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Hi, everyone.

2
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So today we are going to solve this question single no.

3
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OK, so what is the problem statement?

4
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So no statement is very clear.

5
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So we will be given.

6
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So the input will basically be Eddie.

7
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I ready and we have to do so, basically, every element will appear twice inside the area except for

8
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one element, and our task is to basically find that one element.

9
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OK, so let us consider some example.

10
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So this is our input today.

11
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So, see, every element is appearing twice except one element.

12
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So this is orcharding only one time.

13
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So our answer for this input will basically be one.

14
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OK, now C two is updating two times when is appearing two times.

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So forward is the element that we have to find.

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So in this case, our answer will be four, because every element is appearing twice except one element.

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And we have to find this one element.

18
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OK, now let see this example.

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So one is appearing two times, two is appearing two times.

20
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So I have this elementally, so these are being only one time.

21
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So our answer will be three.

22
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OK, so I hope you understood the question.

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Now let's discuss the approaches to solve this question.

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OK, let's discuss approach number one.

25
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OK, so first approach is basically it involves sorting.

26
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OK, so what we will do, so we will sort the area.

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OK.

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So after sorting the area, what will happen?

29
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So let us consider this example.

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OK.

31
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So this area will become one to two.

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So basically what will happen?

33
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So if the elements are slim, they will appear in a decent manner.

34
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OK, so in this case, I will to over the area and I can see when is the different element.

35
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OK, so first I will start the area and then I will iterate over the area.

36
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So what is the time and the space complexity.

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So spaces will be are drawn and time is first.

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I am starting the area so and Logan and then I armatrading out the area which is then.

39
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So basically this is nothing but.

40
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And Logan.

41
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OK, similarly, if you will sort this out.

42
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So what will happen, all the elements which are the same, they will appear in the they will come in.

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They are designed to be OK.

44
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So this idea will become one one, two, two and four.

45
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So basically, when you illustrate they are saying move overhead, they are saying move now.

46
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This is the different element.

47
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OK, so in this way, you can find out the element now for this area, if you will apply the starting

48
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approach, what will happen?

49
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So this will be one one, then again to two and then three.

50
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OK, so these two elements are same.

51
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These tremendous him.

52
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This is occurring one time.

53
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OK, so our answer will be this.

54
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So in this way, what we will do first start theory and then to find out the answer and just outraged

55
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over the error and find the unique element.

56
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OK, so this is the time and the space complexity time and again.

57
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And space complexity is basically one.

58
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OK, so now let us discuss the second approach.

59
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So the second approach, basically it is it uses Hashmat.

60
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OK, so in the second approach, what we do so basically in the second approach we will create a hash

61
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map.

62
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OK, so what we'll do, we will take a map.

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OK, so the key will be element.

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And the value will be basically the account of that element.

65
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OK, so this is the key value that I will totally Hashmat.

66
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OK, so now let's just iterate over the array.

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So let us consider this example.

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OK, four, one, two, one, two.

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So I have for one again, two, one and two.

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OK, so initially my map is empty.

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Now I trade so I have a limit for its count.

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This one.

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Then I have element one.

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So element one, it's count this one, then I have two.

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So I have element to then it's count this one, I have element one.

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So increase the count so it will become two.

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I have this element also includes the count.

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It will become too.

79
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OK, so once you will iterate over the array you hash map is ready.

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I would be ready.

81
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Then what we will do we will trade over the hash map.

82
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OK, we will trade over the hash map.

83
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And if there is an element present whose count is basically one, then our answer will be basically

84
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the key.

85
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OK, so in this case I would answer well before because it's count this one.

86
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OK, now let us consider one more example.

87
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So suppose I have one, two, two, three and one.

88
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Let us prepare our hash map.

89
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OK, so I have this element when so when it's count this one.

90
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OK, now element to its count this one again elemental.

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So I will increase the count.

92
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OK and I have elementary so push inside the hash map.

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Count this one.

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Now I have element one to increase the count.

95
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OK, so after doing this our test prep is ready.

96
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After reading over the area, I would have thought I was ready then I would like to over the map and

97
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Guitry has gone to one, so that's why I would answer will be three.

98
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OK, and similarly with this one, also, if you will prepare the hash map, it will be like element

99
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to count one.

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Then again, element also I will make the count to then I have element one to element one and count

101
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this one.

102
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So this is the Hashmat.

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Now, I trade over the map and you will be able to find the answer, which is one.

104
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OK, so in this case, what is the time and the space complexity?

105
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So basically, time complexities big off, and so we are assuming that the unordered so we lose, we

106
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will use in order to map.

107
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And in order to map the time complexities, basically order of an OK, you can insert, you can search.

