import java.util.*; //We need to handle this challenge from the perspective of understanding the shapes in the top right (positive x axis and y axis) //I have a feeling that once this is established. //I can then use principles of test cases 1-7 in which I had completed the exact overlaps... In this area, I focussed on the //implications of having coordinates in the positive x and negative x axis. I would need to apply the following logic //if ((rect2BottomLeft[1]<0 && rect2TopRight[1]>0)) into my main analysis undertaken in the positive axis. //Since the principles should still hold //When it reaches test case 22 and 23, there are new challenges that have arisen public class Solution { public static int overlappingArea(int[] rect1BottomLeft, int[] rect1TopRight, int[] rect2BottomLeft, int[] rect2TopRight) { int width=0; int height=0; int [] store = new int[4]; int count=0; boolean hasExactOverlap=false; if ((rect1BottomLeft[0]) >= (rect1TopRight[0]) || (rect2BottomLeft[0]) >= (rect2TopRight[0])) { System.out.println("INVALID RECTANGLE CONFIGURATION"); return 0; } //used for test case 11 if (rect1BottomLeft[0]>=rect2TopRight[0] || (rect2BottomLeft[0]>=rect1TopRight[0]) || (rect1BottomLeft[1]>=rect2TopRight[1]) || (rect2BottomLeft[1]>=rect1TopRight[1])) { System.out.println("NO OVERLAP FOUND"); return 0; } if (((rect1BottomLeft[0]) == (rect1TopRight[0])) || ((rect2BottomLeft[0]) == (rect2TopRight[0])) || ((rect1BottomLeft[1]) == (rect1TopRight[1])) || ((rect1BottomLeft[1]) == (rect1TopRight[1]))) { System.out.println("One or more squares are dimensionless.. No overlap/area calculation possible"); return 0; } //now need to focus on exact overlap System.out.println("rect1Topright: " + rect1TopRight[1]); System.out.println("rect2TopRight: " + rect2TopRight[1]); System.out.println("rect1BottomLeft: " + rect1BottomLeft[1]); System.out.println("rect2BottomLeft: " + rect2BottomLeft[1]); if (rect1TopRight[1]==rect2TopRight[1] &&(rect1BottomLeft[1]==rect2BottomLeft[1]) &&(rect1TopRight[0]==rect2TopRight[0]) &&(rect1BottomLeft[0]==rect2BottomLeft[0])) { System.out.println("EXACT overlap"); hasExactOverlap=true; if ((rect2BottomLeft[1]<0 && rect2TopRight[1]>0)) { width = Math.abs(0-rect2BottomLeft[0]) + Math.abs(0-rect2TopRight[0]); height = Math.abs(0-rect2BottomLeft[1]) + Math.abs(0-rect2TopRight[1]); } else { width = Math.abs(0-rect2BottomLeft[0]) - Math.abs(0-rect2TopRight[0]); System.out.println(width); height = Math.abs(0-rect2BottomLeft[1]) - Math.abs(0-rect2TopRight[1]); System.out.println(height); } return (Math.abs(width) * Math.abs(height)); } //this is now if there is violation and overlap spans four quadrants. //it would be quite impossible to calculate if (!hasExactOverlap) { if ((rect1BottomLeft[0]<0) && (rect1TopRight[0]>0) && (rect2BottomLeft[0]<0) && (rect2TopRight[0]>0)) { System.out.println("1CHECKING OVERLAP 4 QUADRANTS"); if ((rect1BottomLeft[1]<0) && (rect1TopRight[1]>0) && (rect2BottomLeft[1]<0) && (rect2TopRight[1]>0)) { System.out.println("VIOLATION"); return 0; } } } //if all in the positive axis or //need to check if any shapes cross an axis, since we know the calculation will //be impacted since we need to either add/subtract the width or height //effectively we need to check //rectangle1 width crosses the x axis //rectangle1 height crosses the y axis //rectangle2 width crosses the x axis //rectangle2 height crosses the y axis /* //all in positive quadrant if ((rect1BottomLeft[1]>=0 && rect2BottomLeft[1]>0 && rect1TopRight[1]>=0 && rect2TopRight[1]>=0) || (rect1BottomLeft[0]<0 && rect1TopRight[0]>0) //rectangle1 width crosses the x axis || (rect1BottomLeft[1]<0 && rect1TopRight[1]>0) //rectangle1 height crosses the y axis || (rect2BottomLeft[0]<0 && rect2TopRight[0]>0) //rectangle2 width crosses the x axis || (rect2BottomLeft[1]<0 && rect2TopRight[1]>0)) //rectangle2 height crosses the y axis */ /* if ((rect1BottomLeft[1]>=0 && rect2BottomLeft[1]>0 && rect1TopRight[1]>=0 && rect2TopRight[1]>=0)) */ //{ //whether the shapes are physically overlapped //looks at x axis //x axis smalller if(rect1BottomLeft[0]rect2TopRight[0]) { //I am in this section due to Test case 58 and 59 //we see the following //rect2=red //if(rect1BottomLeft[0]0) && (rect2BottomLeft[0]<0) && (rect2TopRight[0]>0)) { width = Math.abs(0-rect2BottomLeft[0]) + Math.abs(0-rect2TopRight[0]); height = Math.abs(0-rect1BottomLeft[1]) - Math.abs(0-rect1TopRight[1]); System.out.println(width); System.out.println(height); System.out.println("MUST39"); } else { width = Math.abs(0-rect2BottomLeft[0]) - Math.abs(0-rect2TopRight[0]); height = Math.abs(0-rect1BottomLeft[1]) - Math.abs(0-rect1TopRight[1]); System.out.println("HERE9"); System.out.println(width); System.out.println(height); } return (Math.abs(width) * Math.abs(height)); } else { //This will be new code here to ascertain if a rectangle appears on both sides of the Y axis. //Based on the above statement, we know that rectangle1 top right coordinate is less than rect2 top right //we will know check if rect2 runs on both sides of the Y axis //At moment, we do not know implications if both rectangles cross the Y axis.. But it is ignored at this moment in time //This is in reference to test case 38 and 39 if (rect2BottomLeft[0]<0 && rect2TopRight[0]>0) { //This is in regarding test case 56 and 57 //it does not appear that any form of reference has been made above in relation to both shapes spanning Y axis //the only areas of outer if loops include: //if(rect1BottomLeft[0]0) && (rect2BottomLeft[0]<0) && (rect2TopRight[0]>0)) { //note addition width = Math.abs(0-rect2BottomLeft[0]) + Math.abs(0-rect1TopRight[0]); System.out.println(width); height = Math.abs(0-rect1TopRight[1]) - Math.abs(0-rect2BottomLeft[1]); System.out.println(height); System.out.println("MUST37"); } else { width = Math.abs(0-rect2BottomLeft[0]) - Math.abs(0-rect1TopRight[0]); System.out.println(width); height = Math.abs(0-rect2TopRight[1]) - Math.abs(0-rect1BottomLeft[1]); System.out.println(width); System.out.println("MUST15"); } } //in this else area, we know that rectangle does not cross the Y axis //we know in this area. we know the the blue rectangle is rectangle1..... //We also know that following statement is not true.... rect1TopRight[0]>rect2TopRight[0]) //top right X coordinate of the blue rectangle1 is 8 //top right X coordinate of the red rectangle 2 is also 8 //but in else statement, we need to consider if both are the same.... //in which case the area can be calculated different.. //This is in relation to Test Case 23new1 else { if (rect1TopRight[0]==rect2TopRight[0]) { //red is rectangle2 width = Math.abs(0-rect2BottomLeft[0]) - Math.abs(0-rect2TopRight[0]); System.out.println(width); height = Math.abs(0-rect1TopRight[1]) - Math.abs(0-rect1BottomLeft[1]); System.out.println(width); System.out.println("MUST51"); } else { System.out.println("HERE1"); width = Math.abs(0-rect2BottomLeft[0]) - Math.abs(0-rect1TopRight[0]); System.out.println(width); height = Math.abs(0-rect2BottomLeft[1]) - Math.abs(0-rect1TopRight[1]); System.out.println(height); } } return (Math.abs(width) * Math.abs(height)); } } } //Same code as above, but flipped (test case 9)----------------------------------------------------------- if(rect2BottomLeft[0]rect1TopRight[0]) { width = Math.abs(0-rect1BottomLeft[0]) - Math.abs(0-rect1TopRight[0]); height = Math.abs(0-rect2BottomLeft[1]) - Math.abs(0-rect2TopRight[1]); if ((rect2BottomLeft[0]<0) && (rect2TopRight[0]>0) && (rect1BottomLeft[0]<0) && (rect1TopRight[0]>0)) { width = Math.abs(0-rect1BottomLeft[0]) + Math.abs(0-rect1TopRight[0]); height = Math.abs(0-rect2BottomLeft[1]) - Math.abs(0-rect2TopRight[1]); System.out.println(width); System.out.println(height); System.out.println("MUST40"); } else { System.out.println("HERE10"); System.out.println(width); System.out.println(height); } return (Math.abs(width) * Math.abs(height)); } else { //This will be new code here to ascertain if a rectangle appears on both sides of the Y axis. //Based on the above statement, we know that rectangle1 top right coordinate is less than rect2 top right //we will know check if rect2 runs on both sides of the Y axis //At moment, we do not know implications if both rectangles cross the Y axis.. But it is ignored at this moment in time //This is in reference to test case 38 and 39 if (rect1BottomLeft[0]<0 && rect1TopRight[0]>0) { //This is in regarding test case 56 and 57 //it does not appear that any form of reference has been made above in relation to both shapes spanning Y axis //the only areas of outer if loops include: //if(rect1BottomLeft[0]0) && (rect1BottomLeft[0]<0) && (rect1TopRight[0]>0)) { //note addition width = Math.abs(0-rect1BottomLeft[0]) + Math.abs(0-rect2TopRight[0]); System.out.println(width); height = Math.abs(0-rect2TopRight[1]) - Math.abs(0-rect1BottomLeft[1]); System.out.println(height); System.out.println("MUST38"); } else { width = Math.abs(0-rect1BottomLeft[0]) - Math.abs(0-rect2TopRight[0]); System.out.println(width); height = Math.abs(0-rect1TopRight[1]) - Math.abs(0-rect2BottomLeft[1]); System.out.println(width); System.out.println("MUST16"); } } //in this else area, we know that rectangle does not cross the Y axis //we know in this area. we know the the blue rectangle is rectangle1..... //We also know that following statement is not true.... rect1TopRight[0]>rect2TopRight[0]) //top right X coordinate of the blue rectangle1 is 8 //top right X coordinate of the red rectangle 2 is also 8 //but in else statement, we need to consider if both are the same.... //in which case the area can be calculated different.. //This is in relation to Test Case 23new1 else { if (rect2TopRight[0]==rect1TopRight[0]) { //red is rectangle2 width = Math.abs(0-rect1BottomLeft[0]) - Math.abs(0-rect1TopRight[0]); System.out.println(width); height = Math.abs(0-rect2TopRight[1]) - Math.abs(0-rect2BottomLeft[1]); System.out.println(width); System.out.println("MUST52"); } else { System.out.println("HERE2"); width = Math.abs(0-rect1BottomLeft[0]) - Math.abs(0-rect2TopRight[0]); System.out.println(width); height = Math.abs(0-rect1BottomLeft[1]) - Math.abs(0-rect2TopRight[1]); System.out.println(height); } } return (Math.abs(width) * Math.abs(height)); } } } //whether shape is inside shape //to make it compliant with test case 14 and test case 15, I included less than and equal to as part of equality check if(rect1BottomLeft[0]>=rect2BottomLeft[0] || rect2BottomLeft[0]>=rect1BottomLeft[0]) { //test case 10 if (rect1TopRight[1]<=rect2TopRight[1]) { //needed this as part of test cases 14 and 15 //to exactly identify which is inside the other one //with test case 24 and test case 25 it is performning assumption that shape is inside shape //we can understand from the perspective of the X coordinates, it fits inside //But we have not contemplated the Y coordinate if (rect1BottomLeft[1] >= rect2BottomLeft[1]) { if(rect1BottomLeft[0]>=rect2BottomLeft[0]) { //we are here because of issues in test case 22new2 and 23new2 //at this point we know that rectangle 1 is blue, rectangle 2 is red //overlappingArea(new int[]{4, 2}, new int[]{8, 4},new int[]{4, 0}, new int[]{5, 6})); //we know the blue has a higher or equal Y coordinate than red triangle //we also know that blue has a higher or equal X coordinate than red triangle //but really we are concerned with a different calculation if the x coordinates of the two rectangles are the same //rect2=red if (rect1BottomLeft[0]==rect2BottomLeft[0]) { if (rect1TopRight[1]<=rect2TopRight[1]) { width = Math.abs(0-rect1BottomLeft[0]) - Math.abs(0-rect1TopRight[0]); System.out.println(width); } else { width = Math.abs(0-rect2BottomLeft[0]) - Math.abs(0-rect2TopRight[0]); System.out.println(width); } //CHANGE HERE height = Math.abs(0-rect1BottomLeft[1]) - Math.abs(0-rect1TopRight[1]); System.out.println(height); System.out.println("MUST29"); store[count]=(Math.abs(width) * Math.abs(height)); count++; } else { //We are here because of test case 52. Blue is rectangle1 //overlappingArea(new int[]{-4, -6}, new int[]{4, -3},new int[]{-2, -5}, new int[]{5, -4})) //It is important to understand what we know so far..... //We know the X coordinates of the bottom left of the rectangles are not the same.... //but we know that the rectangle botton left coordinate of rectangle2 (red) is greater than blue //we also know that Y coordinate (top right) of blue rectangle is greater than equal to red.. //we are also here because //x coordinate of blue rectangle(rect2) (bottom left ) is greater than or equal to red(rect1) //OR //x coordinate of red rectangle(rect1) (bottom left ) is greater than or equal to blue(rect2) //In this area of code, we have not touched upon the shapes spanning the Y axis //AND MOST IMPORTANTLY THE OVERLAP SPANNING ACROSS THE Y AXIS //I will get the if expression from above and re-use it here with caution (both rectangles spanning across Y axis) if ((rect2BottomLeft[0]<0) && (rect2TopRight[0]>0) && (rect1BottomLeft[0]<0) && (rect1TopRight[0]>0)) { //note addition of the width //rect1 = red width = Math.abs(0-rect2TopRight[0]) + Math.abs(0-rect1BottomLeft[0]); System.out.println(width); //note since the shape has not crossed height wise across the X axis, //we are not changing this.... //but it will potentially become an issue when I further devise test cases //in which overlap is on both side of the X axis height = Math.abs(0-rect1BottomLeft[1]) - Math.abs(0-rect1TopRight[1]); System.out.println(height); System.out.println("MUST34"); store[count]=(Math.abs(width) * Math.abs(height)); count++; } else { //in here due to test case 82 and 83 //the associated if is due to there being rectangles across the Y axis //this is clearly not the case, yet the logic seems to be lacking massively.... //it needs spliting into if else //we know following at this point: both rectangles have different X coordinate for bottom left //rect2bottomleft X coordinate is greater than or equal rect1bottomleft (red) //rect1topright (red) Y coordinate is greater equal to rect2TopRight //there is every possibility that is requires the exact same code as //red = rect2 if (rect2TopRight[0]>rect1TopRight[0]) { System.out.println("MUST60"); width = Math.abs(0-rect1BottomLeft[0]) - Math.abs(0-rect1TopRight[0]); System.out.println(width); height = Math.abs(0-rect1BottomLeft[1]) - Math.abs(0-rect1TopRight[1]); System.out.println(height); store[count]=(Math.abs(width) * Math.abs(height)); count++; } //this would be shapes not spanning Y axis //it has used the property of one shape.. //this is potentially my first bits of code since it assumes that rect1 and rect2 shares the same X coordinate... //it would be false under any other circumstance.. But I am leaving it intact... if (rect2TopRight[0]==rect1TopRight[0]) { System.out.println("MUST1"); width = Math.abs(0-rect1BottomLeft[0]) - Math.abs(0-rect1TopRight[0]); System.out.println(width); height = Math.abs(0-rect1BottomLeft[1]) - Math.abs(0-rect1TopRight[1]); System.out.println(height); store[count]=(Math.abs(width) * Math.abs(height)); count++; } else { System.out.println("MUST62"); width = Math.abs(0-rect1BottomLeft[0]) - Math.abs(0-rect2TopRight[0]); System.out.println(width); height = Math.abs(0-rect1BottomLeft[1]) - Math.abs(0-rect1TopRight[1]); System.out.println(height); store[count]=(Math.abs(width) * Math.abs(height)); count++; } } } //return (Math.abs(width) * Math.abs(height)); } else { System.out.println("MUST2"); width = Math.abs(0-rect2BottomLeft[0]) - Math.abs(0-rect2TopRight[0]); System.out.println(width); height = Math.abs(0-rect2BottomLeft[1]) - Math.abs(0-rect2TopRight[1]); System.out.println(height); store[count]=(Math.abs(width) * Math.abs(height)); count++; } } //this is taking into consideration of the Y co-ordinates and making a better decision in test case 24 and 25 //we are in this section because Y coordinate of rect1BottomLeft is not greater than or equal to rect2BottomLeft //This is trying to resolve Test Case 14new and 15new //blue is rectangle1 //if (rect1BottomLeft[1] >= rect2BottomLeft[1]) else { //this now tackles test cases 48 and 49 where both rectangles span across the Y axis if ((rect1BottomLeft[0]<0) && (rect1TopRight[0]>0) && (rect2BottomLeft[0]<0) && (rect2TopRight[0]>0)) { //in cases such as test case 14new and test case 15 new we still have where the //x coordinates are same for rect2TopRight for both rectangles... //this has to be handled separately in comparison to having two rectangles spanning across the Y axis //(test case 48 and 49) where the rectTopRight are not same for both rectangles //overlappingArea(new int[]{-5, 1}, new int[]{1, 6},new int[]{-2, 3}, new int[]{1, 6})); //blue is rectangle1 //if they share same x coordinate if (rect1TopRight[0]==rect2TopRight[0]) { //we also need to add width as oppose to subtract width = Math.abs(0-rect2BottomLeft[0]) + Math.abs(0-rect1TopRight[0]); System.out.println(width); height = Math.abs(0-rect2BottomLeft[1]) - Math.abs(0-rect1TopRight[1]); System.out.println(height); System.out.println("MUST31"); } else { //we are here due to test cases 54 and 55 //rect1 = red //we know that top right do not share same x coordinates for the rectangles //we know both rectangles span across Y axis //We know that Y coordinate of blue bottom left is greater than red bottom left //we know that it is not a shape within a shape //But it appears there needs to