How can I find an overlap between two given ranges? [closed]

Question

Is there an effective way to find the overlap between two ranges?

Practically, the two ranges marked as (a - c), and (b - d), and I assume (c > a) && (d > b).

(b <= a <= d) which means if ((a >= b) && (d > a)) (b <= c <= d) which means if ((c >= b) && (d >= c)) (a <= b <= c) which means if ((b > a) && (c > b)) (a <= d <= c) which means if ((d > a) && (c > d)) 

But it never ends, because in this way I can find only one range at the time, and in each if I have to check the other cases as well.

For exemple, if the first condition (1) correct, I know what happening with the start of the range (a) I still need to check the others for the end of the range (c).

Not to mention that all this works in the case that (c > a) && (d > b), and not one of them is equal to another.

Answer 1

Two ranges overlap in one of two basic cases:

• one range contains the other completely (i.e. both the start and end of one range are between the start and end of the other), or
• the start or end of one range is contained within the other range

Conversely, they do not overlap only if neither endpoint of each range is contained within the other range (cases 11 and 12 in your diagram). We can check whether the low end of either range is past the high end of the other, to detect both those cases:

if (a > d || c < b) {     // no overlap } else {     // overlap } 

We can invert the condition and then use DeMorgan's laws to swap the order, if that's preferable:

if (a <= d && c >= b) {     // overlap } else {     // no overlap } 

To find the actual overlap range, you take the maximum of the two low ends, and the minimum of the two high ends:

int e = Math.max(a,b); int f = Math.min(c,d); // overlapping range is [e,f], and overlap exists if e <= f. 

All above assumes that the ranges are inclusive, that is, the range defined by a and c includes both the value of a and the value of c. It is fairly trivial to adjust for exclusive ranges, however.