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Topic: Find a sequence from an array (Read 3934 times) |
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Ved
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Find a sequence from an array
« on: Feb 15th, 2012, 3:26am » |
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Given an int array which might contain duplicates, find if it is a sequence.
Eg. {45,50,47,46,49,48}
is a sequence 45, 46,47,48,49,50
Sorting is an obvious solution. Can this be done in O(n) time and O(1) space
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A
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Re: Find a sequence from an array
« Reply #1 on: Feb 15th, 2012, 8:27am » |
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Using MIN and MAX of the array, we need to check if elements from 1 to N are present.
Assuming array is writable,
if MAX - MIN + 1 NOT EQUAL to size of array => duplicates
A[i] = A[i]-MIN+1 for all i = 1 to n
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Ved
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Re: Find a sequence from an array
« Reply #2 on: Feb 15th, 2012, 9:43pm » |
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I dont get part- A[i] = A[i]-MIN+1 for all i = 1 to n
What that gives is :
1,6,3,2,5,4 - so how do I find the sequence. sorry but I am missing the point here.
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towr
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Re: Find a sequence from an array
« Reply #3 on: Feb 15th, 2012, 10:08pm » |
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Once you know the minimum, you can sort the array in O(n) time if the array forms a sequence. Because you can calculate based on the value were each element should go.
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R
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Re: Find a sequence from an array
« Reply #4 on: Sep 14th, 2012, 9:22pm » |
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[5, 3, 1, 6, 1, 2, 4, 3]
What is the expected answer for the above array? Does it have a sequence? i.e. while looking for a sequence, can I ignore duplicates?
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akasina9
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Re: Find a sequence from an array
« Reply #5 on: Nov 15th, 2012, 9:43am » |
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on Feb 15th, 2012, 8:27am, R0B1N wrote:Using MIN and MAX of the array, we need to check if elements from 1 to N are present.
Assuming array is writable,
if MAX - MIN + 1 NOT EQUAL to size of array => duplicates
A[i] = A[i]-MIN+1 for all i = 1 to n
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Don't you mean:
new_index = A[i] - MIN + 1
in the sorted array?
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