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Commit a2d5c34

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feat: add solutions to lc problem: No.1981
No.1981.Minimize the Difference Between Target and Chosen Elements
1 parent aaf6c36 commit a2d5c34

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8 files changed

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-3
lines changed

8 files changed

+379
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‎solution/1600-1699/1669.Merge In Between Linked Lists/README.md‎

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直接模拟题目中的操作即可。
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在实现上,我们使用两个指针 $p$ 和 $q,ドル初始时均指向链表 `list1` 的头节点。
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然后我们向后移动指针 $p$ 和 $q,ドル直到指针 $p$ 指向链表 `list1` 中第 $a$ 个节点的前一个节点,指针 $q$ 指向链表 `list1` 中第 $b$ 个节点。此时我们将 $p$ 的 `next` 指针指向链表 `list2` 的头节点,将链表 `list2` 的尾节点的 `next` 指针指向 $q$ 的 `next` 指针指向的节点,即可完成题目要求。
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时间复杂度 $O(m + n),ドル空间复杂度 $O(1)$。其中 $m$ 和 $n$ 分别为链表 `list1` 和 `list2` 的长度。
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<!-- tabs:start -->

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

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:动态规划(分组背包)**
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设 $f[i][j]$ 表示前 $i$ 行是否能选出元素和为 $j,ドル则有状态转移方程:
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$$
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f[i][j] = \begin{cases} 1 & \text{如果存在 } x \in row[i] \text{ 使得 } f[i - 1][j - x] = 1 \\ 0 & \text{否则} \end{cases}
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$$
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其中 $row[i]$ 表示第 $i$ 行的元素集合。
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由于 $f[i][j]$ 只与 $f[i - 1][j]$ 有关,因此我们可以使用滚动数组优化空间复杂度。
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最后,遍历 $f$ 数组,找出最小的绝对差即可。
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时间复杂度 $O(m^2 \times n \times C),ドル空间复杂度 $O(m \times C)$。其中 $m$ 和 $n$ 分别为矩阵的行数和列数;而 $C$ 为矩阵元素的最大值。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def minimizeTheDifference(self, mat: List[List[int]], target: int) -> int:
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f = {0}
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for row in mat:
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f = set(a + b for a in f for b in row)
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return min(abs(v - target) for v in f)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int minimizeTheDifference(int[][] mat, int target) {
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Set<Integer> f = new HashSet<>();
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f.add(0);
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for (var row : mat) {
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Set<Integer> g = new HashSet<>();
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for (int a : f) {
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for (int b : row) {
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g.add(a + b);
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}
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}
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f = g;
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}
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int ans = 1 << 30;
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for (int v : f) {
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ans = Math.min(ans, Math.abs(v - target));
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}
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return ans;
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}
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}
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```
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```java
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class Solution {
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public int minimizeTheDifference(int[][] mat, int target) {
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boolean[] f = {true};
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for (var row : mat) {
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int mx = 0;
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for (int x : row) {
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mx = Math.max(mx, x);
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}
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boolean[] g = new boolean[f.length + mx];
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for (int x : row) {
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for (int j = x; j < f.length + x; ++j) {
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g[j] |= f[j - x];
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}
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}
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f = g;
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}
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int ans = 1 << 30;
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for (int j = 0; j < f.length; ++j) {
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if (f[j]) {
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ans = Math.min(ans, Math.abs(j - target));
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int minimizeTheDifference(vector<vector<int>>& mat, int target) {
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vector<int> f = {1};
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for (auto& row : mat) {
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int mx = *max_element(row.begin(), row.end());
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vector<int> g(f.size() + mx);
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for (int x : row) {
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for (int j = x; j < f.size() + x; ++j) {
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g[j] |= f[j - x];
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}
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}
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f = move(g);
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}
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int ans = 1 << 30;
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for (int j = 0; j < f.size(); ++j) {
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if (f[j]) {
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ans = min(ans, abs(j - target));
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func minimizeTheDifference(mat [][]int, target int) int {
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f := []int{1}
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for _, row := range mat {
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mx := 0
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for _, x := range row {
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mx = max(mx, x)
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}
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g := make([]int, len(f)+mx)
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for _, x := range row {
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for j := x; j < len(f)+x; j++ {
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g[j] |= f[j-x]
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}
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}
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f = g
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}
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ans := 1 << 30
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for j, v := range f {
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if v == 1 {
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ans = min(ans, abs(j-target))
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}
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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```
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### **...**

