Sum of similarities of string with all of its suffixes

Given a string str, the task is to find the sum of the similarities of str with each of its suffixes.
The similarity of strings A and B is the length of the longest prefix common to both the strings i.e. the similarity of “aabc” and “aab” is 3 and that of “qwer” and “abc” is 0.

Examples:

Input: str = “ababa”
Output: 9
The suffixes of str are “ababa”, “baba”, “aba”, “ba” and “a”. The similarities of these strings with the original string “ababa” are 5, 0, 3, 0 & 1 respectively.
Thus, the answer is 5 + 0 + 3 + 0 + 1 = 9.

Input: str = “aaabaab”
Output: 13



Approach: Compute Z-array using Z-algorithm – For a string str[0..n-1], Z array is of same length as string. An element Z[i] of Z array stores length of the longest substring starting from str[i] which is also a prefix of str[0..n-1]. The first entry of Z array is the length of the string.
Now, sum all the elements of the Z-array to get the required sum of the similarities.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <iostream>
#include <string>
#include <vector>
using namespace std;
  
// Function to calculate the Z-array for the given string
void getZarr(string str, int n, int Z[])
{
    int L, R, k;
  
    // [L, R] make a window which matches with prefix of s
    L = R = 0;
    for (int i = 1; i < n; ++i) {
  
        // if i>R nothing matches so we will calculate.
        // Z[i] using naive way.
        if (i > R) {
            L = R = i;
  
            // R-L = 0 in starting, so it will start
            // checking from 0'th index. For example,
            // for "ababab" and i = 1, the value of R
            // remains 0 and Z[i] becomes 0. For string
            // "aaaaaa" and i = 1, Z[i] and R become 5
            while (R < n && str[R - L] == str[R])
                R++;
            Z[i] = R - L;
            R--;
        }
        else {
  
            // k = i-L so k corresponds to number which
            // matches in [L, R] interval.
            k = i - L;
  
            // if Z[k] is less than remaining interval
            // then Z[i] will be equal to Z[k].
            // For example, str = "ababab", i = 3, R = 5
            // and L = 2
            if (Z[k] < R - i + 1)
                Z[i] = Z[k];
  
            // For example str = "aaaaaa" and i = 2, R is 5,
            // L is 0
            else {
                // else start from R and check manually
                L = i;
                while (R < n && str[R - L] == str[R])
                    R++;
                Z[i] = R - L;
                R--;
            }
        }
    }
}
  
// Function to return the similarity sum
int sumSimilarities(string s, int n)
{
    int Z[n] = { 0 };
  
    // Compute the Z-array for the given string
    getZarr(s, n, Z);
  
    int total = n;
  
    // Summation of the Z-values
    for (int i = 1; i < n; i++)
        total += Z[i];
  
    return total;
}
  
// Driver code
int main()
{
    string s = "ababa";
    int n = s.length();
  
    cout << sumSimilarities(s, n);
    return 0;
}

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Java














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// Java implementation of the above approach 


  


public class GFG{ 


  


// Function to calculate the Z-array for the given string 


static void getZarr(String str, int n, int Z[]) 


{ 


    int L, R, k; 


  


    // [L, R] make a window which matches with prefix of s 


    L = R = 0; 


    for (int i = 1; i < n; ++i) { 


  


        // if i>R nothing matches so we will calculate. 


        // Z[i] using naive way. 


        if (i > R) { 


            L = R = i; 


  


            // R-L = 0 in starting, so it will start 


            // checking from 0'th index. For example, 


            // for "ababab" and i = 1, the value of R 


            // remains 0 and Z[i] becomes 0. For string 


            // "aaaaaa" and i = 1, Z[i] and R become 5 


            while (R < n && str.charAt(R - L) == str.charAt(R)) 


                R++; 


            Z[i] = R - L; 


            R--; 


        } 


        else { 


  


            // k = i-L so k corresponds to number which 


            // matches in [L, R] interval. 


            k = i - L; 


  


            // if Z[k] is less than remaining interval 


            // then Z[i] will be equal to Z[k]. 


