Skip to content
Related Articles

Related Articles

Improve Article

Sum of similarities of string with all of its suffixes

  • Difficulty Level : Hard
  • Last Updated : 11 Aug, 2021

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:
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++




// 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;
}

Java




// 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
}

Python3




# Python3 implementation 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

C#




//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.
}

Javascript




<script>
 
    // Javascript implementation of
    // the above approach
     
    // Function to calculate the Z-array
    // for the given string
    function getZarr(str, n, Z)
    {
        let L, R, k;
 
        // [L, R] make a window which matches
        // with prefix of s
        L = R = 0;
        for (let 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
    function sumSimilarities(s, n)
    {
        let Z = new Array(n);
        Z.fill(0);
 
        // Compute the Z-array for the given string
        getZarr(s, n, Z);
 
        let total = n;
 
        // Summation of the Z-values
        for (let i = 1; i < n; i++)
            total += Z[i];
 
        return total;
    }
     
    let s = "ababa";
    let n = s.length;
   
    document.write(sumSimilarities(s, n));
     
</script>
Output: 



9

 

Time Complexity: ON) 
Auxiliary Space: O(N) 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :