Reputation: 311
To find the number of occurrences of a given string P ( length m ) in a text T ( length N )
We must use binary search against the suffix array of T.
The issue with using standard binary search ( without the LCP information ) is that in each of the O(log N) comparisons you need to make, you compare P to the current entry of the suffix array, which means a full string comparison of up to m characters. So the complexity is O(m*log N).
The LCP-LR array helps improve this to O(m+log N). know more
How we precompute LCP-LR array from LCP array?
And How does LCP-LR help in finding the number of occurrences of a pattern?
Please Explain the Algorithm with Example
Thank you
Upvotes: 0
Views: 1040
Reputation: 626
Not having enough reps to comment so posting. Is anybody able to create the LCP-LR using @Abhijeet Ashok Muneshwar solution. For ex for text- mississippi the Suffix array-
The LCP array will be
And LCP-LR will be
But the LCP-LR obtained using the code is not same as above. To the method buildLCP_LR i am passing index=0, low=0, high=n
Upvotes: 0
Reputation: 2508
// note that arrSize is O(n)
// int arrSize = 2 * 2 ^ (log(N) + 1) + 1; // start from 1
// LCP = new int[N];
// fill the LCP...
// LCP_LR = new int[arrSize];
// memset(LCP_LR, maxValueOfInteger, arrSize);
//
// init: buildLCP_LR(1, 1, N);
// LCP_LR[1] == [1..N]
// LCP_LR[2] == [1..N/2]
// LCP_LR[3] == [N/2+1 .. N]
// rangeI = LCP_LR[i]
// rangeILeft = LCP_LR[2 * i]
// rangeIRight = LCP_LR[2 * i + 1]
// ..etc
void buildLCP_LR(int index, int low, int high)
{
if(low == high)
{
LCP_LR[index] = LCP[low];
return;
}
int mid = (low + high) / 2;
buildLCP_LR(2*index, low, mid);
buildLCP_LR(2*index+1, mid + 1, high);
LCP_LR[index] = min(LCP_LR[2*index], LCP_LR[2*index + 1]);
}
Reference: https://stackoverflow.com/a/28385677/1428052
Upvotes: 1