be reference to top right of both rectangles being on the same //Y coordinate //So this naturally leads me towards issue that might arise if the botom left of the rectangles share //same X coordinate (this will be potentially test cases 56 and 57) if (rect1TopRight[1]==rect2TopRight[1]) { //again we need to add due to crossing the Y axis //in this area due to test case 66 and 67.. //this is already in an area of code that considers spanning Y axis both rectangles. //so what makes this scenario different to Test case 55 which passes through this area //we know they both have the same Y coordinate for the top right of the rectangle //rect1=blue if (rect2TopRight[0]>=rect1TopRight[0]) { //we are here because of test case 70 and 71 // if (rect2BottomLeft[0]<=rect1BottomLeft[0]) { width = Math.abs(0-rect1BottomLeft[0]) + Math.abs(0-rect1TopRight[0]); height = Math.abs(0-rect2BottomLeft[1]) - Math.abs(0-rect2TopRight[1]); } else { width = Math.abs(0-rect2BottomLeft[0]) + Math.abs(0-rect1TopRight[0]); height = Math.abs(0-rect2BottomLeft[1]) - Math.abs(0-rect2TopRight[1]); System.out.println("MUST42"); } } /* else { //again we need to add due to crossing the Y axis width = Math.abs(0-rect1BottomLeft[0]) + Math.abs(0-rect2TopRight[0]); System.out.println(width); height = Math.abs(0-rect2BottomLeft[1]) - Math.abs(0-rect2TopRight[1]); System.out.println(height); System.out.println("MUST35"); } */ } else { System.out.println("IN THIS SECTION2"); //we are here due to test cases 72 and 73 //we can see it is else statement since the top right X coordinates of rectangles are different.. //but we know that all other principles hold the same as above.... //so it is a case of copying the above code here... //but issue is here that once we do that, we need to copy the equivalent else statement //from the rectangle rotation as part of this code also...... if (rect2TopRight[0]>=rect1TopRight[0]) { //we are here because of test case 70 and 71 // if (rect2BottomLeft[0]<=rect1BottomLeft[0]) { width = Math.abs(0-rect1BottomLeft[0]) + Math.abs(0-rect1TopRight[0]); height = Math.abs(0-rect2BottomLeft[1]) - Math.abs(0-rect2TopRight[1]); } else { width = Math.abs(0-rect2BottomLeft[0]) + Math.abs(0-rect1TopRight[0]); height = Math.abs(0-rect2BottomLeft[1]) - Math.abs(0-rect2TopRight[1]); System.out.println("MUST42"); } } if (rect1TopRight[0]>=rect2TopRight[0]) { //we are here because of test case 70 and 71 // if (rect1BottomLeft[0]<=rect2BottomLeft[0]) { width = Math.abs(0-rect2BottomLeft[0]) + Math.abs(0-rect2TopRight[0]); height = Math.abs(0-rect1BottomLeft[1]) - Math.abs(0-rect1TopRight[1]); } else { width = Math.abs(0-rect1BottomLeft[0]) + Math.abs(0-rect2TopRight[0]); height = Math.abs(0-rect1BottomLeft[1]) - Math.abs(0-rect1TopRight[1]); System.out.println("MUST41"); } } //I am also unsure about the purpose of this code //but I will leave it commented out just incase /* width = Math.abs(0-rect2BottomLeft[0]) - Math.abs(0-rect1TopRight[0]); System.out.println(width); height = Math.abs(0-rect1BottomLeft[1]) - Math.abs(0-rect2TopRight[1]); System.out.println(height); System.out.println("MUST20"); */ } } } else { //when examining test case 14 and test case 15, we are getting incorect width of the larger rectangle. //instead we need the width and height of the nested rectangle //I can only speculate that I need to reference if the rectXBottom (X,Y) coincides with both rectangles //or the rectXTopRight (X,Y) coincides with both rectangles.. //We would then to calculate the overlap area to be the nested rectangle… //I also believe we are in area of code where its confirmed if it nested, so just interested if either top right or bottom left pair up //But right now, it just does not seem so obvious if it will make right decision on with rectangle to process. //But if I do same for both scenarios, this should suffice //In practice this seems very inconsistent with associated if... //if (rect1TopRight[0]==rect2TopRight[0] && rect1TopRight[1]==rect2TopRight[1]) //We now check if the x coordinate (bottom left) of rectangle1 is bigger or smaller than rectangle 2 (if rect1BottomLeft[0]>rect2BottomLeft[0]) //If bigger width = Math.abs(rect1BottomLeft[0]) - Math.abs(rect2TopRight[0]) height = Math.abs(rect1BottomLeft[1]) - Math.abs(rect2TopRight[1]) //else width = Math.abs(rect2BottomLeft[0]) - Math.abs(rect1TopRight[0]) height = Math.abs(rect2BottomLeft[1]) - Math.abs(rect1TopRight[1]) if (rect1TopRight[0]==rect2TopRight[0] && rect1TopRight[1]==rect2TopRight[1]) { System.out.println("MUST26"); //We now check if the x coordinate (bottom left) of rectangle1 is bigger or smaller than rectangle 2 if (rect1BottomLeft[0]>rect2BottomLeft[0]) { System.out.println("MUST27"); width = Math.abs(rect1BottomLeft[0]) - Math.abs(rect1TopRight[0]); height = Math.abs(rect1BottomLeft[1]) - Math.abs(rect1TopRight[1]); } else { System.out.println("MUST28"); width=Math.abs(rect2BottomLeft[0]) - Math.abs(rect2TopRight[0]); height = Math.abs(rect2BottomLeft[1]) - Math.abs(rect2TopRight[1]); } } /* //........................................ if (rect1TopRight[0]==rect2TopRight[0] && rect1TopRight[1]==rect2TopRight[1]) { //We now check if the x coordinate (bottom left) of rectangle1 is bigger or smaller than rectangle 2 if (rect1BottomLeft[0]>rect2BottomLeft[0]) { width = Math.abs(rect1BottomLeft[0]) - Math.abs(rect1TopRight[0]) height = Math.abs(rect1BottomLeft[1]) - Math.abs(rect1TopRight[1]) } else { width=Math.abs(rect2BottomLeft[0]) - Math.abs(rect2TopRight[0]); height = Math.abs(rect2BottomLeft[1]) - Math.abs(rect2TopRight[1]); } } //.................................................................. */ //width = Math.abs(0-rect2BottomLeft[0]) - Math.abs(0-rect1TopRight[0]); //rect1=red //we know this is not the case //if (rect1TopRight[0]==rect2TopRight[0] && rect1TopRight[1]==rect2TopRight[1]) else { //rect2=blue if (rect1BottomLeft[0]= rect1BottomLeft[1]) { if (rect2BottomLeft[0]>=rect1BottomLeft[0]) { //we are here because of issues in test case 22new2 and 23new2 //at this point we know that rectangle 1 is blue, rectangle 2 is red //overlappingArea(new int[]{4, 2}, new int[]{8, 4},new int[]{4, 0}, new int[]{5, 6})); //we know the blue has a higher or equal Y coordinate than red triangle //we also know that blue has a higher or equal X coordinate than red triangle //but really we are concerned with a different calculation if the x coordinates of the two rectangles are the same if (rect2BottomLeft[0]==rect1BottomLeft[0]) { if (rect2TopRight[1]<=rect1TopRight[1]) { width = Math.abs(0-rect2BottomLeft[0]) - Math.abs(0-rect2TopRight[0]); System.out.println(width); } else { width = Math.abs(0-rect1BottomLeft[0]) - Math.abs(0-rect1TopRight[0]); System.out.println(width); } //CHANGE HERE //if (rect2TopRight[1]0) && (rect2BottomLeft[0]<0) && (rect2TopRight[0]>0)) { //note addition of the width //rect1 = blue width = Math.abs(0-rect1TopRight[0]) + Math.abs(0-rect2BottomLeft[0]); System.out.println(width); //note since the shape has not crossed height wise across the X axis, //we are not changing this.... //but it will potentially become an issue when I further devise test cases //in which overlap is on both side of the X axis height = Math.abs(0-rect2BottomLeft[1]) - Math.abs(0-rect2TopRight[1]); System.out.println(height); System.out.println("MUST33"); store[count]=(Math.abs(width) * Math.abs(height)); count++; } else { //in here due to test case 82 and 83 //the associated if is due to there being rectangles across the Y axis //this is clearly not the case, yet the logic seems to be lacking massively.... //it needs spliting into if else //we know following at this point: both rectangles have different X coordinate for bottom left //rect2bottomleft X coordinate is greater than or equal rect1bottomleft (red) //rect1topright (red) Y coordinate is greater equal to rect2TopRight //there is every possibility that is requires the exact same code as //red = rect2 if (rect1TopRight[0]>rect2TopRight[0]) { System.out.println("MUST60"); width = Math.abs(0-rect2BottomLeft[0]) - Math.abs(0-rect2TopRight[0]); System.out.println(width); height = Math.abs(0-rect2BottomLeft[1]) - Math.abs(0-rect2TopRight[1]); System.out.println(height); store[count]=(Math.abs(width) * Math.abs(height)); count++; } //this would be shapes not spanning Y axis //it has used the property of one shape.. //this is potentially my first bits of code since it assumes that rect1 and rect2 shares the same X coordinate... //it would be false under any other circumstance.. But I am leaving it intact... else { if (rect1TopRight[0]==rect2TopRight[0]) { System.out.println("MUST3"); width = Math.abs(0-rect2BottomLeft[0]) - Math.abs(0-rect2TopRight[0]); System.out.println(width); height = Math.abs(0-rect2BottomLeft[1]) - Math.abs(0-rect2TopRight[1]); System.out.println(height); //return (Math.abs(width) * Math.abs(height)); store[count]=(Math.abs(width) * Math.abs(height)); count++; } else { System.out.println("MUST61"); width = Math.abs(0-rect2BottomLeft[0]) - Math.abs(0-rect1TopRight[0]); System.out.println(width); height = Math.abs(0-rect2BottomLeft[1]) - Math.abs(0-rect2TopRight[1]); System.out.println(height); store[count]=(Math.abs(width) * Math.abs(height)); count++; } } } } } else { System.out.println("MUST4"); width = Math.abs(0-rect1BottomLeft[0]) - Math.abs(0-rect1TopRight[0]); System.out.println(width); height = Math.abs(0-rect1BottomLeft[1]) - Math.abs(0-rect1TopRight[1]); System.out.println(height); store[count]=(Math.abs(width) * Math.abs(height)); } } //this is taking into consideration of the Y co-ordinates and making a better decision in test case 24 and 25 //we are here because following is false... //if (rect2BottomLeft[1] >= rect1BottomLeft[1]) else { //this now tackles test cases 48 and 49 where both rectangles span across the Y axis if ((rect2BottomLeft[0]<0) && (rect2TopRight[0]>0) && (rect1BottomLeft[0]<0) && (rect1TopRight[0]>0)) { //in cases such as test case 14new and test case 15 new we still have where the //x coordinates are same for rect2TopRight for both rectangles... //this has to be handled separately in comparison to having two rectangles spanning across the Y axis //(test case 48 and 49) where the rectTopRight are not same for both rectangles //overlappingArea(new int[]{-2, 3}, new int[]{1, 6},new int[]{-5, 1}, new int[]{1, 6})); //red is rectangle1 //if they share same x coordinate if (rect2TopRight[0]==rect1TopRight[0]) { //we also need to add width as oppose to subtract width = Math.abs(0-rect1BottomLeft[0]) + Math.abs(0-rect2TopRight[0]); System.out.println(width); height = Math.abs(0-rect1BottomLeft[1]) - Math.abs(0-rect2TopRight[1]); System.out.println(height); System.out.println("MUST32"); } else { //we are here due to test cases 54 and 55 //rect2 = blue //we know that top right do not share same x coordinates for the rectangles //we know both rectangles span across Y axis //We know that Y coordinate of blue bottom left is greater than red bottom left //we know that it is not a shape within a shape //But it appears there needs to be reference to top right of both rectangles being on the same //Y coordinate //So this naturally leads me towards issue that might arise if the botom left of the rectangles share //same X coordinate (this will be potentially test cases 56 and 57) if (rect2TopRight[1]==rect1TopRight[1]) { //again we need to add due to crossing the Y axis //in this area due to test case 66 and 67.. //this is already in an area of code that considers spanning Y axis both rectangles. //so what makes this scenario different to Test case 55 which passes through this area //we know they both have the same Y coordinate for the top right of the rectangle //rect1=blue if (rect1TopRight[0]>=rect2TopRight[0]) { //we are here because of test case 70 and 71 // if (rect1BottomLeft[0]<=rect2BottomLeft[0]) { width = Math.abs(0-rect2BottomLeft[0]) + Math.abs(0-rect2TopRight[0]); height = Math.abs(0-rect1BottomLeft[1]) - Math.abs(0-rect1TopRight[1]); } else { width = Math.abs(0-rect1BottomLeft[0]) + Math.abs(0-rect2TopRight[0]); height = Math.abs(0-rect1BottomLeft[1]) - Math.abs(0-rect1TopRight[1]); System.out.println("MUST41"); } } } //DOES THIS NEED TO BE IMPROVED else { System.out.println("IN THIS SECTION1"); //we are here due to test cases 72 and 73 //we can see it is else statement since the top right X coordinates of rectangles are different.. //but we know that all other principles hold the same as above.... //so it is a case of copying the above code here... //but issue is here that once we do that, we need to copy the equivalent else statement //from the rectangle rotation as part of this code also...... //I am also unsure about the purpose of this code //but I will leave it commented out just incase /* width = Math.abs(0-rect1BottomLeft[0]) - Math.abs(0-rect2TopRight[0]); System.out.println(width); height = Math.abs(0-rect2BottomLeft[1]) - Math.abs(0-rect1TopRight[1]); System.out.println(height); System.out.println("MUST21"); */ if (rect1TopRight[0]>=rect2TopRight[0]) { //we are here