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Given an array of size N in which every number is between 1
and N, determine if there are any duplicates in it. You are
allowed to destroy the array if you like.

Answers were Sorted based on User's Feedback

Given an array of size N in which every number is between 1 and N, determine if there are any dupli..

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;
}
}

 Is This Answer Correct ? 52 Yes 11 No

Given an array of size N in which every number is between 1 and N, determine if there are any dupli..

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)

 Is This Answer Correct ? 28 Yes 7 No

Given an array of size N in which every number is between 1 and N, determine if there are any dupli..

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

 Is This Answer Correct ? 28 Yes 11 No

Given an array of size N in which every number is between 1 and N, determine if there are any dupli..

#include<stdio.h>
void main()
{
int a,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);
}
}
}

 Is This Answer Correct ? 22 Yes 7 No

Given an array of size N in which every number is between 1 and N, determine if there are any dupli..

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

 Is This Answer Correct ? 21 Yes 10 No

Given an array of size N in which every number is between 1 and N, determine if there are any dupli..

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...

 Is This Answer Correct ? 14 Yes 9 No

Given an array of size N in which every number is between 1 and N, determine if there are any dupli..

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

position and mark it zero.. keep doing so...

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

 Is This Answer Correct ? 18 Yes 14 No

Given an array of size N in which every number is between 1 and N, determine if there are any dupli..

hey scrubs, use a hash map

loop through array {
if(hash[x] == x] return true// is dupe
}

max O(n)

 Is This Answer Correct ? 6 Yes 4 No

Given an array of size N in which every number is between 1 and N, determine if there are any dupli..

@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.

 Is This Answer Correct ? 5 Yes 3 No

Given an array of size N in which every number is between 1 and N, determine if there are any dupli..

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;
--------------------

 Is This Answer Correct ? 4 Yes 3 No

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