1
00:00:01,690 --> 00:00:02,320
Hi, everyone.

2
00:00:02,350 --> 00:00:08,330
So in this lesson, we are going to solve this question, remove duplicates, inanity, so this example.

3
00:00:08,350 --> 00:00:12,060
So this is basically and ready and what we have to do.

4
00:00:12,310 --> 00:00:17,600
So basically we have to return a veteran of integers and in this kind of integers.

5
00:00:18,000 --> 00:00:19,680
So this is the vector.

6
00:00:19,840 --> 00:00:23,170
So in this integers element will appear only once.

7
00:00:23,740 --> 00:00:26,590
They will be all unique elements inside this vector.

8
00:00:26,830 --> 00:00:27,760
So let me show you.

9
00:00:28,060 --> 00:00:30,910
I would put one it here then to.

10
00:00:32,140 --> 00:00:32,800
Then three.

11
00:00:33,890 --> 00:00:39,810
Then, too, is coming again, so, too, is already present inside the Taketo then when so when it

12
00:00:39,830 --> 00:00:41,210
already present, take five.

13
00:00:42,360 --> 00:00:43,350
You're going to take for.

14
00:00:44,520 --> 00:00:49,560
You cannot take one because one is already present and you cannot take three because he's already poisoned,

15
00:00:49,740 --> 00:00:51,120
so this will be my output.

16
00:00:51,180 --> 00:00:54,410
OK, so one, two, three, five and four.

17
00:00:54,690 --> 00:00:55,890
So if you will notice.

18
00:00:55,890 --> 00:00:59,370
So this vector is basically containing all unique elements.

19
00:01:00,610 --> 00:01:05,470
So that is the meaning of removing duplicates, so in this area, duplicates are present, but it will

20
00:01:05,470 --> 00:01:06,160
see the vector.

21
00:01:06,520 --> 00:01:08,170
All elements are unique elements.

22
00:01:08,320 --> 00:01:12,970
And one more thing to notice, elements are appearing in the same order by same order.

23
00:01:13,000 --> 00:01:17,130
I mean, so one is coming before two in the original also.

24
00:01:17,140 --> 00:01:18,540
So one was before two.

25
00:01:18,730 --> 00:01:24,570
Similarly, five is coming before four because in the original Eddie five was coming before four.

26
00:01:24,790 --> 00:01:26,800
So that is the meaning of same order.

27
00:01:27,340 --> 00:01:29,510
So what we have to do so given this integer Eddie.

28
00:01:30,500 --> 00:01:36,170
We have to find this vector, we have to generate this vector that will contain all the unique elements

29
00:01:36,170 --> 00:01:37,610
and in the same order.

30
00:01:38,900 --> 00:01:40,860
Now let's see how we can solve this question.

31
00:01:41,210 --> 00:01:43,380
So when you're solving this question is very simple.

32
00:01:43,400 --> 00:01:44,240
So what do you do?

33
00:01:44,480 --> 00:01:48,540
You pick one element and then you will search whether this element or not.

34
00:01:48,620 --> 00:01:50,330
If it is not present, then you will push.

35
00:01:50,600 --> 00:01:54,410
Similarly, the second element check it is present inside the vector.

36
00:01:54,410 --> 00:01:58,820
Not if it is not present, then you can push a device, not push similarly.

37
00:01:59,180 --> 00:02:00,200
But the third element.

38
00:02:00,200 --> 00:02:03,530
And then I read all the vector and check whether it is present or not present.

39
00:02:03,770 --> 00:02:09,590
So in this we will be able to solve this question now with the time complexity, if we will use this

40
00:02:09,590 --> 00:02:09,970
approach.

41
00:02:10,580 --> 00:02:11,060
So.

42
00:02:13,030 --> 00:02:16,690
So this is my indigeneity, this input, so this is an area.

43
00:02:17,860 --> 00:02:23,560
And if you will follow the above approach, so whatever time complexity, so this is my vector, so

44
00:02:23,560 --> 00:02:25,480
initiate the vector is containing zero elements.

45
00:02:25,780 --> 00:02:30,170
So you pick the first element, you will check that is empty so you can push this element.

46
00:02:30,430 --> 00:02:34,200
So when work is done now, pick the second element.

47
00:02:34,330 --> 00:02:37,870
So you will like to order Vector to check whether this element is present or not.

48
00:02:38,140 --> 00:02:38,980
So basically.

49
00:02:40,260 --> 00:02:45,270
The amount of Ogborn, similarly, you will pick the third element, so you will agree to whatever it

50
00:02:45,270 --> 00:02:46,090
is, pleasant or not.

