CareerCup

S1>e2 || s2>e1 -->success

s1>s2 ? (s1>e2?1:0) : (s2>e1?1:0)

where 1 is Success

if( S2 <S1 < E2 || S1< S2 < E1)
return success

((s2 > s1) && (s2 < e1)) || ((e2 < e1) && (e2 > s1))

!((S2 > E1) || E2 < S1))

Consider the case that two does not overlap and then find NOT

the case two does not overlap is

suppose s1 <e1 && s2<e2

1. s1<s2 && e1<s2
2. s2<s1 && e2<s1

so we have

return s1<s2?(e1<s2?1:0):(e2<s1?1:0);

And then NOT in front of the result I just typed above

case1.
s1<---------s2-->e1---->e2
case2.
s1<---------s2------>e2-->e1
case3.
s2<---------s1------>e2-->e1
case4.
s2<---------s1------>e1-->e2

case5.
s1<---->e1 s2<--->e2
case6.
s2<---->e2 s1<--->e1

J's approach is pretty good but the answer is not neat.

!(e1<s2 || e2 < s1)

hi, usafzz, thanks to give all the situations. just want to point out that total number of combination is p(4,4)/p(2,2)^2=6.

brijesh kumar jaiswal's answer is perfect.

DrFunFrock's way it the right way to do it and is actually correct.. its simple and eligant!

!((e2 < s1) || (e1 < s2))

s1 > e2 && s2 > e1