108
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So all this time complexities are basically ordered offline.

109
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OK, so the time complexity is just started off and we are trading over the edge and preparing the hash

110
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map and the space complexities basically all that often.

111
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OK, so this is the time and the space complexity for using the hash map.

112
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OK, so you can see we are optimizing on time, but here we are using extra space.

113
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Here we are using extra space to prepare the map.

114
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OK, so now let us discuss the best approach for solving this question.

115
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OK, so let's discuss approach number three.

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So we have our approach.

117
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So excited, overrated.

118
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OK, so this is the symbol.

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So what is also basically the timetable of SARS, so you have two values, and B, so if this is one

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one one zero zero one and zero zero.

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And this is basically the output.

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OK, so what they are doing, why is basically A, X or B, OK, so when you will apply the operator,

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what will happen?

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So if the elements are same, then the output will be zero.

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So in this case, the elements are same, the output is zero.

126
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And if the elements are different and output is one.

127
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OK, so this is what, Exodus?

128
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OK, so suppose is basically two and basically three, so what will happen?

129
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So this is a bitwise operator.

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So how bitwise operators will work, so it will convert to into its binary form, so which is basically

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one in zero, will convert three into its binary form, which is nothing but one and one, and then

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we will apply the exact operator.

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So the elements are different.

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Output is one element, the same.

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So the output.

136
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And now we will convert this binary into its decimal form.

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So it will become one.

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OK, so basically two zah two exactly.

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Is nothing but one.

140
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OK, so this is how it works.

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Now there are some properties of XOL.

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OK, so what are those properties.

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So in order to solve this question you must know those properties.

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So the first properties supposed to have an element A.

145
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So the first properties, if you will, or two same elements, you will get zero second property, so

146
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you have an element and if you will take the order with zero, then you will get itself.

147
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OK, so this is the first property.

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This is the second property.

149
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So third property is basically the XOR operator is basically associative.

150
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OK, so if you will write like this A, X or B, X or A, then basically and this, and this will be

151
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zero.

152
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OK, you can see if the elements are same, they are zero.

153
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So what you can write so e x or B x ray.

154
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This is nothing but a X or A, X or B.

155
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OK, and this will become zero.

156
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So I have zero X or B and then we will use the second property.

157
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So this is nothing but B.

158
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OK, so you can write A, X or B, X or so, and they will cancel out each other, you will have B,

159
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OK, so these are the three properties that you should know.

160
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OK, so these are the first two properties and this one is the third property.

161
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Now, let's see how we can use this extra body to solve our question.

162
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OK, so let us take some examples.

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So for one, two, one, two, OK, so I have this idea for one, two, one, two.

164
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So this is for this is one.

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This is two.

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This is one.

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And this is two.

168
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OK, what we will do so we will take the XOR off all the elements, take X out of all the elements.

169
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So what it will look like for X or one X or two X or one expert to.

170
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OK, so what will happen.

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This one and this one, they will cancel out each other, these two and this, they will cancel out

172
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each other.

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Why.

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They will cancel out each other.

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Because I know e if the elements are sim and I'll take the third output will be zero and if you will

176
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take the Zord with zero you will take you will get itself.

177
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OK, so what will happen.

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Only four will be the remaining element.

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So our answer will be for.

180
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OK, let us consider one more example, so.

181
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I have one, then I have two, I have two and three and one.

182
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Now, let's take XOR of all the elements of the.

183
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So what it will look like when X to X to excite three Akhavan.

184
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OK, so this two and these two will cancel out each other.

185
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This one will cancel out this one.

186
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So our output will be three.

187
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OK, so very simple.

188
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So what did be the time this was complexity of using this approach, so space complexity will be a rough

189
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one.

190
00:10:50,380 --> 00:10:53,470
OK, and the time complexity will be what we are doing.

191
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We are just outrating over the.

192
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So the time complexity will be big of an.

193
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OK, so this is the best approach.

194
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OK, this is the best approach.

195
00:11:03,640 --> 00:11:06,790
So what we will do, we will take a variable.

196
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Initially that variable will be zero and then we will take the zoref old element.

197
00:11:12,250 --> 00:11:13,870
OK, this has to be zero.

198
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OK, this has to be zero because what will what do you will write.

199
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You will write like something.

200
00:11:18,520 --> 00:11:24,480
So answer equals answer x or let's say the name of that is a aof.

201
00:11:25,620 --> 00:11:32,340
OK, so basically when the value of zero, that is the first element, so that's why you have to initialize

202
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with the zero because you can see this is the property zero X or any element is basically that element

203
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only.

204
00:11:41,100 --> 00:11:43,960
OK, so you have to initialize your answer with zero.