51
00:02:46,350 --> 00:02:48,420
Similarly for the net element, so.

52
00:02:49,600 --> 00:02:56,170
For the entertainment, what do you have to read over the computer to check whether the element is present

53
00:02:56,170 --> 00:03:01,210
or not so invoke, you have to perform and work because you have to write it or the computer actor to

54
00:03:01,210 --> 00:03:03,100
check whether this element is present or not.

55
00:03:03,520 --> 00:03:05,050
So what does the time complexity.

56
00:03:05,070 --> 00:03:06,460
So if you will add all this.

57
00:03:08,060 --> 00:03:13,930
So and plus and minus one, plus and minus two, it will go until one and we know.

58
00:03:13,980 --> 00:03:17,470
So this is basically off and square and big off and square.

59
00:03:17,690 --> 00:03:19,280
So time complexities.

60
00:03:19,610 --> 00:03:20,780
Big off and square.

61
00:03:22,260 --> 00:03:24,750
Now, how can we optimize our time complexity?

62
00:03:25,470 --> 00:03:31,050
So basically what we are doing, we are picking one element and then we are searching over the computer

63
00:03:31,050 --> 00:03:33,840
vector to check whether the element is present or not.

64
00:03:34,050 --> 00:03:36,810
So if we can perform the searching in Question Time.

65
00:03:38,030 --> 00:03:43,820
If he can perform this routine constant, then our time complexity will become so picking every element

66
00:03:43,820 --> 00:03:47,540
and then checking whether it is present or not so and multiply one.

67
00:03:47,570 --> 00:03:53,600
So that means the time complexity will become linear if we can do the searching in the vector in constant

68
00:03:53,600 --> 00:03:53,830
time.

69
00:03:54,440 --> 00:03:55,870
So here, what is it?

70
00:03:55,870 --> 00:03:56,590
Time, complexity.

71
00:03:56,600 --> 00:04:02,040
So you are picking each and every element and then you have to search inside the vector so and multiply.

72
00:04:02,060 --> 00:04:06,550
And so this is for picking every element and this is for searching inside the vector.

73
00:04:06,920 --> 00:04:14,210
But I am saying if we can perform the search in Question Time so the time complexity will become linear

74
00:04:15,170 --> 00:04:21,410
because and and how we can check in Question Time, basically we can use we can take the help of MAP

75
00:04:21,410 --> 00:04:25,580
because in map searching is constraint time and certainly search.

76
00:04:25,580 --> 00:04:29,420
All three operations are constant in an order to map.

77
00:04:29,990 --> 00:04:36,530
If you will use the map, then insert, delete, search, all your patients will take longer time.

78
00:04:36,740 --> 00:04:38,360
So in this question, what do we lose?

79
00:04:38,810 --> 00:04:40,110
We will lose in order to map.

80
00:04:41,180 --> 00:04:43,850
Now many people can say, why does it matter?

81
00:04:43,880 --> 00:04:48,230
We can also solve this question with the help of Eddie, so how we can solve this question with the

82
00:04:48,230 --> 00:04:48,770
help of Eddie.

83
00:04:49,010 --> 00:04:50,570
So consider this example only.

84
00:04:51,970 --> 00:04:56,150
So what do you do, you will find out the maximum, so five is the maximum element.

85
00:04:56,380 --> 00:05:01,720
And now what you'll do after find out the maximum element, you will make an array of size five plus

86
00:05:01,720 --> 00:05:01,960
one.

87
00:05:02,560 --> 00:05:05,740
So five is the maximum element to make an area of size five plus one.

88
00:05:06,040 --> 00:05:08,170
So the size of that, it will be five, six.

89
00:05:08,770 --> 00:05:09,700
This is five.

90
00:05:09,700 --> 00:05:10,450
And this is zero.

91
00:05:11,050 --> 00:05:13,920
And basically this area will be a bowling alley roof area.

92
00:05:14,440 --> 00:05:15,390
Now, what do you do?

93
00:05:15,850 --> 00:05:17,350
You will check when it's present.

94
00:05:17,620 --> 00:05:21,990
So inside when initially this area will be initialized with false.

95
00:05:22,030 --> 00:05:23,590
So I will check when I go to one.

96
00:05:23,770 --> 00:05:24,520
It is false.

97
00:05:24,730 --> 00:05:27,440
So basically you can push it inside the vector and make it through.

