How do you solve Image Overlap on LeetCode?

Image Overlap (LeetCode #835) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n²) space complexity. Key insight: Understanding translations is crucial for overlap calculation.

What is the brute force solution for Image Overlap?

The brute force approach for Image Overlap is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible translation of one image over the other. For each translation, we count the overlapping 1s to find the maximum overlap. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Image Overlap?

Image Overlap uses the following concepts: Array, Matrix. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Image Overlap?

Explain your thought process clearly while coding. Mention the brute-force approach before jumping to optimizations. Practice translating concepts into code to improve clarity.

What are common mistakes when solving Image Overlap?

Not considering all possible translations. Overlooking edge cases where images have no overlap.

#835

Image Overlap

Medium

ArrayMatrixHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n²)

💡

Intuition

Time O(n²)Space O(n²)

The optimal solution leverages a hashmap to count the translations that yield overlaps. By using the positions of 1s in both matrices, we can efficiently calculate the maximum overlap without redundant checks.

⚙️

Algorithm

4 steps

1Step 1: Create a hashmap to count overlaps for each translation vector derived from 1s in both images.
2Step 2: For each 1 in img1, calculate its translation vector to each 1 in img2.
3Step 3: Increment the count in the hashmap for each translation vector.
4Step 4: The maximum value in the hashmap is the largest overlap.

solution.py14 lines

1from collections import defaultdict
2
3def largestOverlap(img1, img2):
4    n = len(img1)
5    count = defaultdict(int)
6    for i in range(n):
7        for j in range(n):
8            if img1[i][j] == 1:
9                for x in range(n):
10                    for y in range(n):
11                        if img2[x][y] == 1:
12                            dx, dy = i - x, j - y
13                            count[(dx, dy)] += 1
14    return max(count.values(), default=0)

ℹ

Complexity note: The time complexity remains O(n²) as we still check all pairs of 1s, but we use a hashmap to store translation counts, which requires O(n²) space in the worst case.

1Understanding translations is crucial for overlap calculation.
2Using a hashmap can significantly reduce redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.