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feat: add solutions to lc problem: No.2387
No.2387.Median of a Row Wise Sorted Matrix
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‎solution/0400-0499/0402.Remove K Digits/Solution.py

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Original file line numberDiff line numberDiff line change
@@ -1,9 +1,9 @@
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class Solution:
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def removeKdigits(self, num: str, k: int) -> str:
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stack, remain = [], len(num)-k
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stack, remain = [], len(num) - k
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for value in num:
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while k and stack and stack[-1] > value:
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k = k-1
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k = k - 1
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stack.pop()
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stack.append(value)
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return "".join(stack[:remain]).lstrip('0') or '0'
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return "".join(stack[:remain]).lstrip('0') or '0'

‎solution/2300-2399/2387.Median of a Row Wise Sorted Matrix/README.md

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@@ -43,22 +43,137 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：两次二分查找**
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中位数实际上是排序后第 $target = \left \lceil \frac{m\times n}{2} \right \rceil $ 个数。
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我们二分枚举矩阵的元素 $x$，统计网格中大于该元素的个数 $cnt$，如果 $cnt \ge target$，说明中位数在 $x$ 的左侧（包含 $x$），否则在右侧。
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时间复杂度 $O(m\times \log n \times log M)$，其中 $M$ 为矩阵中的最大值。
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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 matrixMedian(self, grid: List[List[int]]) -> int:
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def count(x):
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return sum(bisect_right(row, x) for row in grid)
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m, n = len(grid), len(grid[0])
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target = (m * n + 1) >> 1
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return bisect_left(range(10**6 + 1), target, key=count)
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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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private int[][] grid;
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public int matrixMedian(int[][] grid) {
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this.grid = grid;
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int m = grid.length, n = grid[0].length;
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int target = (m * n + 1) >> 1;
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int left = 0, right = 1000010;
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while (left < right) {
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int mid = (left + right) >> 1;
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if (count(mid) >= target) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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return left;
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}
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private int count(int x) {
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int cnt = 0;
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for (var row : grid) {
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int left = 0, right = row.length;
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while (left < right) {
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int mid = (left + right) >> 1;
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if (row[mid] > x) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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cnt += left;
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}
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return cnt;
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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 matrixMedian(vector<vector<int>>& grid) {
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int m = grid.size(), n = grid[0].size();
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int left = 0, right = 1e6 + 1;
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int target = (m * n + 1) >> 1;
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auto count = [&](int x) {
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int cnt = 0;
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for (auto& row : grid) {
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cnt += (upper_bound(row.begin(), row.end(), x) - row.begin());
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}
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return cnt;
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};
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while (left < right) {
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int mid = (left + right) >> 1;
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if (count(mid) >= target) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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return left;
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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 matrixMedian(grid [][]int) int {
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m, n := len(grid), len(grid[0])
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count := func(x int) int {
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cnt := 0
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for _, row := range grid {
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left, right := 0, n
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for left < right {
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mid := (left + right) >> 1
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if row[mid] > x {
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right = mid
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} else {
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left = mid + 1
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}
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}
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cnt += left
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}
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return cnt
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}
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left, right := 0, 1000010
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target := (m*n + 1) >> 1
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for left < right {
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mid := (left + right) >> 1
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if count(mid) >= target {
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right = mid
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} else {
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left = mid + 1
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}
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}
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return left
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}
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```
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### **TypeScript**

