Chapter 5, Problem 16RE

Explanation of Solution

Test cases for the algorithm given in “R5.12”:

Test case #1:

Consider the values:

“start1” = 5

“end1” = 7

“start2” = 6

“end2” = 8

In the above case:

• The first “if” condition fails because, “start1” is not greater than “start2” (5 is not greater than 6) and therefore, it goes to the “else” part and assigns the value “start2” to “s”.
  • Now “s” = 6.
• The second “if” condition becomes true because, “end1” is less than “end2” (7 is less than 8). The value of “end1” will be assigned to “e”.
  • Now “e” = 7.
• The third “if” condition becomes true because, “s” is less than “e” (6 is less than 7). The string “The appointments overlap” will be printed...