205
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This is compulsory.

206
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OK.

207
00:11:47,340 --> 00:11:48,570
Now that that's right, the good.

208
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So basically, this is a legitimate problem.

209
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OK.

210
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So now let us write the code.

211
00:11:56,920 --> 00:12:03,560
OK, so let's change the name of that to be a so we have to return that integer that is appearing only

212
00:12:03,640 --> 00:12:04,210
one time.

213
00:12:05,820 --> 00:12:08,270
So first, let us create a variable answer.

214
00:12:09,530 --> 00:12:13,430
Which will contain that unique element and this has to be initialized with zero.

215
00:12:14,210 --> 00:12:18,230
OK, so now what we have to do, we are just iterating over the Eddie.

216
00:12:19,240 --> 00:12:20,590
So Edward says.

217
00:12:24,290 --> 00:12:31,190
Now you just have to take the XOR of all the elements, OK, you just have to take the eggs out of all

218
00:12:31,190 --> 00:12:31,700
the elements.

219
00:12:34,950 --> 00:12:37,620
And finally, what you look, you will return, I would answer.

220
00:12:39,010 --> 00:12:40,430
OK, so what I'm doing here.

221
00:12:42,090 --> 00:12:44,890
So let us consider this example for one, two, one, two.

222
00:12:44,940 --> 00:12:47,220
OK, so I have four.

223
00:12:47,640 --> 00:12:51,450
One, two, one, and then two.

224
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So this is one.

225
00:12:53,870 --> 00:12:56,710
OK, so my answer is initially zero.

226
00:12:58,250 --> 00:13:03,680
OK, our answer is initially zero, then I will take the action with the first element so it will become

227
00:13:04,250 --> 00:13:05,980
zero or four.

228
00:13:06,290 --> 00:13:13,400
So this is basically for then I will take the I will take the Tzavela one element.

229
00:13:13,550 --> 00:13:16,170
So this is for X or one.

230
00:13:16,220 --> 00:13:21,380
So this is basically for X or Y and only then I will take the Zoref two.

231
00:13:21,980 --> 00:13:26,630
So this is for X or one and then two.

232
00:13:26,630 --> 00:13:28,550
OK, so this is the value of one.

233
00:13:28,560 --> 00:13:33,010
So you can see answer operator error element.

234
00:13:33,020 --> 00:13:34,400
So this is the answer.

235
00:13:35,890 --> 00:13:38,090
This is the operator and this is the element.

236
00:13:38,430 --> 00:13:42,270
OK, so then I will take XOR with one.

237
00:13:42,680 --> 00:13:44,200
So what is the answer?

238
00:13:44,220 --> 00:13:46,630
The answer is for actual one X or two.

239
00:13:47,230 --> 00:13:54,490
OK, so this is the answer, then the operator and then the area element, which is when now what will

240
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happen.

241
00:13:55,180 --> 00:13:57,730
This one and this one, they will cancel out each other.

242
00:13:58,060 --> 00:14:03,670
So this is basically for X or two, OK, and then this element to.

243
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So our answer will be, which is the current answer to our current answer is, for example.

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Then the operator, Exide operator, and then the element of the too, so this two and these two, they

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will cancel out each other.

246
00:14:19,970 --> 00:14:22,950
So basically when we will come out, so I will reach here.

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So basically, we will come out of this for loop surveillance that will contain food, which is the

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right answer, and we will return that answer.

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OK, very simple.

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Now let us submit our code.

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OK.

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OK, so our goal is working fine, so let us summarize, so basically.

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So basically we have discussed three approaches for solving this question.

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First approach was basically desalting approach.

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That time, complexity was basically and Logan.

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And this space complexity was hard enough when.

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The second approach that we discussed, it involves the use of Hashmat.

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OK, the time complexity we discussed was big off and and the space complexity was also a big off.

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And so we can see here we are improving all the time, but we are taking more space.

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OK, now, the third way that the third way to solve this question is basically with the help of our

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operator.

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OK, the time complexity is basically big off and and the space complexity is basically big of one.

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00:15:32,590 --> 00:15:35,980
OK, so this is the best approach for solving this question.

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00:15:36,860 --> 00:15:40,300
OK, and what are the properties of X or so any element.

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Exide with the same element, output will be zero.

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Any element Exurbia zero basically will be that element only, and the third basically is the associative

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property.

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So A, X or B, X or A is basically nothing.

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It is same as a X or a Asabi.

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So this will become zero.

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So I have zero X or B and this is nothing but B, OK, so in short you can ride this and disable, cancel

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out each other.

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Your output will be.

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00:16:17,290 --> 00:16:19,660
OK, so I hope you understood this question.

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Thank you.