WEBVTT 00:00.180 --> 00:06.090 Hello again! In this video, we are going to look at multiset and multimap. We have already looked 00:06.090 --> 00:08.820 at set and map. In those containers, 00:08.820 --> 00:14.100 the elements have to have distinct keys, so we cannot have different elements which have the same key. 00:14.940 --> 00:16.710 With multiset and multimap. 00:16.710 --> 00:21.180 This requirement is relaxed, so we can have different elements which have the same key. 00:22.350 --> 00:27.900 And apart from that, they all work more or less the same way as set and map. The only real difference is 00:27.900 --> 00:30.420 that multimap does not support subscripting. 00:32.160 --> 00:35.520 So let's try this with a multiset. We still include the 00:35.560 --> 00:35.920 header. 00:37.260 --> 00:44.190 Then we create a multiset object, and we insert some elements. And then we try to insert an element 00:44.190 --> 00:47.910 with the key 6, even though we already have an element with the key 6. 00:49.020 --> 00:50.070 So let's see what happens. 00:51.540 --> 00:57.540 So we call insert three times, with the argument 6, and we end up with three elements with the key 00:57.540 --> 00:57.930 6. 00:58.360 --> 01:00.270 So insert will always succeed. 01:00.720 --> 01:05.820 If you call insert, it will always create a new element, regardless of whether there are any elements 01:05.820 --> 01:06.930 which already have that key. 01:08.270 --> 01:11.990 And the other thing to note is that these are all in order, as you would expect. 01:13.940 --> 01:16.520 And the same with a multimap. We include . 01:17.150 --> 01:22.880 We create an object and add some elements, and then we add some more map elements which have the same 01:22.880 --> 01:23.240 key. 01:23.630 --> 01:24.290 So, "Graham". 01:27.150 --> 01:32.490 And we end up with three elements with the key "Graham". And again, they are all sorted in order of 01:32.490 --> 01:33.150 their keys. 01:35.180 --> 01:41.120 So let's find out what happens if we call erase(). (And the comment rather gives it away!) 01:44.270 --> 01:49.640 So erase() with argument "Graham" will remove all the elements which have key "Graham." And that 01:49.640 --> 01:50.660 may be what we want. 01:50.990 --> 01:54.440 Or perhaps, you may just want to remove one of the elements with key "Graham". 01:56.500 --> 02:02.860 So, a quick recap. insert() will always succeed for multimap and multiset. If we call erase(), 02:02.860 --> 02:05.830 then that will remove all the elements which have that key. 02:06.580 --> 02:13.050 There is a form of erase() which takes a single iterator, so we can use that to remove a single element. 02:13.660 --> 02:17.020 We just have to find an iterator to the element we want to remove. 02:17.710 --> 02:18.910 And how do we do that? 02:21.330 --> 02:28.140 With a map, it is very easy, because only one element can have a given key. With a multimap, there could 02:28.140 --> 02:33.990 be several elements which have the same key. And we need some way to manage all these potential matches. 02:36.710 --> 02:41.060 One thing that does help us, is that the elements in a multimap are sorted. 02:41.300 --> 02:42.740 They are in the order of the keys. 02:43.070 --> 02:46.940 And that means that all the elements which have the same key will be next to each other. 02:47.390 --> 02:51.980 So this means we have a kind of sub-container within the map, which will contain all the elements with 02:51.980 --> 02:52.640 the same key. 02:53.420 --> 02:56.380 And we can use an iterator range to represent those elements. 02:58.170 --> 03:03.630 So there are several ways to do this. One way, which we already know about, is to use find() and count(). 03:06.820 --> 03:12.370 If we call find() for a multimap, this will return an iterator to the first element which has 03:12.370 --> 03:12.940 that key. 03:14.580 --> 03:16.710 And that will be the first element in our sub-range. 03:17.400 --> 03:19.980 If there is no element which matches, then this will return 03:19.980 --> 03:23.220 the last-plus-one iterator, and our range will be empty. 03:24.340 --> 03:29.270 The count() member function will return the number of elements which have that key. And that gives 03:29.270 --> 03:31.030 us the number of elements in our range. 03:31.570 --> 03:37.210 So, now we know enough to loop over all the elements. We can go to the first element and iterate over 03:37.210 --> 03:37.930 every element. 03:40.060 --> 03:45.550 So let's try this with the multimap we had before. We call find() with "Graham" as the key. 03:46.600 --> 03:50.380 Then this will return an iterator to the first element with key "Graham". 03:52.880 --> 04:00.130 And then we can get the number of elements which have key "Graham", and then we can leap over these elements. 04:00.140 --> 04:02.690 So we just keep incrementing the iterator each time. 04:05.360 --> 04:06.080 And there we are. 04:06.290 --> 04:08.540 We get all the elements with key "Graham". 04:11.300 --> 04:17.240 And if we want to find an element with a particular value, then we can just check that inside the 04:17.240 --> 04:19.040 loop, so like that. So, 04:19.280 --> 04:22.040 is there an element with key "Graham" and value 66? 04:22.490 --> 04:23.090 Yes, there is. 04:24.260 --> 04:25.670 Okay, so that is it for this video. 04:26.090 --> 04:29.060 I will see you next time, but until then, keep coding!