feat: solve No.338,2707

Author: BaffinLee

2701-2800/2707. Extra Characters in a String.md

@@ -0,0 +1,83 @@
+# 2707. Extra Characters in a String
+
+- Difficulty: Medium.
+- Related Topics: Array, Hash Table, String, Dynamic Programming, Trie.
+- Similar Questions: Word Break.
+
+## Problem
+
+You are given a **0-indexed** string `s` and a dictionary of words `dictionary`. You have to break `s` into one or more **non-overlapping** substrings such that each substring is present in `dictionary`. There may be some **extra characters** in `s` which are not present in any of the substrings.
+
+Return **the **minimum** number of extra characters left over if you break up **`s`** optimally.**
+
+ 
+Example 1:
+
+```
+Input: s = "leetscode", dictionary = ["leet","code","leetcode"]
+Output: 1
+Explanation: We can break s in two substrings: "leet" from index 0 to 3 and "code" from index 5 to 8. There is only 1 unused character (at index 4), so we return 1.
+
+```
+
+Example 2:
+
+```
+Input: s = "sayhelloworld", dictionary = ["hello","world"]
+Output: 3
+Explanation: We can break s in two substrings: "hello" from index 3 to 7 and "world" from index 8 to 12. The characters at indices 0, 1, 2 are not used in any substring and thus are considered as extra characters. Hence, we return 3.
+```
+
+ 
+**Constraints:**
+
+
+	
+- `1 <= s.length <= 50`
+	
+- `1 <= dictionary.length <= 50`
+	
+- `1 <= dictionary[i].length <= 50`
+	
+- `dictionary[i]` and `s` consists of only lowercase English letters
+	
+- `dictionary` contains distinct words
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {string} s
+ * @param {string[]} dictionary
+ * @return {number}
+ */
+var minExtraChar = function(s, dictionary) {
+    var map = dictionary.reduce((res, item) => {
+        res[item] = 1;
+        return res;
+    }, {});
+    var dp = Array(s.length);
+    for (var i = s.length - 1; i >= 0; i--) {
+        dp[i] = (dp[i + 1] || 0) + 1;
+        var str = '';
+        for (var j = i; j < s.length; j++) {
+            str += s[j];
+            if (map[str]) {
+                dp[i] = Math.min(dp[i], dp[j + 1] || 0);
+            }
+        }
+    }
+    return dp[0];
+};
+```
+
+**Explain:**
+
+Bottom-up dynamic programming.
+
+**Complexity:**
+
+* Time complexity : O(n ^ 2).
+* Space complexity : O(n).

301-400/338. Counting Bits.md

@@ -0,0 +1,91 @@
+# 338. Counting Bits
+
+- Difficulty: Easy.
+- Related Topics: Dynamic Programming, Bit Manipulation.
+- Similar Questions: Number of 1 Bits.
+
+## Problem
+
+Given an integer `n`, return **an array **`ans`** of length **`n + 1`** such that for each **`i`** **(`0 <= i <= n`)**, **`ans[i]`** is the **number of ****`1`****'s** in the binary representation of **`i`.
+
+ 
+Example 1:
+
+```
+Input: n = 2
+Output: [0,1,1]
+Explanation:
+0 --> 0
+1 --> 1
+2 --> 10
+```
+
+Example 2:
+
+```
+Input: n = 5
+Output: [0,1,1,2,1,2]
+Explanation:
+0 --> 0
+1 --> 1
+2 --> 10
+3 --> 11
+4 --> 100
+5 --> 101
+```
+
+ 
+**Constraints:**
+
+
+	
+- `0 <= n <= 105`
+
+
+ 
+**Follow up:**
+
+
+	
+- It is very easy to come up with a solution with a runtime of `O(n log n)`. Can you do it in linear time `O(n)` and possibly in a single pass?
+	
+- Can you do it without using any built-in function (i.e., like `__builtin_popcount` in C++)?
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number} n
+ * @return {number[]}
+ */
+var countBits = function(n) {
+    var res = Array(n + 1);
+    var num = 2;
+    var nextNum = 4;
+    for (var i = 0; i <= n; i++) {
+        if (i === 0) {
+            res[i] = 0;
+        } else if (i === 1) {
+            res[i] = 1;
+        } else {
+            if (i === nextNum) {
+                num = nextNum;
+                nextNum *= 2;
+            }
+            res[i] = 1 + res[i - num];
+        }
+    }
+    return res;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(1).