Create Image_Overlap.cpp

6 CP/LEETCODE/Image_Overlap.cpp

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+// Image Overlap
+// Problem link:
+// https://leetcode.com/problems/image-overlap/
+
+/*
+Intution:
+Same overlapping zone would share the same relative positional differnece(linear transformation vector).
+**Vab = (Xb - Xa, Yb - Ya)**
+```
+
+  IMG 1       IMG 2
+  1 1 0       0 0 0
+  0 1 0       0 1 1
+  0 1 0       0 0 1
+
+// Subtracting the co-ordinates
+
+{0,0}  - {1,1} = { -1 , -1 } 
+{0,0}  - {1,2} = { -1,  -2 }
+{0,0} - {2, 2} = { -2, -2 }
+
+{0,1}  - {1,1} = { -1 , 0 } 
+{0,1}  - {1,2} = { -1 , -1 } 
+{0,1} - {2, 2} = { -2, -1 }
+
+{0,2}  - {1,1} = { -1 , 1 } 
+{0,2}  - {1,2} = { -1 , 0 } 
+{0,2}  - {2,2} = { -2 , 0 } 
+
+and count of { -1 , -1 } is the max overlaping zone = 3 .
+*/
+
+// Implementation:
+
+#include<bits/stdc++.h>
+using namespace std;
+
+class Solution {
+public:
+    vector<pair<int, int>> nonZeroCells(vector<vector<int>>& img){
+        // filter out those non-zero cells in each matrix respectively.
+        vector<pair<int, int>> res;
+        for(int i = 0; i < img.size(); i++){
+            for(int j = 0; j < img.size(); j++){
+                if(img[i][j] == 1)
+                    res.push_back({i,j});
+            }
+        }
+        return res;
+    }
+    int largestOverlap(vector<vector<int>>& img1, vector<vector<int>>& img2) {
+        vector<pair<int, int>> AOnes = nonZeroCells(img1); // {row, col}
+        vector<pair<int, int>> BOnes = nonZeroCells(img2);
+        int maxOverlaps = 0;
+        // calculate the corresponding relative positional differnece(linear transformation vector)
+        // as Vab = (Xb - Xa, Yb - Ya)
+        // (row, col) -> count
+        map<pair<int, int>, int> groupCount;
+        for(auto a: AOnes){
+            for(auto b: BOnes){
+                pair<int, int> val = {b.first-a.first, b.second-a.second};
+                // same overlapping zone would share the same relative positional differnece(linear transformation vector).
+                groupCount[val]++;
+                maxOverlaps = max(maxOverlaps, groupCount[val]);
+            }
+        }
+        return maxOverlaps;
+    }
+};