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feat: add solutions to lc problem: No.1183
No.1183.Maximum Number of Ones
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‎solution/1100-1199/1183.Maximum Number of Ones/README.md

Lines changed: 110 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -50,22 +50,131 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：统计等效位置**
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为了方便说明，我们不妨令 `x = sideLength`。
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考虑一个 $x\times x$ 的正方形，我们需要在正方形里面取最多 `maxOnes` 个点，将其置为 1。注意到当坐标 $(i, j)$ 处的点被选取后，所有坐标为 $(i\pm k_1 \times x, j\pm k_2 \times x)$ 的点都可以等效地置为 1。因此，我们算出坐标 $(i, j)$ 在矩阵中的等效位置的数量，取数量最多的前 `maxOnes` 个即可。
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时间复杂度 $O(m\times n)$。其中 $m$, $n$ 分别是矩阵的行数和列数。
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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 maximumNumberOfOnes(
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self, width: int, height: int, sideLength: int, maxOnes: int
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) -> int:
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x = sideLength
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cnt = [0] * (x * x)
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for i in range(width):
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for j in range(height):
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k = (i % x) * x + (j % x)
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cnt[k] += 1
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cnt.sort(reverse=True)
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return sum(cnt[:maxOnes])
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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 maximumNumberOfOnes(int width, int height, int sideLength, int maxOnes) {
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int x = sideLength;
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int[] cnt = new int[x * x];
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for (int i = 0; i < width; ++i) {
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for (int j = 0; j < height; ++j) {
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int k = (i % x) * x + (j % x);
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++cnt[k];
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}
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}
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Arrays.sort(cnt);
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int ans = 0;
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for (int i = 0; i < maxOnes; ++i) {
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ans += cnt[cnt.length - i - 1];
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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 maximumNumberOfOnes(int width, int height, int sideLength, int maxOnes) {
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int x = sideLength;
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vector<int> cnt(x * x);
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for (int i = 0; i < width; ++i) {
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for (int j = 0; j < height; ++j) {
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int k = (i % x) * x + (j % x);
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++cnt[k];
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}
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}
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sort(cnt.rbegin(), cnt.rend());
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int ans = 0;
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for (int i = 0; i < maxOnes; ++i) {
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ans += cnt[i];
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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 maximumNumberOfOnes(width int, height int, sideLength int, maxOnes int) int {
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x := sideLength
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cnt := make([]int, x*x)
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for i := 0; i < width; i++ {
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for j := 0; j < height; j++ {
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k := (i%x)*x + (j % x)
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cnt[k]++
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}
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}
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sort.Ints(cnt)
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ans := 0
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for i := range cnt[:maxOnes] {
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ans += cnt[len(cnt)-i-1]
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}
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return ans
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}
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```
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### **JavaScript**
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```js
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/**
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* @param {number} width
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* @param {number} height
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* @param {number} sideLength
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* @param {number} maxOnes
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* @return {number}
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*/
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var maximumNumberOfOnes = function (width, height, sideLength, maxOnes) {
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const x = sideLength;
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const cnt = new Array(x * x).fill(0);
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for (let i = 0; i < width; ++i) {
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for (let j = 0; j < height; ++j) {
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const k = (i % x) * x + (j % x);
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++cnt[k];
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}
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}
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cnt.sort((a, b) => b - a);
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let ans = 0;
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for (let i = 0; i < maxOnes; ++i) {
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ans += cnt[i];
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}
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return ans;
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};
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```
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### **...**

