add: 删除字符使频率相同

.vscode/settings.json

@@ -4,10 +4,12 @@
     "aaab",
     "aaabcbc",
     "AABC",
+    "aazz",
     "abca",
     "abcabcabcabc",
     "abcabcbb",
     "abcabccdcdcdef",
+    "abcc",
     "abcda",
     "abcdc",
     "abcdefg",
@@ -37,6 +39,7 @@
     "eleetminicoworoep",
     "emibcn",
     "esac",
+    "Freqs",
     "Gitpod",
     "heapify",
     "holasss",

README.md

@@ -760,6 +760,10 @@ TypeScript / JavaScript 基础算法、数据结构练习，包含 LeetCode 或
 
 ### 其它
 
+- [删除字符使频率相同](src/map/remove-letter-to-equalize-frequency.ts)  [哈希表, 字符串, 计数]
+
+  - LeetCode 2423. 删除字符使频率相同 <https://leetcode.cn/problems/remove-letter-to-equalize-frequency>
+
 - [仅执行一次字符串交换能否使两个字符串相等](src/map/check-if-one-string-swap-can-make-strings-equal.ts)  [哈希表, 字符串, 计数]
 
   - LeetCode 1790. 仅执行一次字符串交换能否使两个字符串相等 <https://leetcode.cn/problems/check-if-one-string-swap-can-make-strings-equal>

src/map/remove-letter-to-equalize-frequency.ts

@@ -0,0 +1,40 @@
+// 删除字符使频率相同
+// https://leetcode.cn/problems/remove-letter-to-equalize-frequency
+// INLINE  ../../images/map/remove-letter-to-equalize-frequency.jpeg
+
+export function equalFrequency (word: string): boolean {
+  const charCount = new Array(26).fill(0)
+  for (const c of word) {
+    charCount[c.charCodeAt(0) - 'a'.charCodeAt(0)]++
+  }
+  const freqCount = new Map()
+  for (const c of charCount) {
+    if (c > 0) {
+      freqCount.set(c, (freqCount.get(c) || 0) + 1)
+    }
+  }
+  for (const c of charCount) {
+    if (c == 0) {
+      continue
+    }
+    freqCount.set(c, freqCount.get(c) - 1)
+    if (freqCount.get(c) == 0) {
+      freqCount.delete(c)
+    }
+    if (c - 1 > 0) {
+      freqCount.set(c - 1, (freqCount.get(c - 1) || 0) + 1)
+    }
+    if (freqCount.size == 1) {
+      return true
+    }
+    if (c - 1 > 0) {
+      freqCount.set(c - 1, freqCount.get(c - 1) - 1)
+      if (freqCount.get(c - 1) == 0) {
+        freqCount.delete(c - 1)
+      }
+    }
+    /* istanbul ignore next */
+    freqCount.set(c, (freqCount.get(c) || 0) + 1)
+  }
+  return false
+}

test/map/remove-letter-to-equalize-frequency.test.ts

@@ -0,0 +1,17 @@
+import { equalFrequency } from '../../src/map/remove-letter-to-equalize-frequency'
+
+test('删除字符使频率相同', () => {
+  // 示例 1：
+  // 输入：word = "abcc"
+  // 输出：true
+  // 解释：选择下标 3 并删除该字母，word 变成 "abc" 且每个字母出现频率都为 1 。
+  const word = 'abcc'
+  expect(equalFrequency(word)).toBeTruthy()
+
+  // 示例 2：
+  // 输入：word = "aazz"
+  // 输出：false
+  // 解释：我们必须删除一个字母，所以要么 "a" 的频率变为 1 且 "z" 的频率为 2 ，要么两个字母频率反过来。所以不可能让剩余所有字母出现频率相同。
+  const word2 = 'aazz'
+  expect(equalFrequency(word2)).toBeFalsy()
+})