‎solution/2300-2399/2387.Median of a Row Wise Sorted Matrix/README_EN.md

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@@ -44,13 +44,120 @@
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### **Python3**
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```python
47-
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class Solution:
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def matrixMedian(self, grid: List[List[int]]) -> int:
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def count(x):
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return sum(bisect_right(row, x) for row in grid)
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m, n = len(grid), len(grid[0])
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target = (m * n + 1) >> 1
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return bisect_left(range(10**6 + 1), target, key=count)
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```
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### **Java**
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```java
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class Solution {
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private int[][] grid;
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public int matrixMedian(int[][] grid) {
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this.grid = grid;
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int m = grid.length, n = grid[0].length;
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int target = (m * n + 1) >> 1;
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int left = 0, right = 1000010;
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while (left < right) {
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int mid = (left + right) >> 1;
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if (count(mid) >= target) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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return left;
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}
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private int count(int x) {
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int cnt = 0;
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for (var row : grid) {
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int left = 0, right = row.length;
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while (left < right) {
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int mid = (left + right) >> 1;
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if (row[mid] > x) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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cnt += left;
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}
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return cnt;
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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:
103+
int matrixMedian(vector<vector<int>>& grid) {
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int m = grid.size(), n = grid[0].size();
105+
int left = 0, right = 1e6 + 1;
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int target = (m * n + 1) >> 1;
107+
auto count = [&](int x) {
108+
int cnt = 0;
109+
for (auto& row : grid) {
110+
cnt += (upper_bound(row.begin(), row.end(), x) - row.begin());
111+
}
112+
return cnt;
113+
};
114+
while (left < right) {
115+
int mid = (left + right) >> 1;
116+
if (count(mid) >= target) {
117+
right = mid;
118+
} else {
119+
left = mid + 1;
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}
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}
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return left;
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}
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};
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```
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### **Go**
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```go
130+
func matrixMedian(grid [][]int) int {
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m, n := len(grid), len(grid[0])
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count := func(x int) int {
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cnt := 0
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for _, row := range grid {
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left, right := 0, n
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for left < right {
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mid := (left + right) >> 1
139+
if row[mid] > x {
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right = mid
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} else {
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left = mid + 1
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}
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}
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cnt += left
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}
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return cnt
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}
149+
left, right := 0, 1000010
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target := (m*n + 1) >> 1
151+
for left < right {
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mid := (left + right) >> 1
153+
if count(mid) >= target {
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right = mid
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} else {
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left = mid + 1
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}
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}
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return left
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}
54161
```
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### **TypeScript**

‎solution/2300-2399/2387.Median of a Row Wise Sorted Matrix/Solution.cpp

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class Solution {
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public:
3+
int matrixMedian(vector<vector<int>>& grid) {
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int m = grid.size(), n = grid[0].size();
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int left = 0, right = 1e6 + 1;
6+
int target = (m * n + 1) >> 1;
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auto count = [&](int x) {
8+
int cnt = 0;
9+
for (auto& row : grid) {
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cnt += (upper_bound(row.begin(), row.end(), x) - row.begin());
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}
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return cnt;
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};
14+
while (left < right) {
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int mid = (left + right) >> 1;
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if (count(mid) >= target) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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return left;
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}
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};

‎solution/2300-2399/2387.Median of a Row Wise Sorted Matrix/Solution.go

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func matrixMedian(grid [][]int) int {
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m, n := len(grid), len(grid[0])
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count := func(x int) int {
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cnt := 0
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for _, row := range grid {
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left, right := 0, n
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for left < right {
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mid := (left + right) >> 1
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if row[mid] > x {
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right = mid
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} else {
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left = mid + 1
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}
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}
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cnt += left
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}
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return cnt
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}
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left, right := 0, 1000010
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target := (m*n + 1) >> 1
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for left < right {
23+
mid := (left + right) >> 1
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if count(mid) >= target {
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right = mid
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} else {
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left = mid + 1
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}
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}
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return left
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}

‎solution/2300-2399/2387.Median of a Row Wise Sorted Matrix/Solution.java

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class Solution {
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private int[][] grid;
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4+
public int matrixMedian(int[][] grid) {
5+
this.grid = grid;
6+
int m = grid.length, n = grid[0].length;
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int target = (m * n + 1) >> 1;
8+
int left = 0, right = 1000010;
9+
while (left < right) {
10+
int mid = (left + right) >> 1;
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if (count(mid) >= target) {
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right = mid;
13+
} else {
14+
left = mid + 1;
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}
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}
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return left;
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}
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private int count(int x) {
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int cnt = 0;
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for (var row : grid) {
23+
int left = 0, right = row.length;
24+
while (left < right) {
25+
int mid = (left + right) >> 1;
26+
if (row[mid] > x) {
27+
right = mid;
28+
} else {
29+
left = mid + 1;
30+
}
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}
32+
cnt += left;
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}
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return cnt;
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}
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}

‎solution/2300-2399/2387.Median of a Row Wise Sorted Matrix/Solution.py

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class Solution:
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def matrixMedian(self, grid: List[List[int]]) -> int:
3+
def count(x):
4+
return sum(bisect_right(row, x) for row in grid)
5+
6+
m, n = len(grid), len(grid[0])
7+
target = (m * n + 1) >> 1
8+
return bisect_left(range(10**6 + 1), target, key=count)

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