98
00:05:28,000 --> 00:05:32,570
No, if the next time when will come, you will see here you will check or Wennerstrom.

99
00:05:32,590 --> 00:05:34,360
So do not push inside the vector.

100
00:05:34,600 --> 00:05:36,490
So definitely you can use this approach.

101
00:05:37,600 --> 00:05:41,680
Let's take one more example, for example, three, so I will check, you will go to three.

102
00:05:41,830 --> 00:05:46,310
So initially, this building that is initialized with force, you will check all three is false.

103
00:05:46,310 --> 00:05:47,260
So you will push three.

104
00:05:47,500 --> 00:05:50,790
But the next time three will appear and we will also make it through.

105
00:05:51,490 --> 00:05:54,610
We will also make it through after pushing inside the vector, we will make it through.

106
00:05:54,640 --> 00:05:59,080
So when the next time things are coming, you will go to the next three and you will check No.

107
00:05:59,180 --> 00:05:59,800
Three, Drew.

108
00:06:00,250 --> 00:06:02,920
So basically you will not push inside the vector again.

109
00:06:03,220 --> 00:06:05,070
So we can definitely use this approach.

110
00:06:05,080 --> 00:06:13,000
We can definitely take the help of Belinelli, but the problem is considered you had in error is basically

111
00:06:13,000 --> 00:06:13,810
of this size.

112
00:06:14,470 --> 00:06:21,370
It is containing elements of this type then then then to power three then then to power six then then

113
00:06:21,370 --> 00:06:22,730
to power seven and so on.

114
00:06:23,200 --> 00:06:24,130
So what you have to do.

115
00:06:25,220 --> 00:06:30,200
If you will follow this approach, what is the maximum limit, so 10 to the power seven is the maximum

116
00:06:30,200 --> 00:06:32,390
limit and obviously the elements are debating.

117
00:06:32,720 --> 00:06:36,200
So then the power three, then to power six, then again, tend to power seven.

118
00:06:36,560 --> 00:06:39,710
Obviously, there are duplicates now in order to you.

119
00:06:40,010 --> 00:06:43,260
So I'll just caution using this method, this is the maximum limit.

120
00:06:43,520 --> 00:06:48,660
So basically the size of the area, the billionaire will be able to sustain to power seven plus one.

121
00:06:48,980 --> 00:06:50,960
So this will be your alley now.

122
00:06:50,990 --> 00:06:52,460
This is very huge.

123
00:06:52,460 --> 00:06:55,720
Speace, very, very huge, huge space.

124
00:06:56,600 --> 00:06:57,840
OK, so this is wrong.

125
00:06:58,880 --> 00:06:59,750
Now, one more thing.

126
00:07:00,470 --> 00:07:02,120
The elements can be negative also.

127
00:07:03,820 --> 00:07:06,430
Integer can be negative, so I can have minus 10.

128
00:07:06,580 --> 00:07:12,770
So how can you store minus 10 here, minus 10 index does not exist, minus 10 index will not exist.

129
00:07:12,940 --> 00:07:16,270
So that means this approach has a lot of limitations.

130
00:07:16,960 --> 00:07:22,120
If that is basically off, if there is containing elements of these types and also if that is containing

131
00:07:22,120 --> 00:07:25,160
negative element, then you cannot use this approach.

132
00:07:25,870 --> 00:07:28,400
So basically we have to use map in this case.

133
00:07:29,080 --> 00:07:31,930
Now, if you are using map, you can definitely do something like this.

134
00:07:32,620 --> 00:07:36,070
My map of minus five is true.

135
00:07:36,340 --> 00:07:37,840
You can definitely do something like this.

136
00:07:37,860 --> 00:07:46,150
Similarly, you can do my map of to power seven is true, not my map of 10000.

137
00:07:46,150 --> 00:07:49,180
Doesn't mean that your map is often to power seven sites.

138
00:07:50,050 --> 00:07:56,830
We will see the implementation map will take very, very small space maps will take very, very small

139
00:07:56,830 --> 00:07:57,190
space.

140
00:07:57,500 --> 00:08:00,840
So it doesn't mean the size of the map will return to power.

141
00:08:00,850 --> 00:08:04,590
So no, it will be very, very, very, very large as compared to the areas.

142
00:08:06,160 --> 00:08:07,290
And one more thing.

143
00:08:07,330 --> 00:08:09,060
So this is in digit area.

144
00:08:09,880 --> 00:08:14,740
So we are talking about indigeneity now in the question if there was connectivity.