            // For example, str = "ababab", i = 3, R = 5 


            // and L = 2 


            if (Z[k] < R - i + 1) 


                Z[i] = Z[k]; 


  


            // For example str = "aaaaaa" and i = 2, R is 5, 


            // L is 0 


            else { 


                // else start from R and check manually 


                L = i; 


                while (R < n && str.charAt(R - L) == str.charAt(R)) 


                    R++; 


                Z[i] = R - L; 


                R--; 


            } 


        } 


    } 


} 


  


// Function to return the similarity sum 


static int sumSimilarities(String s, int n) 


{ 


    int Z[] = new int[n] ; 


  


    // Compute the Z-array for the given string 


    getZarr(s, n, Z); 


  


    int total = n; 


  


    // Summation of the Z-values 


    for (int i = 1; i < n; i++) 


        total += Z[i]; 


  


    return total; 


} 


  


// Driver code 


public static void main(String []args) 


{ 


    String s = "ababa"; 


    int n = s.length(); 


  


    System.out.println(sumSimilarities(s, n)); 


} 


// This code is contributed by Ryuga 


} 



















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Python3





# Python3 implmentation of the approach

def getZarr(s, n, Z):

	L, R, k = 0, 0, 0


	# [L, R] make a window which matches

	# with prefix of s

	for i in range(n):

		# if i>R nothing matches so we will

		# calculate Z[i] using naive way.

		if i > R:

			L, R = i, i


			”’

			R-L = 0 in starting, so it will start

			checking from 0’th index. For example,

			for “ababab” and i = 1, the value of R

			remains 0 and Z[i] becomes 0. For string

			“aaaaaa” and i = 1, Z[i] and R become 5

			”’

			while R < n and s[R - L] == s[R]:
				R += 1
			Z[i] = R - L
			R -= 1
		else:
		    
			# k = i-L so k corresponds to number 
			# which matches in [L, R] interval.
			k = i - L
			
			# if Z[k] is less than remaining interval 
			# then Z[i] will be equal to Z[k]. 
			# For example, str = "ababab", i = 3, R = 5 
			# and L = 2 
			if Z[k] < R - i + 1:
				Z[i] = Z[k]
			else:
				L = i
				while R < n and s[R - L] == s[R]:
					R += 1
				Z[i] = R - L
				R -= 1
				
def sumSimilarities(s, n):
	Z = [0 for i in range(n)]
	
	# Compute the Z-array for the 
	# given string
	getZarr(s, n, Z)
	
	total = n
	
	# summation of the Z-values
	for i in range(n):
		total += Z[i]
	return total
	
# Driver Code
s = "ababa"

n = len(s)

print(sumSimilarities(s, n))

# This code is contributed 
# by Mohit kumar 29



[tabby title="C#"]


                











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//C# implementation of the above approach 


using System; 


  


public class GFG{ 


    // Function to calculate the Z-array for the given string 


static void getZarr(string str, int n, int []Z) 


{ 


    int L, R, k; 


  


    // [L, R] make a window which matches with prefix of s 


    L = R = 0; 


    for (int i = 1; i < n; ++i) { 


  


        // if i>R nothing matches so we will calculate. 


        // Z[i] using naive way. 


        if (i > R) { 


            L = R = i; 


  


            // R-L = 0 in starting, so it will start 


            // checking from 0'th index. For example, 


            // for "ababab" and i = 1, the value of R 


            // remains 0 and Z[i] becomes 0. For string 


            // "aaaaaa" and i = 1, Z[i] and R become 5 


            while (R < n && str[R - L] == str[R]) 


                R++; 


            Z[i] = R - L; 


            R--; 


        } 


        else { 


  


            // k = i-L so k corresponds to number which 


            // matches in [L, R] interval. 


            k = i - L; 


  


            // if Z[k] is less than remaining interval 


            // then Z[i] will be equal to Z[k]. 


            // For example, str = "ababab", i = 3, R = 5 


            // and L = 2 


            if (Z[k] < R - i + 1) 


                Z[i] = Z[k]; 


  


            // For example str = "aaaaaa" and i = 2, R is 5, 


            // L is 0 


            else { 


                // else start from R and check manually 


                L = i; 


                while (R < n && str[R - L] == str[R]) 


                    R++; 


                Z[i] = R - L; 


                R--; 


            } 


        } 


    } 


} 


  


// Function to return the similarity sum 


static int sumSimilarities(string s, int n) 


{ 


    int []Z = new int[n] ; 


  


    // Compute the Z-array for the given string 


    getZarr(s, n, Z); 


  


    int total = n; 


  


    // Summation of the Z-values 


    for (int i = 1; i < n; i++) 


        total += Z[i]; 


  


    return total; 


} 


  


// Driver code 


    static public void Main (){ 


          


    string s = "ababa"; 


    int n = s.Length; 


  


    Console.WriteLine(sumSimilarities(s, n)); 


} 


// This code is contributed by ajit. 


} 



















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Output:


9






          
          
          
          

            

    

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