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feat: add solutions to lc problem: No.2643 #4279

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63 changes: 38 additions & 25 deletions solution/2600-2699/2643.Row With Maximum Ones/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -32,7 +32,7 @@ tags:
<pre>
<strong>输入：</strong>mat = [[0,1],[1,0]]
<strong>输出：</strong>[0,1]
<strong>解释：</strong>两行中 1 的数量相同。所以返回下标最小的行，下标为 0 。该行 1 的数量为 1 。所以，答案为 [0,1] 。
<strong>解释：</strong>两行中 1 的数量相同。所以返回下标最小的行，下标为 0 。该行 1 的数量为 1 。所以，答案为 [0,1] 。
</pre>

<p><strong>示例 2：</strong></p>
Expand Down Expand Up @@ -69,9 +69,16 @@ tags:

### 方法一：模拟

我们直接遍历矩阵，统计每一行中 $1$ 的个数，更新最大值和对应的行下标。注意，如果当前行的 $1$ 的个数与最大值相等，我们需要选择行下标较小的那一行。
我们初始化一个数组 $\textit{ans} = [0, 0]$，用于记录最多 $1$ 的行的下标和 $1$ 的数量。

时间复杂度 $(m \times n)$，其中 $m$ 和 $n$ 分别为矩阵的行数和列数。空间复杂度 $O(1)$。
然后遍历矩阵的每一行，对于每一行：

- 计算该行 $1$ 的数量 $\textit{cnt}$（由于矩阵中只包含 $0$ 和 $1$，我们可以直接对该行求和）；
- 如果 $\textit{ans}[1] < \textit{cnt}$，则更新 $\textit{ans} = [i, \textit{cnt}]$。

遍历结束后，返回 $\textit{ans}$。

时间复杂度 $O(m \times n)$，其中 $m$ 和 $n$ 分别是矩阵的行数和列数。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand All @@ -82,7 +89,7 @@ class Solution:
def rowAndMaximumOnes(self, mat: List[List[int]]) -> List[int]:
ans = [0, 0]
for i, row in enumerate(mat):
cnt = row.count(1)
cnt = sum(row)
if ans[1] < cnt:
ans = [i, cnt]
return ans
Expand All @@ -97,9 +104,7 @@ class Solution {
for (int i = 0; i < mat.length; ++i) {
int cnt = 0;
for (int x : mat[i]) {
if (x == 1) {
++cnt;
}
cnt += x;
}
if (ans[1] < cnt) {
ans[0] = i;
Expand All @@ -119,13 +124,9 @@ public:
vector<int> rowAndMaximumOnes(vector<vector<int>>& mat) {
vector<int> ans(2);
for (int i = 0; i < mat.size(); ++i) {
int cnt = 0;
for (auto& x : mat[i]) {
cnt += x == 1;
}
int cnt = accumulate(mat[i].begin(), mat[i].end(), 0);
if (ans[1] < cnt) {
ans[0] = i;
ans[1] = cnt;
ans = {i, cnt};
}
}
return ans;
Expand All @@ -137,16 +138,14 @@ public:

```go
func rowAndMaximumOnes(mat [][]int) []int {
ans := make([]int, 2)
ans := []int{0, 0}
for i, row := range mat {
cnt := 0
for _, x := range row {
if x == 1 {
cnt++
}
cnt += x
}
if ans[1] < cnt {
ans[0], ans[1] = i, cnt
ans = []int{i, cnt}
}
}
return ans
Expand All @@ -158,8 +157,8 @@ func rowAndMaximumOnes(mat [][]int) []int {
```ts
function rowAndMaximumOnes(mat: number[][]): number[] {
const ans: number[] = [0, 0];
for (let i = 0; i < mat.length; ++i) {
const cnt = mat[i].reduce((a, b) => a + b);
for (let i = 0; i < mat.length; i++) {
const cnt = mat[i].reduce((sum, num) => sum + num, 0);
if (ans[1] < cnt) {
ans[0] = i;
ans[1] = cnt;
Expand All @@ -175,20 +174,34 @@ function rowAndMaximumOnes(mat: number[][]): number[] {
impl Solution {
pub fn row_and_maximum_ones(mat: Vec<Vec<i32>>) -> Vec<i32> {
let mut ans = vec![0, 0];

for (i, row) in mat.iter().enumerate() {
let cnt = row.iter().filter(|&v| *v == 1).count() as i32;
let cnt = row.iter().sum();
if ans[1] < cnt {
ans[0] = i as i32;
ans[1] = cnt;
ans = vec![i as i32, cnt];
}
}

