feat: add solutions to lc problems: No.1991,1985 (#3990)

Author: yanglbme

solution/1900-1999/1981.Minimize the Difference Between Target and Chosen Elements/README.md

@@ -146,31 +146,6 @@ class Solution {
 }
 ```
 
-#### Java
-
-```java
-class Solution {
-    public int minimizeTheDifference(int[][] mat, int target) {
-        Set<Integer> f = new HashSet<>();
-        f.add(0);
-        for (var row : mat) {
-            Set<Integer> g = new HashSet<>();
-            for (int a : f) {
-                for (int b : row) {
-                    g.add(a + b);
-                }
-            }
-            f = g;
-        }
-        int ans = 1 << 30;
-        for (int v : f) {
-            ans = Math.min(ans, Math.abs(v - target));
-        }
-        return ans;
-    }
-}
-```
-
 #### C++
 
 ```cpp

solution/1900-1999/1981.Minimize the Difference Between Target and Chosen Elements/README_EN.md

@@ -110,36 +110,6 @@ class Solution:
 
 #### Java
 
-```java
-class Solution {
-    public int minimizeTheDifference(int[][] mat, int target) {
-        boolean[] f = {true};
-        for (var row : mat) {
-            int mx = 0;
-            for (int x : row) {
-                mx = Math.max(mx, x);
-            }
-            boolean[] g = new boolean[f.length + mx];
-            for (int x : row) {
-                for (int j = x; j < f.length + x; ++j) {
-                    g[j] |= f[j - x];
-                }
-            }
-            f = g;
-        }
-        int ans = 1 << 30;
-        for (int j = 0; j < f.length; ++j) {
-            if (f[j]) {
-                ans = Math.min(ans, Math.abs(j - target));
-            }
-        }
-        return ans;
-    }
-}
-```
-
-#### Java
-
 ```java
 class Solution {
     public int minimizeTheDifference(int[][] mat, int target) {

solution/1900-1999/1981.Minimize the Difference Between Target and Chosen Elements/Solution.java

@@ -1,20 +1,25 @@
 class Solution {
     public int minimizeTheDifference(int[][] mat, int target) {
-        Set<Integer> f = new HashSet<>();
-        f.add(0);
+        boolean[] f = {true};
         for (var row : mat) {
-            Set<Integer> g = new HashSet<>();
-            for (int a : f) {
-                for (int b : row) {
-                    g.add(a + b);
+            int mx = 0;
+            for (int x : row) {
+                mx = Math.max(mx, x);
+            }
+            boolean[] g = new boolean[f.length + mx];
+            for (int x : row) {
+                for (int j = x; j < f.length + x; ++j) {
+                    g[j] |= f[j - x];
                 }
             }
             f = g;
         }
         int ans = 1 << 30;
-        for (int v : f) {
-            ans = Math.min(ans, Math.abs(v - target));
+        for (int j = 0; j < f.length; ++j) {
+            if (f[j]) {
+                ans = Math.min(ans, Math.abs(j - target));
+            }
         }
         return ans;
     }
-}
+}

solution/1900-1999/1981.Minimize the Difference Between Target and Chosen Elements/Solution2.java

@@ -78,7 +78,11 @@ nums 中的数字按非递减顺序排列为 ["0","0"]
 
 <!-- solution:start -->
 
-### 方法一：自定义排序
+### 方法一：排序或快速选择
+
+我们可以将 $\textit{nums}$ 数组中的字符串按照整数从大到小排序，然后取第 $k$ 个元素即可。也可以使用快速选择算法，找到第 $k$ 大的整数。
+
+时间复杂度 $O(n \times \log n)$ 或 $O(n)$，其中 $n$ 是 $\textit{nums}$ 数组的长度。空间复杂度 $O(\log n)$ 或 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -87,13 +91,7 @@ nums 中的数字按非递减顺序排列为 ["0","0"]
 ```python
 class Solution:
     def kthLargestNumber(self, nums: List[str], k: int) -> str:
-        def cmp(a, b):
-            if len(a) != len(b):
-                return len(b) - len(a)
-            return 1 if b > a else -1
-
-        nums.sort(key=cmp_to_key(cmp))
-        return nums[k - 1]
+        return nlargest(k, nums, key=lambda x: int(x))[k - 1]
 ```
 
 #### Java
@@ -114,8 +112,9 @@ class Solution {
 class Solution {
 public:
     string kthLargestNumber(vector<string>& nums, int k) {
