feat: solve No.3016

Author: BaffinLee

3001-3100/3016. Minimum Number of Pushes to Type Word II.md

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+# 3016. Minimum Number of Pushes to Type Word II
+
+- Difficulty: Medium.
+- Related Topics: Hash Table, String, Greedy, Sorting, Counting.
+- Similar Questions: Letter Combinations of a Phone Number.
+
+## Problem
+
+You are given a string `word` containing lowercase English letters.
+
+Telephone keypads have keys mapped with **distinct** collections of lowercase English letters, which can be used to form words by pushing them. For example, the key `2` is mapped with `["a","b","c"]`, we need to push the key one time to type `"a"`, two times to type `"b"`, and three times to type `"c"` **.**
+
+It is allowed to remap the keys numbered `2` to `9` to **distinct** collections of letters. The keys can be remapped to **any** amount of letters, but each letter **must** be mapped to **exactly** one key. You need to find the **minimum** number of times the keys will be pushed to type the string `word`.
+
+Return **the **minimum** number of pushes needed to type **`word` **after remapping the keys**.
+
+An example mapping of letters to keys on a telephone keypad is given below. Note that `1`, `*`, `#`, and `0` do **not** map to any letters.
+
+![](https://assets.leetcode.com/uploads/2023/12/26/keypaddesc.png)
+
+ 
+Example 1:
+
+![](https://assets.leetcode.com/uploads/2023/12/26/keypadv1e1.png)
+
+```
+Input: word = "abcde"
+Output: 5
+Explanation: The remapped keypad given in the image provides the minimum cost.
+"a" -> one push on key 2
+"b" -> one push on key 3
+"c" -> one push on key 4
+"d" -> one push on key 5
+"e" -> one push on key 6
+Total cost is 1 + 1 + 1 + 1 + 1 = 5.
+It can be shown that no other mapping can provide a lower cost.
+```
+
+Example 2:
+
+![](https://assets.leetcode.com/uploads/2023/12/26/keypadv2e2.png)
+
+```
+Input: word = "xyzxyzxyzxyz"
+Output: 12
+Explanation: The remapped keypad given in the image provides the minimum cost.
+"x" -> one push on key 2
+"y" -> one push on key 3
+"z" -> one push on key 4
+Total cost is 1 * 4 + 1 * 4 + 1 * 4 = 12
+It can be shown that no other mapping can provide a lower cost.
+Note that the key 9 is not mapped to any letter: it is not necessary to map letters to every key, but to map all the letters.
+```
+
+Example 3:
+
+![](https://assets.leetcode.com/uploads/2023/12/27/keypadv2.png)
+
+```
+Input: word = "aabbccddeeffgghhiiiiii"
+Output: 24
+Explanation: The remapped keypad given in the image provides the minimum cost.
+"a" -> one push on key 2
+"b" -> one push on key 3
+"c" -> one push on key 4
+"d" -> one push on key 5
+"e" -> one push on key 6
+"f" -> one push on key 7
+"g" -> one push on key 8
+"h" -> two pushes on key 9
+"i" -> one push on key 9
+Total cost is 1 * 2 + 1 * 2 + 1 * 2 + 1 * 2 + 1 * 2 + 1 * 2 + 1 * 2 + 2 * 2 + 6 * 1 = 24.
+It can be shown that no other mapping can provide a lower cost.
+```
+
+ 
+**Constraints:**
+
+
+	
+- `1 <= word.length <= 105`
+	
+- `word` consists of lowercase English letters.
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {string} word
+ * @return {number}
+ */
+var minimumPushes = function(word) {
+    var frequencyMap = Array(26).fill(0);
+    var a = 'a'.charCodeAt(0);
+    for (var i = 0; i < word.length; i++) {
+        frequencyMap[word[i].charCodeAt(0) - a] += 1;
+    }
+    var arr = frequencyMap.sort((a, b) => b - a);
+    var res = 0;
+    for (var i = 0; i < arr.length; i++) {
+        res += Math.ceil((i + 1) / 8) * arr[i];
+    }
+    return res;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(1).