‎solution/1100-1199/1183.Maximum Number of Ones/README_EN.md

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@@ -49,13 +49,114 @@ The best solution that has 4 ones is:
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### **Python3**
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```python
52-
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class Solution:
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def maximumNumberOfOnes(
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self, width: int, height: int, sideLength: int, maxOnes: int
55+
) -> int:
56+
x = sideLength
57+
cnt = [0] * (x * x)
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for i in range(width):
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for j in range(height):
60+
k = (i % x) * x + (j % x)
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cnt[k] += 1
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cnt.sort(reverse=True)
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return sum(cnt[:maxOnes])
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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 maximumNumberOfOnes(int width, int height, int sideLength, int maxOnes) {
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int x = sideLength;
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int[] cnt = new int[x * x];
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for (int i = 0; i < width; ++i) {
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for (int j = 0; j < height; ++j) {
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int k = (i % x) * x + (j % x);
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++cnt[k];
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}
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}
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Arrays.sort(cnt);
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int ans = 0;
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for (int i = 0; i < maxOnes; ++i) {
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ans += cnt[cnt.length - i - 1];
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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 maximumNumberOfOnes(int width, int height, int sideLength, int maxOnes) {
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int x = sideLength;
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vector<int> cnt(x * x);
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for (int i = 0; i < width; ++i) {
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for (int j = 0; j < height; ++j) {
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int k = (i % x) * x + (j % x);
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++cnt[k];
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}
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}
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sort(cnt.rbegin(), cnt.rend());
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int ans = 0;
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for (int i = 0; i < maxOnes; ++i) {
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ans += cnt[i];
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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 maximumNumberOfOnes(width int, height int, sideLength int, maxOnes int) int {
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x := sideLength
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cnt := make([]int, x*x)
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for i := 0; i < width; i++ {
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for j := 0; j < height; j++ {
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k := (i%x)*x + (j % x)
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cnt[k]++
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}
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}
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sort.Ints(cnt)
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ans := 0
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for i := range cnt[:maxOnes] {
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ans += cnt[len(cnt)-i-1]
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}
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return ans
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}
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```
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### **JavaScript**
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```js
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/**
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* @param {number} width
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* @param {number} height
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* @param {number} sideLength
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* @param {number} maxOnes
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* @return {number}
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*/
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var maximumNumberOfOnes = function (width, height, sideLength, maxOnes) {
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const x = sideLength;
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const cnt = new Array(x * x).fill(0);
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for (let i = 0; i < width; ++i) {
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for (let j = 0; j < height; ++j) {
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const k = (i % x) * x + (j % x);
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++cnt[k];
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}
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}
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cnt.sort((a, b) => b - a);
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let ans = 0;
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for (let i = 0; i < maxOnes; ++i) {
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ans += cnt[i];
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}
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return ans;
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};
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```
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### **...**

‎solution/1100-1199/1183.Maximum Number of Ones/Solution.cpp

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,19 @@
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class Solution {
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public:
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int maximumNumberOfOnes(int width, int height, int sideLength, int maxOnes) {
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int x = sideLength;
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vector<int> cnt(x * x);
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for (int i = 0; i < width; ++i) {
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for (int j = 0; j < height; ++j) {
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int k = (i % x) * x + (j % x);
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++cnt[k];
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}
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}
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sort(cnt.rbegin(), cnt.rend());
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int ans = 0;
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for (int i = 0; i < maxOnes; ++i) {
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ans += cnt[i];
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}
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return ans;
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}
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};

‎solution/1100-1199/1183.Maximum Number of Ones/Solution.go

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,16 @@
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func maximumNumberOfOnes(width int, height int, sideLength int, maxOnes int) int {
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x := sideLength
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cnt := make([]int, x*x)
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for i := 0; i < width; i++ {
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for j := 0; j < height; j++ {
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k := (i%x)*x + (j % x)
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cnt[k]++
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}
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}
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sort.Ints(cnt)
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ans := 0
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for i := range cnt[:maxOnes] {
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ans += cnt[len(cnt)-i-1]
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}
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return ans
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}

‎solution/1100-1199/1183.Maximum Number of Ones/Solution.java

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,18 @@
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class Solution {
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public int maximumNumberOfOnes(int width, int height, int sideLength, int maxOnes) {
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int x = sideLength;
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int[] cnt = new int[x * x];
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for (int i = 0; i < width; ++i) {
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for (int j = 0; j < height; ++j) {
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int k = (i % x) * x + (j % x);
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++cnt[k];
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}
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}
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Arrays.sort(cnt);
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int ans = 0;
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for (int i = 0; i < maxOnes; ++i) {
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ans += cnt[cnt.length - i - 1];
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}
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return ans;
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}
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}

‎solution/1100-1199/1183.Maximum Number of Ones/Solution.js

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,23 @@
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/**
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* @param {number} width
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* @param {number} height
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* @param {number} sideLength
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* @param {number} maxOnes
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* @return {number}
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*/
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var maximumNumberOfOnes = function (width, height, sideLength, maxOnes) {
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const x = sideLength;
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const cnt = new Array(x * x).fill(0);
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for (let i = 0; i < width; ++i) {
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for (let j = 0; j < height; ++j) {
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const k = (i % x) * x + (j % x);
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++cnt[k];
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}
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}
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cnt.sort((a, b) => b - a);
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let ans = 0;
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for (let i = 0; i < maxOnes; ++i) {
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ans += cnt[i];
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}
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return ans;
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};

‎solution/1100-1199/1183.Maximum Number of Ones/Solution.py

Lines changed: 12 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,12 @@
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class Solution:
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def maximumNumberOfOnes(
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self, width: int, height: int, sideLength: int, maxOnes: int
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) -> int:
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x = sideLength
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cnt = [0] * (x * x)
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for i in range(width):
8+
for j in range(height):
9+
k = (i % x) * x + (j % x)
10+
cnt[k] += 1
11+
cnt.sort(reverse=True)
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return sum(cnt[:maxOnes])

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