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feat: add solutions to lc problem: No.2643 (#4279)
No.2643.Row With Maximum Ones
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‎solution/2600-2699/2643.Row With Maximum Ones/README.md

+38-25
Original file line numberDiff line numberDiff line change
@@ -32,7 +32,7 @@ tags:
3232
<pre>
3333
<strong>输入：</strong>mat = [[0,1],[1,0]]
3434
<strong>输出：</strong>[0,1]
35-
<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] 。
35+
<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] 。
3636
</pre>
3737

3838
<p><strong>示例 2：</strong></p>
@@ -69,9 +69,16 @@ tags:
6969

7070
### 方法一：模拟
7171

72-
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。
72+
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。
7373

74-
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 $O(1)$。
74+
然后遍历矩阵的每一行，对于每一行：
75+
76+
- 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）；
77+
- 如果 $\textit{ans}[1] < \textit{cnt}$，则更新 $\textit{ans} = [i, \textit{cnt}]$。
78+
79+
遍历结束后，返回 $\textit{ans}$。
80+
81+
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 $O(1)$。
7582

7683
<!-- tabs:start -->
7784

@@ -82,7 +89,7 @@ class Solution:
8289
def rowAndMaximumOnes(self, mat: List[List[int]]) -> List[int]:
8390
ans = [0, 0]
8491
for i, row in enumerate(mat):
85-
cnt = row.count(1)
92+
cnt = sum(row)
8693
if ans[1] < cnt:
8794
ans = [i, cnt]
8895
return ans
@@ -97,9 +104,7 @@ class Solution {
97104
for (int i = 0; i < mat.length; ++i) {
98105
int cnt = 0;
99106
for (int x : mat[i]) {
100-
if (x == 1) {
101-
++cnt;
102-
}
107+
cnt += x;
103108
}
104109
if (ans[1] < cnt) {
105110
ans[0] = i;
@@ -119,13 +124,9 @@ public:
119124
vector<int> rowAndMaximumOnes(vector<vector<int>>& mat) {
120125
vector<int> ans(2);
121126
for (int i = 0; i < mat.size(); ++i) {
122-
int cnt = 0;
123-
for (auto& x : mat[i]) {
124-
cnt += x == 1;
125-
}
127+
int cnt = accumulate(mat[i].begin(), mat[i].end(), 0);
126128
if (ans[1] < cnt) {
127-
ans[0] = i;
128-
ans[1] = cnt;
129+
ans = {i, cnt};
129130
}
130131
}
131132
return ans;
@@ -137,16 +138,14 @@ public:
137138
138139
```go
139140
func rowAndMaximumOnes(mat [][]int) []int {
140-
ans := make([]int, 2)
141+
ans := []int{0, 0}
141142
for i, row := range mat {
142143
cnt := 0
143144
for _, x := range row {
144-
if x == 1 {
145-
cnt++
146-
}
145+
cnt += x
147146
}
148147
if ans[1] < cnt {
149-
ans[0], ans[1] = i, cnt
148+
ans = []int{i, cnt}
150149
}
151150
}
152151
return ans
@@ -158,8 +157,8 @@ func rowAndMaximumOnes(mat [][]int) []int {
158157
```ts
159158
function rowAndMaximumOnes(mat: number[][]): number[] {
160159
const ans: number[] = [0, 0];
161-
for (let i = 0; i < mat.length; ++i) {
162-
const cnt = mat[i].reduce((a, b) => a + b);
160+
for (let i = 0; i < mat.length; i++) {
161+
const cnt = mat[i].reduce((sum, num) => sum + num, 0);
163162
if (ans[1] < cnt) {
164163
ans[0] = i;
165164
ans[1] = cnt;
@@ -175,20 +174,34 @@ function rowAndMaximumOnes(mat: number[][]): number[] {
175174
impl Solution {
176175
pub fn row_and_maximum_ones(mat: Vec<Vec<i32>>) -> Vec<i32> {
177176
let mut ans = vec![0, 0];
178-
179177
for (i, row) in mat.iter().enumerate() {
180-
let cnt = row.iter().filter(|&v| *v == 1).count() as i32;
178+
let cnt = row.iter().sum();
181179
if ans[1] < cnt {
182-
ans[0] = i as i32;
183-
ans[1] = cnt;
180+
ans = vec![i as i32, cnt];
184181
}
185182
}
186-
187183
ans
188184
}
189185
}
190186
```
191187

188+
#### C#
189+
190+
```cs
191+
public class Solution {
192+
public int[] RowAndMaximumOnes(int[][] mat) {
193+
int[] ans = new int[2];
194+
for (int i = 0; i < mat.Length; i++) {
195+
int cnt = mat[i].Sum();
196+
if (ans[1] < cnt) {
197+
ans = new int[] { i, cnt };
198+
}
199+
}
200+
return ans;
201+
}
202+
}
203+
```
204+
192205
<!-- tabs:end -->
