Discussion:
Missing Number in an array
TUSHAR_MCA
2011-07-18 11:31:52 UTC
Given an array of size n. It contains numbers in the range 1 to n.
Each number is present at least once except for 2 numbers. Find the
missing numbers ?
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varun pahwa
2011-07-18 11:57:32 UTC
make two equations as .
suppose numbers to be x,y
x + y = p = (n*(n+1))/2 - (sum of all elements of array).

x^2 + y^2 = q = (n*(n+1)*(2n+1))/6 - (sum of square of all elements of
array).

so 2*x*y can be calculated as (p^2 - q);

so, a quad equation is formed as you now (x + y) and (2*xy).

P.S. :: overflow is not handled.

Post by TUSHAR_MCA
Given an array of size n. It contains numbers in the range 1 to n.
Each number is present at least once except for 2 numbers. Find the
missing numbers ?
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Varun Pahwa
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ashish mann
2011-07-18 12:09:54 UTC
A little better way would be
x + y = p = (n*(n+1))/2 - (sum of all elements of array).
x * y = q = n! / (product of all elements of the array).

z^2 - (sum of the roots)z + (product of the roots) = 0
Post by varun pahwa
make two equations as .
suppose numbers to be x,y
x + y = p = (n*(n+1))/2 - (sum of all elements of array).
x^2 + y^2 = q = (n*(n+1)*(2n+1))/6 - (sum of square of all elements of
array).
so 2*x*y can be calculated as (p^2 - q);
so, a quad equation is formed as you now (x + y) and (2*xy).
P.S. :: overflow is not handled.
Post by TUSHAR_MCA
Given an array of size n. It contains numbers in the range 1 to n.
Each number is present at least once except for 2 numbers. Find the
missing numbers ?
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Varun Pahwa
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Aashish Mann
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ankit sambyal
2011-07-18 12:48:49 UTC
@Ashish : Cud u plz highlight how to write code to solve a quadratic equation ?
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Dumanshu
2011-07-18 13:08:28 UTC
Quadratic equation solver: just solve it the way we do on paper, take
coefficients as input and apply formula x = (-b+sqrt(b^2 - 4*a*c))/
2*a...
anything wrong with this?

And for the given problem, it says "atleast" once and not "exactly"
once. So, Ankit's solution of bit vector is the best one.
Post by ankit sambyal
@Ashish : Cud u plz highlight how to write code to solve a quadratic equation ?
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ankit sambyal
2011-07-18 12:01:24 UTC
1. Initialize a bit vector of size n.
2. For every no. set the corresponding bit vector.
3. Now scan through the bit vectors and get the missing numbers
corressponding to the unset bits in the bit vector.

Time complexity : O(n)
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varun pahwa
2011-07-18 12:11:19 UTC
The above solution will work if each other number is exactly once present.
but if that 's not true.
then 4 equations can be formed.
Assuming a,b repeated number where a may or may be equal to b.

