feat: add solutions to lc problems: No.1980,1981 (doocs#2448)

* No.1980.Find Unique Binary String
* No.1981.Minimize the Difference Between Target and Chosen Elements

Author: yanglbme

solution/1900-1999/1980.Find Unique Binary String/README.md

@@ -57,7 +57,7 @@
 
 然后我们从 $0$ 开始枚举长度为 $n$ 的二进制字符串中 `'1'` 出现的次数 $i$，如果 $mask$ 的第 $i$ 位为 $0$，则说明长度为 $n$ 的二进制字符串中 `'1'` 出现次数为 $i$ 的字符串不存在，我们可以将这个字符串作为答案返回。
 
-时间复杂度 $O(L)$，空间复杂度 $O(1)$。其中 $L$ 为 `nums` 中字符串的总长度。
+时间复杂度 $O(L)$，其中 $L$ 为 `nums` 中字符串的总长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -127,6 +127,21 @@ func findDifferentBinaryString(nums []string) string {
 }
 ```
 
+```ts
+function findDifferentBinaryString(nums: string[]): string {
+    let mask = 0;
+    for (let x of nums) {
+        const cnt = x.split('').filter(c => c === '1').length;
+        mask |= 1 << cnt;
+    }
+    for (let i = 0; ; ++i) {
+        if (((mask >> i) & 1) === 0) {
+            return '1'.repeat(i) + '0'.repeat(nums.length - i);
+        }
+    }
+}
+```
+
 ```cs
 public class Solution {
     public string FindDifferentBinaryString(string[] nums) {

solution/1900-1999/1980.Find Unique Binary String/README_EN.md

@@ -46,7 +46,15 @@
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Counting + Enumeration
+
+Since the number of occurrences of '1' in a binary string of length $n$ can be $0, 1, 2, \cdots, n$ (there are $n + 1$ possibilities), we can certainly find a new binary string that has a different number of '1's from every string in `nums`.
+
+We can use an integer $mask$ to record the occurrence of '1' in all strings, i.e., the $i$-th bit of $mask$ is $1$ indicates that there is a string of length $n$ in which '1' appears $i$ times, otherwise it does not exist.
+
+Then we start to enumerate the number of times '1' appears in a binary string of length $n$ from $0$. If the $i$-th bit of $mask$ is $0$, it means that there is no string of length $n$ in which '1' appears $i$ times. We can return this string as the answer.
+
+The time complexity is $O(L)$, where $L$ is the total length of the strings in `nums`. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -116,6 +124,21 @@ func findDifferentBinaryString(nums []string) string {
 }
 ```
 
+```ts
+function findDifferentBinaryString(nums: string[]): string {
+    let mask = 0;
+    for (let x of nums) {
+        const cnt = x.split('').filter(c => c === '1').length;
+        mask |= 1 << cnt;
+    }
+    for (let i = 0; ; ++i) {
+        if (((mask >> i) & 1) === 0) {
+            return '1'.repeat(i) + '0'.repeat(nums.length - i);
+        }
+    }
+}
+```
+
 ```cs
 public class Solution {
     public string FindDifferentBinaryString(string[] nums) {

solution/1900-1999/1980.Find Unique Binary String/Solution.ts

@@ -0,0 +1,12 @@
+function findDifferentBinaryString(nums: string[]): string {
+    let mask = 0;
+    for (let x of nums) {
+        const cnt = x.split('').filter(c => c === '1').length;
+        mask |= 1 << cnt;
+    }
+    for (let i = 0; ; ++i) {
+        if (((mask >> i) & 1) === 0) {
+            return '1'.repeat(i) + '0'.repeat(nums.length - i);
+        }
+    }
+}

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

@@ -98,6 +98,34 @@ class Solution:
         return min(abs(v - target) for v in f)
 ```
 
+```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
 class Solution {
     public int minimizeTheDifference(int[][] mat, int target) {
@@ -179,38 +207,4 @@ func abs(x int) int {
 
 <!-- tabs:end -->
 
-### 方法二
-
-<!-- tabs:start -->
-
-```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;
-    }
-}
-```
-
-<!-- tabs:end -->
-
 <!-- end -->

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

@@ -61,7 +61,21 @@ The absolute difference is 1.
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Dynamic Programming (Grouped Knapsack)
+
+Let $f[i][j]$ represent whether it is possible to select elements from the first $i$ rows with a sum of $j$. Then we have the state transition equation:
+
+$$
+f[i][j] = \begin{cases} 1 & \text{if there exists } x \in row[i] \text{ such that } f[i - 1][j - x] = 1 \\ 0 & \text{otherwise} \end{cases}
+$$
+
+where $row[i]$ represents the set of elements in the $i$-th row.
+
+Since $f[i][j]$ is only related to $f[i - 1][j]$, we can use a rolling array to optimize the space complexity.
+
+Finally, we traverse the $f$ array to find the smallest absolute difference.
+
+The time complexity is $O(m^2 \times n \times C)$ and the space complexity is $O(m \times C)$. Here, $m$ and $n$ are the number of rows and columns of the matrix, respectively, and $C$ is the maximum value of the matrix elements.
 
 <!-- tabs:start -->
 
@@ -74,6 +88,34 @@ class Solution:
         return min(abs(v - target) for v in f)
 ```
 
+```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
 class Solution {
     public int minimizeTheDifference(int[][] mat, int target) {
@@ -155,38 +197,4 @@ func abs(x int) int {
 
 <!-- tabs:end -->
 
-### Solution 2
-
-<!-- tabs:start -->
-
-```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;
-    }
-}
-```
-
-<!-- tabs:end -->
-
 <!-- end -->