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194207
<!-- solution:end -->

‎solution/2600-2699/2643.Row With Maximum Ones/README_EN.md

+38-25
Original file line numberDiff line numberDiff line change
@@ -31,7 +31,7 @@ tags:
3131
<pre>
3232
<strong>Input:</strong> mat = [[0,1],[1,0]]
3333
<strong>Output:</strong> [0,1]
34-
<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].
34+
<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].
3535
</pre>
3636

3737
<p><strong class="example">Example 2:</strong></p>
@@ -68,9 +68,16 @@ tags:
6868

6969
### Solution 1: Simulation
7070

71-
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.
71+
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.
7272

73-
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 $O(1)$.
73+
Then, we iterate through each row of the matrix:
74+
75+
- 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).
76+
- If $\textit{ans}[1] < \textit{cnt}$, update $\textit{ans} = [i, \textit{cnt}]$.
77+
78+
After finishing the iteration, we return $\textit{ans}$.
79+
80+
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 $O(1)$.
7481

7582
<!-- tabs:start -->
7683

@@ -81,7 +88,7 @@ class Solution:
8188
def rowAndMaximumOnes(self, mat: List[List[int]]) -> List[int]:
8289
ans = [0, 0]
8390
for i, row in enumerate(mat):
84-
cnt = row.count(1)
91+
cnt = sum(row)
8592
if ans[1] < cnt:
8693
ans = [i, cnt]
8794
return ans
@@ -96,9 +103,7 @@ class Solution {
96103
for (int i = 0; i < mat.length; ++i) {
97104
int cnt = 0;
98105
for (int x : mat[i]) {
99-
if (x == 1) {
100-
++cnt;
101-
}
106+
cnt += x;
102107
}
103108
if (ans[1] < cnt) {
104109
ans[0] = i;
@@ -118,13 +123,9 @@ public:
118123
vector<int> rowAndMaximumOnes(vector<vector<int>>& mat) {
119124
vector<int> ans(2);
120125
for (int i = 0; i < mat.size(); ++i) {
121-
int cnt = 0;
122-
for (auto& x : mat[i]) {
123-
cnt += x == 1;
124-
}
126+
int cnt = accumulate(mat[i].begin(), mat[i].end(), 0);
125127
if (ans[1] < cnt) {
126-
ans[0] = i;
127-
ans[1] = cnt;
128+
ans = {i, cnt};
128129
}
129130
}
130131
return ans;
@@ -136,16 +137,14 @@ public:
136137
137138
```go
138139
func rowAndMaximumOnes(mat [][]int) []int {
139-
ans := make([]int, 2)
140+
ans := []int{0, 0}
140141
for i, row := range mat {
141142
cnt := 0
142143
for _, x := range row {
143-
if x == 1 {
144-
cnt++
145-
}
144+
cnt += x
146145
}
147146
if ans[1] < cnt {
148-
ans[0], ans[1] = i, cnt
147+
ans = []int{i, cnt}
149148
}
150149
}
151150
return ans
@@ -157,8 +156,8 @@ func rowAndMaximumOnes(mat [][]int) []int {
157156
```ts
158157
function rowAndMaximumOnes(mat: number[][]): number[] {
159158
const ans: number[] = [0, 0];
160-
for (let i = 0; i < mat.length; ++i) {
161-
const cnt = mat[i].reduce((a, b) => a + b);
159+
for (let i = 0; i < mat.length; i++) {
160+
const cnt = mat[i].reduce((sum, num) => sum + num, 0);
162161
if (ans[1] < cnt) {
163162
ans[0] = i;
164163
ans[1] = cnt;
@@ -174,20 +173,34 @@ function rowAndMaximumOnes(mat: number[][]): number[] {
174173
impl Solution {
175174
pub fn row_and_maximum_ones(mat: Vec<Vec<i32>>) -> Vec<i32> {
176175
let mut ans = vec![0, 0];
177-
178176
for (i, row) in mat.iter().enumerate() {
179-
let cnt = row.iter().filter(|&v| *v == 1).count() as i32;
177+
let cnt = row.iter().sum();
180178
if ans[1] < cnt {
181-
ans[0] = i as i32;
182-
ans[1] = cnt;
179+
ans = vec![i as i32, cnt];
183180
}
184181
}
185-
186182
ans
187183
}
188184
}
189185
```
190186

187+
#### C#
188+
189+
```cs
190+
public class Solution {
191+
public int[] RowAndMaximumOnes(int[][] mat) {
192+
int[] ans = new int[2];
193+
for (int i = 0; i < mat.Length; i++) {
194+
int cnt = mat[i].Sum();
195+
if (ans[1] < cnt) {
196+
ans = new int[] { i, cnt };
197+
}
198+
}
199+
return ans;
200+
}
201+
}
202+
```
203+
191204
<!-- tabs:end -->
192205

