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feat: add solutions to lc problem: No.0835 (doocs#1365)
No.0835.Image Overlap
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‎solution/0800-0899/0835.Image Overlap/README.md

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Original file line numberDiff line numberDiff line change
@@ -61,22 +61,145 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：枚举**
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我们可以枚举 $img1$ 和 $img2$ 的每个 $1$ 的位置，分别记为 $(i, j)$ 和 $(h, k)$。然后我们计算得到偏移量 $(i - h, j - k)$，记为 $(dx, dy)$，用哈希表 $cnt$ 记录每个偏移量出现的次数。最后我们遍历哈希表 $cnt$，找到出现次数最多的偏移量，即为答案。
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时间复杂度 $O(n^4)$，空间复杂度 $O(n^2)$。其中 $n$ 是 $img1$ 的边长。
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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 largestOverlap(self, img1: List[List[int]], img2: List[List[int]]) -> int:
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n = len(img1)
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cnt = Counter()
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for i in range(n):
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for j in range(n):
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if img1[i][j]:
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for h in range(n):
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for k in range(n):
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if img2[h][k]:
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cnt[(i - h, j - k)] += 1
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return max(cnt.values()) if cnt else 0
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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 largestOverlap(int[][] img1, int[][] img2) {
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int n = img1.length;
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Map<List<Integer>, Integer> cnt = new HashMap<>();
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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for (int j = 0; j < n; ++j) {
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if (img1[i][j] == 1) {
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for (int h = 0; h < n; ++h) {
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for (int k = 0; k < n; ++k) {
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if (img2[h][k] == 1) {
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List<Integer> t = List.of(i - h, j - k);
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ans = Math.max(ans, cnt.merge(t, 1, Integer::sum));
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}
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}
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}
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}
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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 largestOverlap(vector<vector<int>>& img1, vector<vector<int>>& img2) {
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int n = img1.size();
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map<pair<int, int>, int> cnt;
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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for (int j = 0; j < n; ++j) {
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if (img1[i][j]) {
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for (int h = 0; h < n; ++h) {
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for (int k = 0; k < n; ++k) {
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if (img2[h][k]) {
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ans = max(ans, ++cnt[{i - h, j - k}]);
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}
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}
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}
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}
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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 largestOverlap(img1 [][]int, img2 [][]int) (ans int) {
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type pair struct{ x, y int }
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cnt := map[pair]int{}
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for i, row1 := range img1 {
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for j, x1 := range row1 {
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if x1 == 1 {
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for h, row2 := range img2 {
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for k, x2 := range row2 {
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if x2 == 1 {
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t := pair{i - h, j - k}
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cnt[t]++
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ans = max(ans, cnt[t])
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}
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}
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}
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}
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}
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}
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return
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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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```
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### **TypeScript**
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```ts
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function largestOverlap(img1: number[][], img2: number[][]): number {
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const n = img1.length;
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const cnt: Map<number, number> = new Map();
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let ans = 0;
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for (let i = 0; i < n; ++i) {
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for (let j = 0; j < n; ++j) {
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if (img1[i][j] === 1) {
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for (let h = 0; h < n; ++h) {
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for (let k = 0; k < n; ++k) {
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if (img2[h][k] === 1) {
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const t = (i - h) * 200 + (j - k);
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cnt.set(t, (cnt.get(t) ?? 0) + 1);
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ans = Math.max(ans, cnt.get(t)!);
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}
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}
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}
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}
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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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### **...**