ans
}
}
```

#### C#

```cs
public class Solution {
public int[] RowAndMaximumOnes(int[][] mat) {
int[] ans = new int[2];
for (int i = 0; i < mat.Length; i++) {
int cnt = mat[i].Sum();
if (ans[1] < cnt) {
ans = new int[] { i, cnt };
}
}
return ans;
}
}
```

<!-- tabs:end -->

<!-- solution:end -->
Expand Down
63 changes: 38 additions & 25 deletions solution/2600-2699/2643.Row With Maximum Ones/README_EN.md
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Expand Up @@ -31,7 +31,7 @@ tags:
<pre>
<strong>Input:</strong> mat = [[0,1],[1,0]]
<strong>Output:</strong> [0,1]
<strong>Explanation:</strong> Both rows have the same number of 1&#39;s. So we return the index of the smaller row, 0, and the maximum count of ones (1<code>)</code>. So, the answer is [0,1].
<strong>Explanation:</strong> Both rows have the same number of 1&#39;s. So we return the index of the smaller row, 0, and the maximum count of ones (1<code>)</code>. So, the answer is [0,1].
</pre>

<p><strong class="example">Example 2:</strong></p>
Expand Down Expand Up @@ -68,9 +68,16 @@ tags:

### Solution 1: Simulation

We directly traverse the matrix, count the number of $1$s in each row, and update the maximum value and the corresponding row index. Note that if the number of $1$s in the current row is equal to the maximum value, we need to choose the row with the smaller index.
We initialize an array $\textit{ans} = [0, 0]$ to store the index of the row with the most $1$s and the count of $1$s.

The time complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns of the matrix, respectively. The space complexity is $O(1)$.
Then, we iterate through each row of the matrix:

- Compute the number of $1$s in the current row, denoted as $\textit{cnt}$ (since the matrix contains only $0$s and $1$s, we can directly sum up the row).
- If $\textit{ans}[1] < \textit{cnt}$, update $\textit{ans} = [i, \textit{cnt}]$.

After finishing the iteration, we return $\textit{ans}$.

The time complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns in the matrix, respectively. The space complexity is $O(1)$.

<!-- tabs:start -->

Expand All @@ -81,7 +88,7 @@ class Solution:
def rowAndMaximumOnes(self, mat: List[List[int]]) -> List[int]:
ans = [0, 0]
for i, row in enumerate(mat):
cnt = row.count(1)
cnt = sum(row)
if ans[1] < cnt:
ans = [i, cnt]
return ans
Expand All @@ -96,9 +103,7 @@ class Solution {
for (int i = 0; i < mat.length; ++i) {
int cnt = 0;
for (int x : mat[i]) {
if (x == 1) {
++cnt;
}
cnt += x;
}
if (ans[1] < cnt) {
ans[0] = i;
Expand All @@ -118,13 +123,9 @@ public:
vector<int> rowAndMaximumOnes(vector<vector<int>>& mat) {
vector<int> ans(2);
for (int i = 0; i < mat.size(); ++i) {
int cnt = 0;
for (auto& x : mat[i]) {
cnt += x == 1;
}
int cnt = accumulate(mat[i].begin(), mat[i].end(), 0);
if (ans[1] < cnt) {
ans[0] = i;
ans[1] = cnt;
ans = {i, cnt};
}
}
return ans;
Expand All @@ -136,16 +137,14 @@ public:

```go
func rowAndMaximumOnes(mat [][]int) []int {
ans := make([]int, 2)
ans := []int{0, 0}
for i, row := range mat {
cnt := 0
for _, x := range row {
if x == 1 {
cnt++
}
cnt += x
}
if ans[1] < cnt {
ans[0], ans[1] = i, cnt
ans = []int{i, cnt}
}
}
return ans
Expand All @@ -157,8 +156,8 @@ func rowAndMaximumOnes(mat [][]int) []int {
```ts
function rowAndMaximumOnes(mat: number[][]): number[] {
const ans: number[] = [0, 0];
for (let i = 0; i < mat.length; ++i) {
const cnt = mat[i].reduce((a, b) => a + b);
for (let i = 0; i < mat.length; i++) {
const cnt = mat[i].reduce((sum, num) => sum + num, 0);
if (ans[1] < cnt) {
ans[0] = i;
ans[1] = cnt;
Expand All @@ -174,20 +173,34 @@ function rowAndMaximumOnes(mat: number[][]): number[] {
impl Solution {
pub fn row_and_maximum_ones(mat: Vec<Vec<i32>>) -> Vec<i32> {
let mut ans = vec![0, 0];

for (i, row) in mat.iter().enumerate() {
let cnt = row.iter().filter(|&v| *v == 1).count() as i32;
let cnt = row.iter().sum();
if ans[1] < cnt {
ans[0] = i as i32;
ans[1] = cnt;
ans = vec![i as i32, cnt];
}
}

ans
}
}
```

#### C#

```cs
public class Solution {
public int[] RowAndMaximumOnes(int[][] mat) {
int[] ans = new int[2];
for (int i = 0; i < mat.Length; i++) {
int cnt = mat[i].Sum();
if (ans[1] < cnt) {
ans = new int[] { i, cnt };
}
}
return ans;
}
}
```

<!-- tabs:end -->

<!-- solution:end -->
Expand Down
10 changes: 3 additions & 7 deletions solution/2600-2699/2643.Row With Maximum Ones/Solution.cpp
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Original file line number Diff line number Diff line change
Expand Up @@ -3,15 +3,11 @@ class Solution {
vector<int> rowAndMaximumOnes(vector<vector<int>>& mat) {
vector<int> ans(2);
for (int i = 0; i < mat.size(); ++i) {
int cnt = 0;
for (auto& x : mat[i]) {
cnt += x == 1;
}
int cnt = accumulate(mat[i].begin(), mat[i].end(), 0);
if (ans[1] < cnt) {
ans[0] = i;
ans[1] = cnt;
ans = {i, cnt};
}
}
return ans;
}
};
};
12 changes: 12 additions & 0 deletions solution/2600-2699/2643.Row With Maximum Ones/Solution.cs
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Original file line number Diff line number Diff line change
@@ -0,0 +1,12 @@
public class Solution {
public int[] RowAndMaximumOnes(int[][] mat) {
int[] ans = new int[2];
for (int i = 0; i < mat.Length; i++) {
int cnt = mat[i].Sum();
if (ans[1] < cnt) {
ans = new int[] { i, cnt };
}
}
return ans;
}
}
10 changes: 4 additions & 6 deletions solution/2600-2699/2643.Row With Maximum Ones/Solution.go
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Original file line number Diff line number Diff line change
@@ -1,15 +1,13 @@
func rowAndMaximumOnes(mat [][]int) []int {
ans := make([]int, 2)
ans := []int{0, 0}
for i, row := range mat {
cnt := 0
for _, x := range row {
if x == 1 {
cnt++
}
cnt += x
}
if ans[1] < cnt {
ans[0], ans[1] = i, cnt
ans = []int{i, cnt}
}
}
return ans
}
}
6 changes: 2 additions & 4 deletions solution/2600-2699/2643.Row With Maximum Ones/Solution.java
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Original file line number Diff line number Diff line change
Expand Up @@ -4,9 +4,7 @@ public int[] rowAndMaximumOnes(int[][] mat) {
for (int i = 0; i < mat.length; ++i) {
int cnt = 0;
for (int x : mat[i]) {
if (x == 1) {
++cnt;
}
cnt += x;
}
if (ans[1] < cnt) {
ans[0] = i;
Expand All @@ -15,4 +13,4 @@ public int[] rowAndMaximumOnes(int[][] mat) {
}
return ans;
}
}
}
2 changes: 1 addition & 1 deletion solution/2600-2699/2643.Row With Maximum Ones/Solution.py
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Original file line number Diff line number Diff line change
Expand Up @@ -2,7 +2,7 @@ class Solution:
def rowAndMaximumOnes(self, mat: List[List[int]]) -> List[int]:
ans = [0, 0]
for i, row in enumerate(mat):
cnt = row.count(1)
cnt = sum(row)
if ans[1] < cnt:
ans = [i, cnt]
return ans
7 changes: 2 additions & 5 deletions solution/2600-2699/2643.Row With Maximum Ones/Solution.rs
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Original file line number Diff line number Diff line change
@@ -1,15 +1,12 @@
impl Solution {
pub fn row_and_maximum_ones(mat: Vec<Vec<i32>>) -> Vec<i32> {
let mut ans = vec![0, 0];

for (i, row) in mat.iter().enumerate() {
let cnt = row.iter().filter(|&v| *v == 1).count() as i32;
let cnt = row.iter().sum();
if ans[1] < cnt {
ans[0] = i as i32;
ans[1] = cnt;
ans = vec![i as i32, cnt];
}
}

ans
}
}
4 changes: 2 additions & 2 deletions solution/2600-2699/2643.Row With Maximum Ones/Solution.ts
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Original file line number Diff line number Diff line change
@@ -1,7 +1,7 @@
function rowAndMaximumOnes(mat: number[][]): number[] {
const ans: number[] = [0, 0];
for (let i = 0; i < mat.length; ++i) {
const cnt = mat[i].reduce((a, b) => a + b);
for (let i = 0; i < mat.length; i++) {
const cnt = mat[i].reduce((sum, num) => sum + num, 0);
if (ans[1] < cnt) {
ans[0] = i;
ans[1] = cnt;
Expand Down