-        auto cmp = [](const string& a, const string& b) { return a.size() == b.size() ? a > b : a.size() > b.size(); };
-        sort(nums.begin(), nums.end(), cmp);
+        nth_element(nums.begin(), nums.begin() + k - 1, nums.end(), [](const string& a, const string& b) {
+            return a.size() == b.size() ? a > b : a.size() > b.size();
+        });
         return nums[k - 1];
     }
 };

solution/1900-1999/1985.Find the Kth Largest Integer in the Array/README.md

@@ -76,7 +76,11 @@ The 2<sup>nd</sup> largest integer in nums is &quot;0&quot;.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Sorting or Quickselect
+
+We can sort the strings in the $\textit{nums}$ array in descending order as integers, and then take the $k$-th element. Alternatively, we can use the quickselect algorithm to find the $k$-th largest integer.
+
+The time complexity is $O(n \times \log n)$ or $O(n)$, where $n$ is the length of the $\textit{nums}$ array. The space complexity is $O(\log n)$ or $O(1)$.
 
 <!-- tabs:start -->
 
@@ -85,13 +89,7 @@ The 2<sup>nd</sup> largest integer in nums is &quot;0&quot;.
 ```python
 class Solution:
     def kthLargestNumber(self, nums: List[str], k: int) -> str:
-        def cmp(a, b):
-            if len(a) != len(b):
-                return len(b) - len(a)
-            return 1 if b > a else -1
-
-        nums.sort(key=cmp_to_key(cmp))
-        return nums[k - 1]
+        return nlargest(k, nums, key=lambda x: int(x))[k - 1]
 ```
 
 #### Java
@@ -112,8 +110,9 @@ class Solution {
 class Solution {
 public:
     string kthLargestNumber(vector<string>& nums, int k) {
-        auto cmp = [](const string& a, const string& b) { return a.size() == b.size() ? a > b : a.size() > b.size(); };
-        sort(nums.begin(), nums.end(), cmp);
+        nth_element(nums.begin(), nums.begin() + k - 1, nums.end(), [](const string& a, const string& b) {
+            return a.size() == b.size() ? a > b : a.size() > b.size();
+        });
         return nums[k - 1];
     }
 };

solution/1900-1999/1985.Find the Kth Largest Integer in the Array/README_EN.md

@@ -1,8 +1,9 @@
 class Solution {
 public:
     string kthLargestNumber(vector<string>& nums, int k) {
-        auto cmp = [](const string& a, const string& b) { return a.size() == b.size() ? a > b : a.size() > b.size(); };
-        sort(nums.begin(), nums.end(), cmp);
+        nth_element(nums.begin(), nums.begin() + k - 1, nums.end(), [](const string& a, const string& b) {
+            return a.size() == b.size() ? a > b : a.size() > b.size();
+        });
         return nums[k - 1];
     }
-};
+};

solution/1900-1999/1985.Find the Kth Largest Integer in the Array/Solution.cpp

@@ -1,9 +1,3 @@
 class Solution:
     def kthLargestNumber(self, nums: List[str], k: int) -> str:
-        def cmp(a, b):
-            if len(a) != len(b):
-                return len(b) - len(a)
-            return 1 if b > a else -1
-
-        nums.sort(key=cmp_to_key(cmp))
-        return nums[k - 1]
+        return nlargest(k, nums, key=lambda x: int(x))[k - 1]

solution/1900-1999/1985.Find the Kth Largest Integer in the Array/Solution.py

@@ -81,7 +81,17 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: State Compression Dynamic Programming + Subset Enumeration
+
+We note that $n$ does not exceed $14$, so we can consider using state compression dynamic programming to solve this problem.
+
+We use a binary number $i$ of length $n$ to represent the current task state, where the $j$-th bit of $i$ is $1$ if and only if the $j$-th task is completed. We use $f[i]$ to represent the minimum number of work sessions needed to complete all tasks with state $i$.
+
+We can enumerate all subsets $j$ of $i$, where each bit of the binary representation of $j$ is a subset of the corresponding bit of the binary representation of $i$, i.e., $j \subseteq i$. If the tasks corresponding to $j$ can be completed in one work session, then we can update $f[i]$ using $f[i \oplus j] + 1$, where $i \oplus j$ represents the bitwise XOR of $i$ and $j$.
