code>. Therefore, <code>word</code> becomes equal to &quot;bdcbdcdcd&quot; where <code>freq(&#39;b&#39;) == 2</code>, <code>freq(&#39;c&#39;) == 3</code>, and <code>freq(&#39;d&#39;) == 4</code>.</p>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
+<p><strong>Input: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">word = &quot;aaabaaa&quot;, k = 2</span></p>
+
+<p><strong>Output: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">1</span></p>
+
+<p><strong>Explanation:</strong> We can make <code>word</code> <code>2</code>-special by deleting <code>1</code> occurrence of <code>&quot;b&quot;</code>. Therefore, <code>word</code> becomes equal to <code>&quot;aaaaaa&quot;</code> where each letter&#39;s frequency is now uniformly <code>6</code>.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= word.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;= k &lt;= 10<sup>5</sup></code></li>
+	<li><code>word</code> consists only of lowercase English letters.</li>
+</ul>

3360-minimum-deletions-to-make-string-k-special/solution.py

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+# Approach: Hash Table + Enumeration
+
+# n = len(word), c = size of character set, here 26
+# Time: O(n + c^2)
+# Space: O(c)
+
+from collections import defaultdict
+
+class Solution:
+    def minimumDeletions(self, word: str, k: int) -> int:
+        cnt = defaultdict(int)
+        for c in word:
+            cnt[c] += 1
+
+        res = len(word)
+        for a in cnt.values():
+            deleted = 0
+            for b in cnt.values():
+                if a > b:
+                    deleted += b
+                elif b > a + k:
+                    deleted += b - (a + k)
+
+            res = min(res, deleted)
+
+        return res
+