‎solution/0800-0899/0835.Image Overlap/README_EN.md

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@@ -56,13 +56,130 @@ The number of positions that have a 1 in both images is 3 (shown in red).
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### **Python3**
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```python
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class Solution:
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def largestOverlap(self, img1: List[List[int]], img2: List[List[int]]) -> int:
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n = len(img1)
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cnt = Counter()
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for i in range(n):
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for j in range(n):
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if img1[i][j]:
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for h in range(n):
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for k in range(n):
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if img2[h][k]:
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cnt[(i - h, j - k)] += 1
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return max(cnt.values()) if cnt else 0
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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 largestOverlap(int[][] img1, int[][] img2) {
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int n = img1.length;
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Map<List<Integer>, Integer> cnt = new HashMap<>();
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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for (int j = 0; j < n; ++j) {
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if (img1[i][j] == 1) {
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for (int h = 0; h < n; ++h) {
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for (int k = 0; k < n; ++k) {
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if (img2[h][k] == 1) {
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List<Integer> t = List.of(i - h, j - k);
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ans = Math.max(ans, cnt.merge(t, 1, Integer::sum));
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}
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}
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}
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}
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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 largestOverlap(vector<vector<int>>& img1, vector<vector<int>>& img2) {
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int n = img1.size();
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map<pair<int, int>, int> cnt;
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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for (int j = 0; j < n; ++j) {
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if (img1[i][j]) {
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for (int h = 0; h < n; ++h) {
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for (int k = 0; k < n; ++k) {
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if (img2[h][k]) {
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ans = max(ans, ++cnt[{i - h, j - k}]);
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}
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}
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}
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}
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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 largestOverlap(img1 [][]int, img2 [][]int) (ans int) {
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type pair struct{ x, y int }
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cnt := map[pair]int{}
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for i, row1 := range img1 {
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for j, x1 := range row1 {
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if x1 == 1 {
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for h, row2 := range img2 {
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for k, x2 := range row2 {
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if x2 == 1 {
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t := pair{i - h, j - k}
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cnt[t]++
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ans = max(ans, cnt[t])
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}
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}
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}
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}
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}
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}
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return
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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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```
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### **TypeScript**
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```ts
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function largestOverlap(img1: number[][], img2: number[][]): number {
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const n = img1.length;
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const cnt: Map<number, number> = new Map();
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let ans = 0;
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for (let i = 0; i < n; ++i) {
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for (let j = 0; j < n; ++j) {
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if (img1[i][j] === 1) {
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for (let h = 0; h < n; ++h) {
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for (let k = 0; k < n; ++k) {
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if (img2[h][k] === 1) {
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const t = (i - h) * 200 + (j - k);
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cnt.set(t, (cnt.get(t) ?? 0) + 1);
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ans = Math.max(ans, cnt.get(t)!);
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}
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}
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}
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}
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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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### **...**

‎solution/0800-0899/0835.Image Overlap/Solution.cpp

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class Solution {
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public:
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int largestOverlap(vector<vector<int>>& img1, vector<vector<int>>& img2) {
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int n = img1.size();
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map<pair<int, int>, int> cnt;
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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for (int j = 0; j < n; ++j) {
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if (img1[i][j]) {
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for (int h = 0; h < n; ++h) {
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for (int k = 0; k < n; ++k) {
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if (img2[h][k]) {
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ans = max(ans, ++cnt[{i - h, j - k}]);
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}
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}
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}
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}
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}
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}
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return ans;
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}
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};

‎solution/0800-0899/0835.Image Overlap/Solution.go

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func largestOverlap(img1 [][]int, img2 [][]int) (ans int) {
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type pair struct{ x, y int }
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cnt := map[pair]int{}
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for i, row1 := range img1 {
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for j, x1 := range row1 {
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if x1 == 1 {
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for h, row2 := range img2 {
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for k, x2 := range row2 {
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if x2 == 1 {
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t := pair{i - h, j - k}
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cnt[t]++
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ans = max(ans, cnt[t])
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}
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}
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}
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}
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}
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}
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return
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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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}

‎solution/0800-0899/0835.Image Overlap/Solution.java

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class Solution {
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public int largestOverlap(int[][] img1, int[][] img2) {
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int n = img1.length;
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Map<List<Integer>, Integer> cnt = new HashMap<>();
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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for (int j = 0; j < n; ++j) {
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if (img1[i][j] == 1) {
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for (int h = 0; h < n; ++h) {
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for (int k = 0; k < n; ++k) {
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if (img2[h][k] == 1) {
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List<Integer> t = List.of(i - h, j - k);
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ans = Math.max(ans, cnt.merge(t, 1, Integer::sum));
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}
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}
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}
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}
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}
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}
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return ans;
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}
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}

‎solution/0800-0899/0835.Image Overlap/Solution.py

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Original file line numberDiff line numberDiff line change
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class Solution:
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def largestOverlap(self, img1: List[List[int]], img2: List[List[int]]) -> int:
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n = len(img1)
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cnt = Counter()
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for i in range(n):
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for j in range(n):
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if img1[i][j]:
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for h in range(n):
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for k in range(n):
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if img2[h][k]:
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cnt[(i - h, j - k)] += 1
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return max(cnt.values()) if cnt else 0

‎solution/0800-0899/0835.Image Overlap/Solution.ts

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Original file line numberDiff line numberDiff line change
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function largestOverlap(img1: number[][], img2: number[][]): number {
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const n = img1.length;
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const cnt: Map<number, number> = new Map();
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let ans = 0;
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for (let i = 0; i < n; ++i) {
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for (let j = 0; j < n; ++j) {
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if (img1[i][j] === 1) {
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for (let h = 0; h < n; ++h) {
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for (let k = 0; k < n; ++k) {
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if (img2[h][k] === 1) {
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const t = (i - h) * 200 + (j - k);
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cnt.set(t, (cnt.get(t) ?? 0) + 1);
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ans = Math.max(ans, cnt.get(t)!);
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}
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}
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}
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}
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}
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}
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return ans;
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}

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