Sunday, July 8, 2007

Finding a duplicated integer problem.

Given an array of n integers from 1 to n with one integer repeated..

Here is the simplest of C programs (kind of dumb)


#include < stdio.h >
#include < stdlib.h >
#include < conio.h >

int i,j=0,k,a1[10];

main()
{
printf("Enter the array of numbers between 1 and 100(you can repeat the numbers):");
for(i=0;i <= 9;i++)
{
scanf("%d",&a1[i]);
}

while(j < 10)
{
for(k=0;k < 10;k++)
{
if(a1[j]==a1[k] && j!=k)
{
printf("Duplicate found!");
printf("The duplicate is %d\n",a1[j]);
getch();
}
}
j=j+1;
}

getch();
return(0);
}

6 comments:

Anonymous said...

sum of 1 to n = n(n+1) /2

duplicate number = (sum of array - sum of 1 to n)

Anonymous said...

YOu need to modify this to
duplicate number= total_no- (sumof 1 to n -sum of array).

This would give the correct result

mmc said...

what is "total_no"?
duplicate number = (sum of array - sum of 1 to n) is only correct

Gangadhar Mylapuram said...

The above solutions are wrong.
This can be solved in o(n) using two mathematical equations

SU= Sum of given elements;
ASU= Actual sum of 1-n integers

MU = Multiplication of given numbers
AMU = Multiplication of 1-n integers

R= Repeated number
M= Missing number
SU-R+M=ASU
MU * M/R = AMU

Substitute the values of SU,ASU,MU,AMU and solve the equations. You will get the answers.

sandeep kumar said...

Thanks for the info. It is very helpful.
Regards
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