# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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. ` `} `

Output:

```9
``` My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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