145
00:08:16,860 --> 00:08:23,760
If there was crack berry, for example, if there was characters like it, then C, C, then D, B,

146
00:08:24,030 --> 00:08:28,790
so if instead of a degenerate character it was given, then definitely what can do.

147
00:08:28,800 --> 00:08:31,080
You can make an array of size 256.

148
00:08:32,500 --> 00:08:34,510
And obviously, this area will be a billionaire.

149
00:08:35,630 --> 00:08:37,850
And you can use this approach, for example.

150
00:08:38,020 --> 00:08:41,850
So go to index 97, so initially this was false.

151
00:08:41,960 --> 00:08:44,330
What you will do, so you will see false.

152
00:08:44,450 --> 00:08:47,300
You will push inside the vector and then you will make it through.

153
00:08:47,720 --> 00:08:50,860
Similarly, when you will come again, you will go to index 97.

154
00:08:50,870 --> 00:08:51,640
So this is true.

155
00:08:51,650 --> 00:08:53,970
Then do not push it again, see?

156
00:08:54,320 --> 00:08:58,820
So go to index ninety seven ninety nine and it will check.

157
00:08:58,820 --> 00:08:59,450
It is false.

158
00:08:59,480 --> 00:09:02,370
So you will make it through, you will push AC and so on.

159
00:09:02,630 --> 00:09:06,710
So basically if the correct area was given you can definitely use this boolean area approach.

160
00:09:06,980 --> 00:09:11,670
But since we have in beta that an integer can be very big and indeed it can be negative also.

161
00:09:11,690 --> 00:09:17,480
So we have to use map in this question so that our time complexity can be linear.

162
00:09:19,000 --> 00:09:20,590
OK, so how can we use map?

163
00:09:20,600 --> 00:09:24,550
So it is very, very simple, so considered a very small example.

164
00:09:24,730 --> 00:09:26,290
So consider above an example.

165
00:09:27,190 --> 00:09:30,370
So what we will do so I have this goes like this in your area.

166
00:09:30,380 --> 00:09:35,770
So one, two, three, then two, then one, then let's say five for one and three.

167
00:09:36,430 --> 00:09:37,190
So what will do?

168
00:09:37,540 --> 00:09:39,130
I am going to construct a map.

169
00:09:39,130 --> 00:09:41,200
So let's say the name of map is my map.

170
00:09:41,860 --> 00:09:47,080
Give a key will be element and there will be basically true or false.

171
00:09:48,100 --> 00:09:50,650
So what we will do so pick the first element.

172
00:09:50,650 --> 00:09:51,940
Not so initially.

173
00:09:51,940 --> 00:09:52,840
My map is empty.

174
00:09:52,870 --> 00:09:53,680
So this is map.

175
00:09:55,960 --> 00:10:00,660
So initially, members is empty, no check, so one percent inside the map, no, it is not present.

176
00:10:00,670 --> 00:10:04,320
So what you will do, you will push one and one or two.

177
00:10:04,360 --> 00:10:08,830
So the value of one is true and it will push one inside the vector now, too.

178
00:10:09,310 --> 00:10:10,930
So to present inside the map.

179
00:10:10,930 --> 00:10:11,240
No.

180
00:10:11,410 --> 00:10:14,730
So push two and make it through now three.

181
00:10:14,740 --> 00:10:16,150
So three plus inside the map.

182
00:10:16,150 --> 00:10:16,440
No.

183
00:10:16,780 --> 00:10:18,790
Then push inside the map and make it through.

184
00:10:19,270 --> 00:10:23,710
So obviously when you will push inside the map we will also push inside the vector.

185
00:10:23,710 --> 00:10:25,510
So you will also push inside the vector.

186
00:10:26,260 --> 00:10:28,780
So this is the vector that we have to return.

187
00:10:28,810 --> 00:10:31,150
OK, now again, two is coming.

188
00:10:31,330 --> 00:10:33,280
So two is already present inside the map.

189
00:10:33,280 --> 00:10:34,990
So do not do anything one.

190
00:10:35,170 --> 00:10:37,060
So one is already inside the map.

191
00:10:37,060 --> 00:10:37,850
Do not do anything.

192
00:10:38,290 --> 00:10:39,820
Five five is not president.

193
00:10:39,850 --> 00:10:46,870
Then you will push five inside the map and you will also push five inside the vector then four.

194
00:10:47,020 --> 00:10:52,630
So far it is not pleasant inside the map, so you will make it through and they will push for inside

195
00:10:52,630 --> 00:10:53,140
the vector.

