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)
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)
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