193206
<!-- solution:end -->

‎solution/2600-2699/2643.Row With Maximum Ones/Solution.cpp

+3-7
Original file line numberDiff line numberDiff line change
@@ -3,15 +3,11 @@ class Solution {
33
vector<int> rowAndMaximumOnes(vector<vector<int>>& mat) {
44
vector<int> ans(2);
55
for (int i = 0; i < mat.size(); ++i) {
6-
int cnt = 0;
7-
for (auto& x : mat[i]) {
8-
cnt += x == 1;
9-
}
6+
int cnt = accumulate(mat[i].begin(), mat[i].end(), 0);
107
if (ans[1] < cnt) {
11-
ans[0] = i;
12-
ans[1] = cnt;
8+
ans = {i, cnt};
139
}
1410
}
1511
return ans;
1612
}
17-
};
13+
};

‎solution/2600-2699/2643.Row With Maximum Ones/Solution.cs

+12
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,12 @@
1+
public class Solution {
2+
public int[] RowAndMaximumOnes(int[][] mat) {
3+
int[] ans = new int[2];
4+
for (int i = 0; i < mat.Length; i++) {
5+
int cnt = mat[i].Sum();
6+
if (ans[1] < cnt) {
7+
ans = new int[] { i, cnt };
8+
}
9+
}
10+
return ans;
11+
}
12+
}

‎solution/2600-2699/2643.Row With Maximum Ones/Solution.go

+4-6
Original file line numberDiff line numberDiff line change
@@ -1,15 +1,13 @@
11
func rowAndMaximumOnes(mat [][]int) []int {
2-
ans := make([]int, 2)
2+
ans := []int{0, 0}
33
for i, row := range mat {
44
cnt := 0
55
for _, x := range row {
6-
if x == 1 {
7-
cnt++
8-
}
6+
cnt += x
97
}
108
if ans[1] < cnt {
11-
ans[0], ans[1] = i, cnt
9+
ans = []int{i, cnt}
1210
}
1311
}
1412
return ans
15-
}
13+
}

‎solution/2600-2699/2643.Row With Maximum Ones/Solution.java

+2-4
Original file line numberDiff line numberDiff line change
@@ -4,9 +4,7 @@ public int[] rowAndMaximumOnes(int[][] mat) {
44
for (int i = 0; i < mat.length; ++i) {
55
int cnt = 0;
66
for (int x : mat[i]) {
7-
if (x == 1) {
8-
++cnt;
9-
}
7+
cnt += x;
108
}
119
if (ans[1] < cnt) {
1210
ans[0] = i;
@@ -15,4 +13,4 @@ public int[] rowAndMaximumOnes(int[][] mat) {
1513
}
1614
return ans;
1715
}
18-
}
16+
}

‎solution/2600-2699/2643.Row With Maximum Ones/Solution.py

+1-1
Original file line numberDiff line numberDiff line change
@@ -2,7 +2,7 @@ class Solution:
22
def rowAndMaximumOnes(self, mat: List[List[int]]) -> List[int]:
33
ans = [0, 0]
44
for i, row in enumerate(mat):
5-
cnt = row.count(1)
5+
cnt = sum(row)
66
if ans[1] < cnt:
77
ans = [i, cnt]
88
return ans

‎solution/2600-2699/2643.Row With Maximum Ones/Solution.rs

+2-5
Original file line numberDiff line numberDiff line change
@@ -1,15 +1,12 @@
11
impl Solution {
22
pub fn row_and_maximum_ones(mat: Vec<Vec<i32>>) -> Vec<i32> {
33
let mut ans = vec![0, 0];
4-
54
for (i, row) in mat.iter().enumerate() {
6-
let cnt = row.iter().filter(|&v| *v == 1).count() as i32;
5+
let cnt = row.iter().sum();
76
if ans[1] < cnt {
8-
ans[0] = i as i32;
9-
ans[1] = cnt;
7+
ans = vec![i as i32, cnt];
108
}
119
}
12-
1310
ans
1411
}
1512
}

‎solution/2600-2699/2643.Row With Maximum Ones/Solution.ts

+2-2
Original file line numberDiff line numberDiff line change
@@ -1,7 +1,7 @@
11
function rowAndMaximumOnes(mat: number[][]): number[] {
22
const ans: number[] = [0, 0];
3-
for (let i = 0; i < mat.length; ++i) {
4-
const cnt = mat[i].reduce((a, b) => a + b);
3+
for (let i = 0; i < mat.length; i++) {
4+
const cnt = mat[i].reduce((sum, num) => sum + num, 0);
55
if (ans[1] < cnt) {
66
ans[0] = i;
77
ans[1] = cnt;

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