196
00:10:53,650 --> 00:10:57,490
Then again, one so one is present then three.

197
00:10:58,000 --> 00:11:00,370
So Ts present and this will be your answer.

198
00:11:01,380 --> 00:11:07,650
This will be, Warren said, OK, so searching in map, so we will use dart count function as discussed

199
00:11:07,650 --> 00:11:08,540
in the previous video.

200
00:11:08,760 --> 00:11:11,650
So this time complexity of current function is constant.

201
00:11:11,700 --> 00:11:16,620
If you are using an order to map and if you are using map, then the time complexity will be log-in.

202
00:11:17,640 --> 00:11:22,950
So we will use on our road map so that time complexity is constrained, so with the help of MAP, we

203
00:11:22,950 --> 00:11:24,580
can solve this question in linear time.

204
00:11:25,590 --> 00:11:27,510
Now, let us write the code for discussion.

205
00:11:27,690 --> 00:11:29,790
A piece of code is going to look very, very simple.

206
00:11:32,410 --> 00:11:39,400
So what we have to do, we have to return our function will return basically vector of integers and

207
00:11:39,400 --> 00:11:42,400
let's say the name of the function is basically remove duplicates.

208
00:11:44,490 --> 00:11:50,910
So what will take so this function will take and that is input and obviously it will take size also.

209
00:11:52,970 --> 00:11:57,290
And then do you have to do so, let us first create our vector of.

210
00:11:58,940 --> 00:12:04,970
So first, let us create our output vector of integers and output so it will contain all the unique

211
00:12:04,970 --> 00:12:06,420
elements and in the same order.

212
00:12:07,010 --> 00:12:09,440
And now let us also create an order to map.

213
00:12:11,360 --> 00:12:12,650
So key will be element.

214
00:12:14,660 --> 00:12:15,750
It will be deja.

215
00:12:17,120 --> 00:12:20,100
And they will be basically boolean value, true or false?

216
00:12:20,690 --> 00:12:22,790
And let's see the name of my is my map.

217
00:12:24,020 --> 00:12:30,950
So you can see yourself, so the structure of my map is very simple, give a loopier.

218
00:12:30,950 --> 00:12:36,510
So she's going to be element and element as integer and this is going to be true or false.

219
00:12:36,530 --> 00:12:40,700
So basically, boolean value key and integer and boolean.

220
00:12:42,410 --> 00:12:45,580
Initially my map is empty, then you have to do so.

221
00:12:45,850 --> 00:12:49,250
Now let us over the so I equals zero.

222
00:12:49,580 --> 00:12:51,140
I understand and I placeless.

223
00:12:53,020 --> 00:12:57,780
Now, first, let us search whether this element is present or not, so I will use common functions

224
00:12:57,780 --> 00:12:59,810
so my map does not count.

225
00:13:00,160 --> 00:13:02,220
So what it will take, it will take as input.

226
00:13:02,800 --> 00:13:04,920
Basically, Miss Element element is integer.

227
00:13:05,290 --> 00:13:07,870
So if if I equals equals zero.

228
00:13:08,620 --> 00:13:13,280
So I told you can't function will only return to things zero one zero means not present.

229
00:13:13,600 --> 00:13:18,080
So if the element is not present inside the map that means it is coming for the first time.

230
00:13:18,370 --> 00:13:24,580
So if it is coming for the first time, you will push it inside your vector because it is coming for

231
00:13:24,580 --> 00:13:25,300
the first time.

232
00:13:25,700 --> 00:13:26,950
So output push pushback.

233
00:13:27,760 --> 00:13:30,210
And also you will push it inside the map.

234
00:13:30,520 --> 00:13:33,310
So my map dart bush.

235
00:13:34,530 --> 00:13:38,280
--, my Amapa, and that's great because so Alphie.

236
00:13:39,840 --> 00:13:40,910
Is basically true.

237
00:13:42,410 --> 00:13:47,690
OK, so this is key and this is value, so Elfy is key and this is value.

238
00:13:47,700 --> 00:13:49,210
So if is true.

239
00:13:49,560 --> 00:13:53,310
And finally, what you can do, you can return your vector.

240
00:13:53,700 --> 00:13:55,140
So return output.

241
00:13:57,700 --> 00:14:01,420
Simple, now let us create an error here and let us do our program.

242
00:14:03,530 --> 00:14:05,590
So let's give the elemental.

243
00:14:09,820 --> 00:14:10,660
So this is the.

244
00:14:11,870 --> 00:14:14,540
Edit And now let us call the function.