then equations will be
x + y = a + b;
x^2 + y^2 = a^2 + b^2.
x.y = a.b
x^3 + y^3 = a^3 + b^3.
now 4 equations 4 variables can be solved.
Post by ankit sambyal
1. Initialize a bit vector of size n.
2. For every no. set the corresponding bit vector.
3. Now scan through the bit vectors and get the missing numbers
corressponding to the unset bits in the bit vector.
Time complexity : O(n)
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Varun Pahwa
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Aakash Johari
2011-07-18 12:13:18 UTC
@varun: can you write the code for one?
Post by varun pahwa
The above solution will work if each other number is exactly once present.
but if that 's not true.
then 4 equations can be formed.
Assuming a,b repeated number where a may or may be equal to b.
then equations will be
x + y = a + b;
x^2 + y^2 = a^2 + b^2.
x.y = a.b
x^3 + y^3 = a^3 + b^3.
now 4 equations 4 variables can be solved.
Post by ankit sambyal
1. Initialize a bit vector of size n.
2. For every no. set the corresponding bit vector.
3. Now scan through the bit vectors and get the missing numbers
corressponding to the unset bits in the bit vector.
Time complexity : O(n)
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Varun Pahwa
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Aakash Johari
2011-07-18 12:14:15 UTC
Yes, you will have to write a quad eq. solver. It's easy to write.
Post by Aakash Johari
@varun: can you write the code for one?
Post by varun pahwa
The above solution will work if each other number is exactly once present.
but if that 's not true.
then 4 equations can be formed.
Assuming a,b repeated number where a may or may be equal to b.
then equations will be
x + y = a + b;
x^2 + y^2 = a^2 + b^2.
x.y = a.b
x^3 + y^3 = a^3 + b^3.
now 4 equations 4 variables can be solved.
Post by ankit sambyal
1. Initialize a bit vector of size n.
2. For every no. set the corresponding bit vector.
3. Now scan through the bit vectors and get the missing numbers
corressponding to the unset bits in the bit vector.
Time complexity : O(n)
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Varun Pahwa
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-Aakash Johari
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varun pahwa
2011-07-18 12:27:18 UTC
sorry that would not work. only it could work if each element is present
exactly once.
Post by Aakash Johari
Yes, you will have to write a quad eq. solver. It's easy to write.
Post by Aakash Johari
@varun: can you write the code for one?
Post by varun pahwa
The above solution will work if each other number is exactly once
present. but if that 's not true.
then 4 equations can be formed.
Assuming a,b repeated number where a may or may be equal to b.
then equations will be
x + y = a + b;
x^2 + y^2 = a^2 + b^2.
x.y = a.b
x^3 + y^3 = a^3 + b^3.
now 4 equations 4 variables can be solved.
Post by ankit sambyal
1. Initialize a bit vector of size n.
2. For every no. set the corresponding bit vector.
3. Now scan through the bit vectors and get the missing numbers
corressponding to the unset bits in the bit vector.
Time complexity : O(n)
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Varun Pahwa
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-Aakash Johari
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Varun Pahwa
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Nishant Mittal
2011-07-18 12:31:24 UTC
1.Calculate XOR of all the array elements.
2.XOR the result with all numbers from 1 to n =>x1
3.After 2nd step, all elements would nullify each other except 2 missing
elements(let x and y) and x1 will contain XOR of x and y
4.All the bits that are set in x1 will be set in either x or y. Get the
rightmost set bit from x1.
5.divide the elements of the array in two sets  one set of elements with
same bit set and other set with same bit not set. By doing so, you will get
x in one set and y in another set.
6.XOR all the elements of 1st set with the numbers between 1 to n which have
same bit set and XOR the 2nd set with the numbers between 1 to n which have
same bit not set. Now result of both set will have the desired result
Post by varun pahwa
sorry that would not work. only it could work if each element is present
exactly once.
Post by Aakash Johari
Yes, you will have to write a quad eq. solver. It's easy to write.
Post by Aakash Johari
@varun: can you write the code for one?
Post by varun pahwa
The above solution will work if each other number is exactly once
present. but if that 's not true.
then 4 equations can be formed.
Assuming a,b repeated number where a may or may be equal to b.
then equations will be
x + y = a + b;
x^2 + y^2 = a^2 + b^2.
x.y = a.b
x^3 + y^3 = a^3 + b^3.
now 4 equations 4 variables can be solved.
Post by ankit sambyal
1. Initialize a bit vector of size n.
2. For every no. set the corresponding bit vector.
3. Now scan through the bit vectors and get the missing numbers
corressponding to the unset bits in the bit vector.
Time complexity : O(n)
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Varun Pahwa
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ankit sambyal
2011-07-18 12:34:45 UTC
@Nishant : If the array contains duplicate elements, ur algo will fail !!!
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ankit sambyal
2011-07-18 12:31:46 UTC
@varun : My solution does not fail if every other number is not
present exactly once.

