# 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

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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 ``#include ``#include ``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

 ``
Output:
`9`

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

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