245
00:14:17,590 --> 00:14:19,090
Selecter and outpoured.

246
00:14:20,140 --> 00:14:22,240
So the name of the function is remove duplicates.

247
00:14:26,700 --> 00:14:30,240
This function will take at as input and what is the size of the area?

248
00:14:30,410 --> 00:14:36,360
So two, one, two, three, four, five, six, seven, eight, nine, 10, 11 and 12.

249
00:14:36,360 --> 00:14:37,670
So they are total 12 element.

250
00:14:37,690 --> 00:14:38,830
So size of that is 12.

251
00:14:39,630 --> 00:14:46,040
And now let us bring our vector so that we can see the output so close to zero.

252
00:14:46,050 --> 00:14:46,770
I listin.

253
00:14:48,440 --> 00:14:49,460
Outpoured says.

254
00:14:51,250 --> 00:14:52,090
I placeless.

255
00:15:03,450 --> 00:15:06,810
No, first, before running this program, let's see what our output should be.

256
00:15:09,190 --> 00:15:14,530
So when when is coming for the first time, then five is going for the first time, two is coming for

257
00:15:14,530 --> 00:15:19,090
the first time, for what is coming for the first time, will not take this four, then three is coming

258
00:15:19,090 --> 00:15:19,840
for the first time.

259
00:15:20,940 --> 00:15:26,850
So we will not take this three, we will take six, we will also take seven, we will not take one and

260
00:15:26,850 --> 00:15:28,800
we will not take one, we will not take two.

261
00:15:29,040 --> 00:15:30,620
So they should be our answer.

262
00:15:32,190 --> 00:15:36,600
When five to four to six, seven one, five to four to six, seven Solarte.

263
00:15:37,570 --> 00:15:38,530
Yurtseven Applebaum.

264
00:15:40,500 --> 00:15:46,860
OK, so basically we have to include Victor, we have to include an order to map, so instead of including

265
00:15:46,860 --> 00:15:48,480
these two, we can write.

266
00:15:53,960 --> 00:15:55,280
So now let's on the program.

267
00:15:58,960 --> 00:16:00,850
OK, so my blood count.

268
00:16:02,170 --> 00:16:02,850
What is the idea?

269
00:16:02,900 --> 00:16:04,510
So if my map.

270
00:16:05,660 --> 00:16:07,230
Not Count Elfy.

271
00:16:07,610 --> 00:16:09,080
OK, so the name of that is.

272
00:16:10,160 --> 00:16:11,660
So let's change the name.

273
00:16:13,810 --> 00:16:14,920
The name is Enno.

274
00:16:19,520 --> 00:16:27,200
So this is the output when five, 246, seven, so our output is right as discussed when five to four

275
00:16:27,200 --> 00:16:27,830
to six, seven.

276
00:16:27,840 --> 00:16:31,220
So all the elements are unique and elements are appearing in the same order.

277
00:16:31,910 --> 00:16:35,060
Now, let us discuss what is the time complexity of a solution.

278
00:16:38,980 --> 00:16:44,470
Now, what we are doing here, so we are just editing all the data and basically the scoring function

279
00:16:44,470 --> 00:16:48,970
will be constrained by this pushback will take time and this inspection will take constant time.

280
00:16:49,210 --> 00:16:52,030
So our time complexity is basically linear.

281
00:16:53,120 --> 00:16:58,370
Now, if you are using map, so instead of using an order to map, if you will use map, then what will

282
00:16:58,370 --> 00:16:58,680
happen?

283
00:16:58,850 --> 00:17:03,920
So this current function will take a long time and social function will also take longer time.

284
00:17:04,069 --> 00:17:06,200
So time complexity will become and login.

285
00:17:08,240 --> 00:17:12,890
So this is the time complexity, if you will, use map, and this is the time complexity, if you will

286
00:17:12,890 --> 00:17:13,970
use an online map.

287
00:17:16,329 --> 00:17:21,280
And this is my output, so all the elements are unique and elements are appearing in the same order.

288
00:17:21,280 --> 00:17:26,940
So you can see when five, then two, then four and then three, six and seven.

289
00:17:27,160 --> 00:17:29,410
So one, five to four, three, six, seven.

290
00:17:29,740 --> 00:17:31,350
So I hope you understood this code.

291
00:17:32,020 --> 00:17:33,730
I hope you understood this question also.

292
00:17:34,960 --> 00:17:38,040
I will see you in the next one, and if you have any doubt, you can ask me.

293
00:17:38,440 --> 00:17:38,920
Thank you.