Java Optimal Solution . Will Help you to build Logic in tackling such problems 
IntuitionIntuition
The problem asks us to find the largest square area that can be formed by the intersection of any two given rectangles. My initial thought is
# Intuition
Intuition
The problem asks us to find the largest square area that can be formed by the intersection of any two given rectangles. My initial thought is to iterate through all possible pairs of rectangles and calculate the overlapping area. Then, determine the largest square that can fit within that intersection area.

# Approach
- Iterate through pairs: Iterate through all possible pairs of rectangles using nested loops.
- Check for intersection: For each pair, check if the rectangles intersect. If they don't, move to the next pair.
- Calculate intersection: If they intersect, calculate the coordinates of the intersection rectangle.
- Determine square side: Calculate the side length of the largest square that can fit within the intersection rectangle. This will be the minimum of the width and height of the intersection.
- Update maximum area: Update the maximum square area found so far.
- Return result: Return the maximum square area.

# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
O(n^2 )

- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
O(1)

# Code
```java []
class Solution {
    public long largestSquareArea(int[][] bl, int[][] tr) {
       long ans =0 , n= bl.length;
       for(int i=0;i<n;i++){
        for(int j=i+1;j<n;j++){
            if(tr[i][0] <= bl[j][0] ||
               tr[i][1] <= bl[j][1] ||
               tr[j][0] <= bl[i][0] ||
               tr[j][1] <= bl[i][1]
            ){
                continue;
            }
            int side = Math.min(
                Math.abs(Math.min(tr[i][0],tr[j][0]) - Math.max(bl[i][0],bl[j][0])),
                Math.abs(Math.min(tr[i][1],tr[j][1]) - Math.max(bl[i][1],bl[j][1]))
            );
            ans = Math.max(side,ans);
        }
       }

        return ans*ans;
    }
}
```
['Java']