+
+The final answer is $f[2^n - 1]$.
+
+The time complexity is $O(n \times 3^n)$, and the space complexity is $O(2^n)$. Here, $n$ is the number of tasks.
 
 <!-- tabs:start -->
 

solution/1900-1999/1986.Minimum Number of Work Sessions to Finish the Tasks/README_EN.md

@@ -54,7 +54,7 @@ The unique good subsequences are &quot;1&quot; and &quot;11&quot;.</pre>
 <pre>
 <strong>Input:</strong> binary = &quot;101&quot;
 <strong>Output:</strong> 5
-<strong>Explanation:</strong> The good subsequences of binary are [&quot;1&quot;, &quot;0&quot;, &quot;1&quot;, &quot;10&quot;, &quot;11&quot;, &quot;101&quot;]. 
+<strong>Explanation:</strong> The good subsequences of binary are [&quot;1&quot;, &quot;0&quot;, &quot;1&quot;, &quot;10&quot;, &quot;11&quot;, &quot;101&quot;].
 The unique good subsequences are &quot;0&quot;, &quot;1&quot;, &quot;10&quot;, &quot;11&quot;, and &quot;101&quot;.
 </pre>
 
@@ -72,7 +72,22 @@ The unique good subsequences are &quot;0&quot;, &quot;1&quot;, &quot;10&quot;, &
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Dynamic Programming
+
+We define $f$ as the number of distinct good subsequences ending with $1$, and $g$ as the number of distinct good subsequences ending with $0$ and starting with $1$. Initially, $f = g = 0$.
+
+For a binary string, we can traverse each bit from left to right. Suppose the current bit is $c$:
+
+-   If $c = 0$, we can append $c$ to the $f$ and $g$ distinct good subsequences, so update $g = (g + f) \bmod (10^9 + 7)$;
+-   If $c = 1$, we can append $c$ to the $f$ and $g$ distinct good subsequences, and also append $c$ alone, so update $f = (f + g + 1) \bmod (10^9 + 7)$.
+
+If the string contains $0$, the final answer is $f + g + 1$, otherwise the answer is $f + g$.
+
+The time complexity is $O(n)$, where $n$ is the length of the string. The space complexity is $O(1)$.
+
+Similar problems:
+
+-   [940. Distinct Subsequences II](https://github.com/doocs/leetcode/blob/main/solution/0900-0999/0940.Distinct%20Subsequences%20II/README_EN.md)
 
 <!-- tabs:start -->
 

solution/1900-1999/1987.Number of Unique Good Subsequences/README_EN.md

@@ -75,11 +75,11 @@ tags:
 
 我们可以用两个指针 $i$ 和 $j$ 指向鬼和非鬼的人，初始时 $i=0$, $j=0$。
 
-然后我们从左到右遍历数组，当遇到鬼时，即 $team[i]=1$ 时，如果此时 $j \lt n$ 并且 $team[j]=1$ 或者 $i - j \gt dist$，则指针 $j$ 循环右移，也即是说，我们要找到第一个不是鬼的人，且 $i$ 和 $j$ 之间的距离不超过 $dist$。如果找到了这样的人，则将指针 $j$ 右移一位，表示我们已经抓住了这个人，同时答案加一。继续遍历数组，直到遍历完整个数组。
+然后我们从左到右遍历数组，当遇到鬼时，即 $team[i]=1$ 时，如果此时 $j \lt n$ 并且 $\textit{team}[j]=1$ 或者 $i - j \gt \textit{dist}$，则指针 $j$ 循环右移，也即是说，我们要找到第一个不是鬼的人，且 $i$ 和 $j$ 之间的距离不超过 $\textit{dist}$。如果找到了这样的人，则将指针 $j$ 右移一位，表示我们已经抓住了这个人，同时答案加一。继续遍历数组，直到遍历完整个数组。
 
 最后返回答案即可。
 
-时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组的长度。
+时间复杂度 $O(n)$，其中 $n$ 为数组 $\textit{team}$ 的长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 

solution/1900-1999/1989.Maximum Number of People That Can Be Caught in Tag/README.md

@@ -69,7 +69,15 @@ There are no people who are not "it" to catch.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Two Pointers
+
+We can use two pointers $i$ and $j$ to point to the ghost and non-ghost people, initially $i=0$, $j=0$.
+
+Then we traverse the array from left to right. When we encounter a ghost, i.e., $team[i]=1$, if $j \lt n$ and $\textit{team}[j]=1$ or $i - j \gt \textit{dist}$, then move pointer $j$ to the right in a loop. This means we need to find the first non-ghost person such that the distance between $i$ and $j$ does not exceed $\textit{dist}$. If such a person is found, move pointer $j$ one step to the right, indicating that we have caught this person, and increment the answer by one. Continue traversing the array until the entire array is processed.
+
+Finally, return the answer.
+
+The time complexity is $O(n)$, where $n$ is the length of the array $\textit{team}$. The space complexity is $O(1)$.
 
 <!-- tabs:start -->