because of test case 70 and 71 // if (rect1BottomLeft[0]<=rect2BottomLeft[0]) { width = Math.abs(0-rect2BottomLeft[0]) + Math.abs(0-rect2TopRight[0]); height = Math.abs(0-rect1BottomLeft[1]) - Math.abs(0-rect1TopRight[1]); } else { width = Math.abs(0-rect1BottomLeft[0]) + Math.abs(0-rect2TopRight[0]); height = Math.abs(0-rect1BottomLeft[1]) - Math.abs(0-rect1TopRight[1]); System.out.println("MUST41"); } } if (rect2TopRight[0]>=rect1TopRight[0]) { //we are here because of test case 70 and 71 // if (rect2BottomLeft[0]<=rect1BottomLeft[0]) { width = Math.abs(0-rect1BottomLeft[0]) + Math.abs(0-rect1TopRight[0]); height = Math.abs(0-rect2BottomLeft[1]) - Math.abs(0-rect2TopRight[1]); } else { width = Math.abs(0-rect2BottomLeft[0]) + Math.abs(0-rect1TopRight[0]); height = Math.abs(0-rect2BottomLeft[1]) - Math.abs(0-rect2TopRight[1]); System.out.println("MUST42"); } } } /* else { width = Math.abs(0-rect1BottomLeft[0]) - Math.abs(0-rect2TopRight[0]); System.out.println(width); height = Math.abs(0-rect2BottomLeft[1]) - Math.abs(0-rect1TopRight[1]); System.out.println(height); System.out.println("MUST21"); } */ } } else { //when examining test case 14 and test case 15, we are getting incorect width of the larger rectangle. //instead we need the width and height of the nested rectangle //I can only speculate that I need to reference if the rectXBottom (X,Y) coincides with both rectangles //or the rectXTopRight (X,Y) coincides with both rectangles.. //We would then to calculate the overlap area to be the nested rectangle… //I also believe we are in area of code where its confirmed if it nested, so just interested if either top right or bottom left pair up //But right now, it just does not seem so obvious if it will make right decision on with rectangle to process. //But if I do same for both scenarios, this should suffice //In practice this seems very inconsistent with associated if... //if (rect1TopRight[0]==rect2TopRight[0] && rect1TopRight[1]==rect2TopRight[1]) //We now check if the x coordinate (bottom left) of rectangle1 is bigger or smaller than rectangle 2 (if rect1BottomLeft[0]>rect2BottomLeft[0]) //If bigger width = Math.abs(rect1BottomLeft[0]) - Math.abs(rect2TopRight[0]) height = Math.abs(rect1BottomLeft[1]) - Math.abs(rect2TopRight[1]) //else width = Math.abs(rect2BottomLeft[0]) - Math.abs(rect1TopRight[0]) height = Math.abs(rect2BottomLeft[1]) - Math.abs(rect1TopRight[1]) if (rect2TopRight[0]==rect1TopRight[0] && rect2TopRight[1]==rect1TopRight[1]) { //We now check if the x coordinate (bottom left) of rectangle1 is bigger or smaller than rectangle 2 System.out.println("MUST23"); if (rect2BottomLeft[0]>rect1BottomLeft[0]) { //System.out.println("The overlapping area is: " + overlappingArea(new int[]{3, -1}, new int[]{9, 6},new int[]{6, -3}, new int[]{9, 6})); System.out.println("MUST24"); width = Math.abs(rect2BottomLeft[0]) - Math.abs(rect2TopRight[0]); height = Math.abs(rect2BottomLeft[1]) - Math.abs(rect2TopRight[1]); } else { System.out.println("MUST25"); width=Math.abs(rect1BottomLeft[0]) - Math.abs(rect1TopRight[0]); height = Math.abs(rect1BottomLeft[1]) - Math.abs(rect1TopRight[1]); } } else { if (rect2BottomLeft[0]>rect1BottomLeft[0]) { width = Math.abs(0-rect1BottomLeft[0]) - Math.abs(0-rect1TopRight[0]); System.out.println(width); height = Math.abs(0-rect1BottomLeft[1]) - Math.abs(0-rect1TopRight[1]); System.out.println(height); System.out.println("MUST8a"); } else { width = Math.abs(0-rect1BottomLeft[0]) - Math.abs(0-rect2TopRight[0]); System.out.println(width); height = Math.abs(0-rect1BottomLeft[1]) - Math.abs(0-rect2TopRight[1]); System.out.println(height); System.out.println("MUST8"); store[count]=(Math.abs(width) * Math.abs(height)); count++; return GetHighestArea(store); } } } return (Math.abs(width) * Math.abs(height)); } } } //} //Not all in the positive axis //else //{ // if ((rect2BottomLeft[1]<0 && rect2TopRight[1]>0)) // { // width = Math.abs(0-rect2BottomLeft[0]) + Math.abs(0-rect2TopRight[0]); //height = Math.abs(0-rect2BottomLeft[1]) + Math.abs(0-rect2TopRight[1]); // } //} //return (Math.abs(width) * Math.abs(height)); return GetHighestArea(store); } public static int GetHighestArea(int []store) { System.out.println("**********************************************"); int min = store[0]; for (int i: store) { System.out.println("AREA: " + i); if (i