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

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@@ -64,13 +64,142 @@ The absolute difference is 1.
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### **Python3**
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```python
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class Solution:
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def minimizeTheDifference(self, mat: List[List[int]], target: int) -> int:
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f = {0}
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for row in mat:
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f = set(a + b for a in f for b in row)
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return min(abs(v - target) for v in f)
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```
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### **Java**
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```java
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class Solution {
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public int minimizeTheDifference(int[][] mat, int target) {
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Set<Integer> f = new HashSet<>();
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f.add(0);
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for (var row : mat) {
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Set<Integer> g = new HashSet<>();
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for (int a : f) {
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for (int b : row) {
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g.add(a + b);
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}
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}
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f = g;
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}
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int ans = 1 << 30;
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for (int v : f) {
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ans = Math.min(ans, Math.abs(v - target));
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}
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return ans;
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}
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}
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```
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```java
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class Solution {
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public int minimizeTheDifference(int[][] mat, int target) {
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boolean[] f = {true};
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for (var row : mat) {
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int mx = 0;
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for (int x : row) {
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mx = Math.max(mx, x);
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}
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boolean[] g = new boolean[f.length + mx];
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for (int x : row) {
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for (int j = x; j < f.length + x; ++j) {
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g[j] |= f[j - x];
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}
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}
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f = g;
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}
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int ans = 1 << 30;
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for (int j = 0; j < f.length; ++j) {
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if (f[j]) {
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ans = Math.min(ans, Math.abs(j - target));
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int minimizeTheDifference(vector<vector<int>>& mat, int target) {
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vector<int> f = {1};
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for (auto& row : mat) {
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int mx = *max_element(row.begin(), row.end());
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vector<int> g(f.size() + mx);
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for (int x : row) {
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for (int j = x; j < f.size() + x; ++j) {
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g[j] |= f[j - x];
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}
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}
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f = move(g);
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}
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int ans = 1 << 30;
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for (int j = 0; j < f.size(); ++j) {
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if (f[j]) {
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ans = min(ans, abs(j - target));
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func minimizeTheDifference(mat [][]int, target int) int {
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f := []int{1}
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for _, row := range mat {
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mx := 0
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for _, x := range row {
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mx = max(mx, x)
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}
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g := make([]int, len(f)+mx)
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for _, x := range row {
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for j := x; j < len(f)+x; j++ {
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g[j] |= f[j-x]
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}
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}
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f = g
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}
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ans := 1 << 30
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for j, v := range f {
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if v == 1 {
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ans = min(ans, abs(j-target))
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}
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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```
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### **...**
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class Solution {
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public:
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int minimizeTheDifference(vector<vector<int>>& mat, int target) {
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vector<int> f = {1};
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for (auto& row : mat) {
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int mx = *max_element(row.begin(), row.end());
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vector<int> g(f.size() + mx);
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for (int x : row) {
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for (int j = x; j < f.size() + x; ++j) {
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g[j] |= f[j - x];
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}
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}
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f = move(g);
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}
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int ans = 1 << 30;
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for (int j = 0; j < f.size(); ++j) {
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if (f[j]) {
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ans = min(ans, abs(j - target));
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}
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}
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return ans;
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}
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};
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func minimizeTheDifference(mat [][]int, target int) int {
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f := []int{1}
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for _, row := range mat {
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mx := 0
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for _, x := range row {
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mx = max(mx, x)
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}
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g := make([]int, len(f)+mx)
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for _, x := range row {
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for j := x; j < len(f)+x; j++ {
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g[j] |= f[j-x]
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}
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}
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f = g
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}
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ans := 1 << 30
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for j, v := range f {
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if v == 1 {
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ans = min(ans, abs(j-target))
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}
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}

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