Skip to main content

Two numbers with minimum difference

Find the two numbers whose difference is minimum among the set of numbers.

For example the sequence is 5, 13, 7, 0, 10, 20, 1, 15, 4, 19

The algorithm should return min diff = 20-19 = 1.

Make an efficient algorithm, yeah best could be O(n).

Not sure if i could use DP here, will post a solution tommorow. Let me know if you find a answer.

Algorithm:MiniMumDiffN
Get the size of array , say n.
Get the range of array, range.
Bucket Sort the above array.
Search the above sorted array for minimum difference.
This will be the minimum difference numbers.

Now question comes where the range of above given array is very big.
Algorithm:DivideArray
Take an random element.
Search for its position in array.
Return the index of element.

Algorithm:MiniMumDiff
if array_range > MAXRANGE
int mid = divideArray (arr,start,end);
d1,max1 = MiniMumDiff( arr,start,mid-1);
d2,max2 = MiniMumDIff(arr,mid+1,end);
d3 = difference max1 and arr[mid]
d4 = difference max2 and arr[mid]
return minimum( d1,d2,d3,d4)
else
return MiniMumDiffN(arr,start,end);


TC MiniMumDiffN - O(n)
TC DivideArray - O(n)
TC MiniMumDIff - O(log(range)n*2n)

And MiniMumDiff will be dividing range n in range 2 power (i) = n/range;
So MiniMumDiff will be if TC O(n)

Comments

  1. really stupid, big Oh of bucket is mn where n is number of elements and m is max no of digits in any number

    ReplyDelete

Post a Comment

Popular posts from this blog

Car Parking Problem

There is n parking slots and n-1 car already parked. Lets say car parked with initial arrangement and we want to make the car to be parked to some other arrangement. Lets say n = 5, inital = free, 3, 4, 1, 2 desired = 1, free, 2, 4 ,3 Give an algorithm with minimum steps needed to get desired arrangement. Told by one of my friend and after a lot of search i really got a nice solution. I will post solution in comment part

DEShaw Interview Questions

ther are N numbers frm 1 to N and starting from index 1 we will keep deleting every alternate going in cyclic order with array. Only one element will be left at the end. Tell us the index of element in array we started. e.g. there are 5 nums 1 2 3 4 5 then after 1st iteration 1 3 5 will be remained. .. then 1 will be next to be elliminated and then 5 3 will remain alone... give sum efficient algorithm to calculate which numer will remain at the end Answer: 2*(n-2^p)+1 where p=floor(log2 n)

One way : Heap or Stack allocation

Took my time to find solution for this and it actually made me realize how much i have to learn. Now question is how can we enforce the creation of objects whether on heap or stack. Creating object on Heap only. Make your destructor private and create a destroy function. Class A {  ~A(){} public:  A(){} void destroy() { delete this; } } Now if you can not create your object: A a; // error: 'A::~A()' is private A  a = new A(); Great. Now all the objects will be allocated to heap only. Making destructor private a better choice than constructor as we don't know number of constructor so early in programming. Creating object on stack While i knew that it has to do something with operator new, but could not figure out how can i create object on stack and return the memory back to class initialization. Though answer was simple, i had to google for that: Overload operator new, new[], operator delete and delete[] and make them private ;) class A { pr...