{ 1: (“Paul”, 20, “Male”){ 2: (“Bob”, 22, “Male”) { 3: (“Steve”, 24, “Male”) { 4: (“Stan”, 23, “Male”) { 5: (“Matt”, 25, “Male”) { 6: (“Frank”, 25, “Male”) { 7: (“Amy”, 22, “Female”) { 8: (“Jim”, 23, “Male”) { 9: (“Dan”, 21, “Male”) { 10: (“Phil”, 25, “Male”) { 11: (“Katy”, 27, “Female”) { 12: (“Jacob”, 26, “Male”) { 13: (“Josh”, 21, “Male”) { 14: (“Ben”, 34, “Male”) { 15: (“Shawn”, 23, “Male”) day1=[1,3,4,7,15];
day2=[8,7,4,14,11];
day3=[5,7,3,14,11];
The police has rounded up a group of individuals that could robbed stores. Everyone cell phone number gives off a key number(Ex. Dan key number is 9). Help the police catch the culprit who must appear in the lists, day1, day2 and day3. Using Phyton 3
![search(arr, #searching x from the key for i in range(len (arr)): def x): 4 if arr[i]x: return i return -1 9 def main_function (day1 , day2 day3): 10 searching in the given lists for same value for i in range(en (day1)): if search (day2, day1[i])1: 12 13 if search (day3 , day1li])1 return i 15 return -1 17 Dictionary 1 (Paul, 20, Male), 2 (Bob, 22, Male) 3 (Steve, 24, Male), 4: (Stan, 23, Male), 5 (Matt, 25, Male), 6: (Frank, 25, Male) , 7(Amy, 22, Female), 8(Jim, 23, Male), 9Dan, 21, Male), 10 (Phil, 25, Male), 11 (Katy, 27, Female), 12 (Jacob, 26, Male), 13 (Josh, 21, Male), 14 (Ben, 34, Male), 15 (Shawn, 23, Male) 19 20 21 23 24 25 26 28 29 31 32 34 day1-[1,3,4,7,15] 35 day2-[8,7,4,14,11] 36 day3-[5,7,3,14,11] 38 if main_function(day1 ,day2 , day3)-1: print (InThere is no one culprit here who is present all three daysn) 40 else: 41 42 index- main_function(day1, day2 , day3) print (nCulprit ist : t%d , Dictionary[day1 [index]] )](https://media.cheggcdn.com/media%2Fa17%2Fa1767eac-44e6-47f4-bfdc-8b11b735f91b%2FphpWzCvdR.png)