If a bit is already set, if we again set it, there's no harm.. The
algo works perfectly f9..
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SAMMM
2011-07-18 12:37:48 UTC
#include<stdio.h>
int main()
{

int a[]={1,2,3,5,2}; // 2 isrepeated and 4 is missing
int i=0,x=0,j=0,bit;
while(i<5)
{
j^=i+1;j^=a[i];
i++;
}

bit=j&~(j-1); //set bits

i=0;j=0;

while(i<5)
{
if(bit&(i+1)) x^=(i+1); //two set are needed to iterate
else j^=(i+1); //two set are needed to iterate
i++;
}
i=0;
while(i<5)
{
if(bit&a[i]) x^=a[i]; //two set are needed to iterate
else j^=a[i]; //two set are needed to iterate
i++;
}

printf("%d %d",x,j);
return 0;
}

Hav a look .... The trick is in the set bit [ bit=j&~(j-1); //set
bits]
Post by TUSHAR_MCA
Given an array of size n. It contains numbers in the range 1 to n.
Each number is present at least once except for 2 numbers. Find the
missing numbers ?
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Nishant Mittal
2011-07-18 12:42:10 UTC
@ankit...i m assuming that array of size n contains n-2 elements with 2
elements missing
Post by SAMMM
#include<stdio.h>
int main()
{
int a[]={1,2,3,5,2}; // 2 isrepeated and 4 is missing
int i=0,x=0,j=0,bit;
while(i<5)
{
j^=i+1;j^=a[i];
i++;
}
bit=j&~(j-1); //set bits
i=0;j=0;
while(i<5)
{
if(bit&(i+1)) x^=(i+1); //two set are needed to iterate
else j^=(i+1); //two set are needed to iterate
i++;
}
i=0;
while(i<5)
{
if(bit&a[i]) x^=a[i]; //two set are needed to iterate
else j^=a[i]; //two set are needed to iterate
i++;
}
printf("%d %d",x,j);
return 0;
}
Hav a look .... The trick is in the set bit [ bit=j&~(j-1); //set
bits]
Post by TUSHAR_MCA
Given an array of size n. It contains numbers in the range 1 to n.
Each number is present at least once except for 2 numbers. Find the
missing numbers ?
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ankit sambyal
2011-07-18 12:43:52 UTC
@Nishant: Read the question carefully. It says "Each number is present
at least once except for 2 numbers".
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ankit sambyal
2011-07-18 12:45:52 UTC
Sry for the typo in my previous mail.
@Nishant: Read the question carefully. It says "Each number is present
at least once except for 2 numbers".
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Nishant Mittal
2011-07-18 12:53:28 UTC
ok then 1st remove duplicates from the array in O(n) then apply my algo...
Post by ankit sambyal
@Nishant: Read the question carefully. It says "Each number is present
at least once except for 2 numbers".
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SkRiPt KiDdIe
2011-07-18 13:03:07 UTC
Nishant. Your algorithm works for finding repeated nos. in a [n+2] array
where [1-n] are present atleast once and the same number is not being
repeated twice.

Reanalyze the question.
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Nishant Mittal
2011-07-18 13:10:24 UTC
@SkRiPt KiDdIe... I think you need to analyze my algo again.... it will work
for both the cases...
Post by SkRiPt KiDdIe
Nishant. Your algorithm works for finding repeated nos. in a [n+2] array
where [1-n] are present atleast once and the same number is not being
repeated twice.
Reanalyze the question.
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ankit sambyal
2011-07-18 13:06:25 UTC
@nishant : Time complexity ?????? will increase too much
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Nishant Mittal
2011-07-18 13:11:05 UTC
@ankit.... for removing duplicates=O(n)+O(n)
Post by ankit sambyal
@nishant : Time complexity ?????? will increase too much
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Nishant Mittal
2011-07-18 13:12:44 UTC
and for finding missing 2 elements= O(n)+O(n)+O(n)+O(n)
so total will be O(n)
but there will be no overflow

On Mon, Jul 18, 2011 at 6:41 PM, Nishant Mittal
Post by Nishant Mittal
@ankit.... for removing duplicates=O(n)+O(n)
Post by ankit sambyal
@nishant : Time complexity ?????? will increase too much
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SkRiPt KiDdIe
2011-07-18 13:26:02 UTC
@Nishant:

1 4 5 4 5 n=5

1 2 3 4 5

included in xor as well along with repeating elements.

