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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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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Improved By : Ryuga