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image overlap | leetcode solution

835. Image Overlap

You are given two images, img1 and img2, represented as binary, square matrices of size n x n. A binary matrix has only 0s and 1s as values.

We translate one image however we choose by sliding all the 1 bits left, right, up, and/or down any number of units. We then place it on top of the other image. We can then calculate the overlap by counting the number of positions that have a 1 in both images.

Note also that a translation does not include any kind of rotation. Any 1 bits that are translated outside of the matrix borders are erased.

Return the largest possible overlap.

 

Example 1:

Input: img1 = [[1,1,0],[0,1,0],[0,1,0]], img2 = [[0,0,0],[0,1,1],[0,0,1]]
Output: 3
Explanation: We translate img1 to right by 1 unit and down by 1 unit.

The number of positions that have a 1 in both images is 3 (shown in red).

Example 2:

Input: img1 = [[1]], img2 = [[1]]
Output: 1

Example 3:

Input: img1 = [[0]], img2 = [[0]]
Output: 0

 

Constraints:

  • n == img1.length == img1[i].length
  • n == img2.length == img2[i].length
  • 1 <= n <= 30
  • img1[i][j] is either 0 or 1.
  • img2[i][j] is either 0 or 1.

solution:

class Solution {
public:
    int largestOverlap(vector<vector<int>>& img1, vector<vector<int>>& img2) {
        int n=img1.size();
        vector<pair<int,int>>vec_a;
        vector<pair<int,int>>vec_b;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(img1[i][j]==1){
                    vec_a.push_back({i,j});
                }
                if(img2[i][j]==1){
                    vec_b.push_back({i,j});
                }
            }
        }
        int ans=0;
        map<pair<int,int>,int>mp;
        for(auto [i1,j1]:vec_a){
            for(auto [i2,j2]:vec_b){
                mp[{i1-i2,j1-j2}]++;
                ans=max(ans,mp[{i1-i2,j1-j2}]);
            }
        }
        return ans;
    }
};

 

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