Answer / zhixingren

The following algorithm works. It is O(n) and no extra
storage required
public static bool HaveDuplicate(int[] nums)
{
for (int i = 0; i < nums.Length; i++)
{
int tmp = nums[i];
if (tmp < 0)
tmp = -tmp;

if (nums[tmp - 1] < 0)
{
return true;
}
else
{
nums[tmp - 1] = -nums[tmp - 1];
}
}
return false;
}
}

Answer / ashutosh k

You can use intermediate step for count sort.

Take a new array B of size N, initialize all values to 0.
and then
for each i from 1 to N,
if((++B[A[i]]) > 1) - there are duplicates.

--- Time complexity O(n)

Answer / monika

the formula given for summation shud be (n*(n+1))/2

but it will solve the problem only if we have one duplicate.
if we have many duplicates then we can still get the
correct summation
for ex:
1 2 2 5 5

it has two duplicates but still total is 5*(5+1)/2=15

Answer / shri

#include<stdio.h>
void main()
{
int a[3],j,i,n;
clrscr();
printf("enter n");
scanf("%d",&n);
printf("enter els");
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
for(i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
if(a[i]==a[j])
printf("elements %d and %d are same",i+1,j+1);
}
}
}

Answer / raghuram

guys..sort the array using quicksort..and check adjacent
elements..u can know there are duplicates or
not..complexity-O(nlogn)+n-1

Answer / miraj

When its given you are allowed to destroy the array..do
think in that direction..


Read the first element go to that position, read that
position and mark it zero.. keep doing so...

If at any place we find zero.. that means array contains
duplicate.

Answer / anyone

If you mark places with zero, you will lose the original
value in that place which, eventually, leads to errors...

Assume an array starting as: 1-5-5-...

After first step array would be: 1-0-5-...

You will get stuck at the second step since every element
that you marked with zero will lead to the position 0.

Converting to negative makes more sense...

Answer / michael

hey scrubs, use a hash map

loop through array {
if(hash[x] == x] return true// is dupe
else hash.add(x,x);
}

max O(n)

Answer / somebody

@Michael

Moron you are using extra space there.that is obvious.using
extra space many tough problems can be solved.without extra
space tell some better methods.

Answer / ronstar luan

We need not sort this array and we need save space.
{2, 4, 5, 1, 3} m=2;
{2, 2, 5, 1, 3} m=4;
{2, 2, 5, 4, 3} m=1;
{1, 2, 5, 4, 3} m=2;
--------------------
{1, 2, 5, 4, 3} m=5;
{1, 2, 5, 4, 5} m=3;
{1, 2, 3, 4, 5} m=5;
--------------------

main() { char *p; printf("%d %d ",sizeof(*p),sizeof(p)); }

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