Hope now you got it. You are giving solution for a question which i have
defined in previous post. and your algo will fail when
the final xor has no set bit i.e. same number is being repeated twice.
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SkRiPt KiDdIe
2011-07-18 13:26:41 UTC
Post by TUSHAR_MCA
Post by SkRiPt KiDdIe
1 4 5 4 5 n=5
1 2 3 4 5
included in xor as well along with repeating elements.
Hope now you got it. You are giving solution for a question which i have
defined in previous post. and your algo will fail when
the final xor has no set bit i.e. same number is being repeated twice.
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Nishant Mittal
2011-07-18 13:57:26 UTC
@SkRiPt KiDdIe... see my 3rd post...i've mentioned that 1st we have to
remove duplicate numbers....
Post by TUSHAR_MCA
Post by TUSHAR_MCA
Post by SkRiPt KiDdIe
1 4 5 4 5 n=5
1 2 3 4 5
are included in xor as well along with repeating elements.
Hope now you got it. You are giving solution for a question which i have
defined in previous post. and your algo will fail when
the final xor has no set bit i.e. same number is being repeated twice.
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KRISHNA
2011-07-18 20:23:08 UTC
Algorithm:

1. Given array A[], Set B[], sum=0,repeatedNumb=0;
2. for index i to n
B[ A[i] -1 ] = B[ A[i] -1]+1
sum=sum+A[i]
3. for index i to n
if B[i]=2
repeatedNumb=i+1
break;

4. Nsum=n*(n+1)/2
5. missingNumb=Nsum+repeatedNumb-sum

Step-2 to3, finding the repeatedNum, complexity=O(n)+O(n)
Step4: finding the Nsum, complexity=O(n) if formula is not used

Thanks,
Krishna
Post by TUSHAR_MCA
1 4 5 4 5    n=5
1 2 3 4 5
included in xor as well along with repeating elements.
Hope now you got it. You are giving solution for a question which i have
defined in previous post. and your algo will fail when
the final xor has no set bit i.e.  same number is being repeated twice.
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2011-07-19 01:00:08 UTC
a[]={1,2,3,4,5} //real a[]

a[]={1,2,3,4,3} //unreal a[]...as 2 numbers not present once...3
is 2 times and 5 is 0 times

step 1: sum=n*(n+1)/2

sum real a[]=5*6/2=15

sum unreal a[]=1+2+3+4+3=13

this means if a is 1 number nd b is 2nd then we know a-b=sum
real a[]- sum unreal a[]

step 2: now square both arrays

repeat step 1

now we know a^2-b^2=sum real a[]- sum unreal a[]

step 3: we know a-b & a^2-b^2....solve for a and b
Post by KRISHNA
1. Given array A[], Set B[], sum=0,repeatedNumb=0;
2. for index i to n
B[ A[i] -1 ] = B[ A[i] -1]+1
sum=sum+A[i]
3. for index i to n
if B[i]=2
repeatedNumb=i+1
break;
4. Nsum=n*(n+1)/2
5. missingNumb=Nsum+repeatedNumb-sum
Step-2 to3, finding the repeatedNum, complexity=O(n)+O(n)
Step4: finding the Nsum, complexity=O(n) if formula is not used
Thanks,
Krishna
Post by TUSHAR_MCA
Post by SkRiPt KiDdIe
1 4 5 4 5 n=5
1 2 3 4 5
are
Post by TUSHAR_MCA
Post by SkRiPt KiDdIe
included in xor as well along with repeating elements.
Hope now you got it. You are giving solution for a question which i
have
Post by TUSHAR_MCA
Post by SkRiPt KiDdIe
defined in previous post. and your algo will fail when
the final xor has no set bit i.e. same number is being repeated
twice.
Post by TUSHAR_MCA
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