Sum of similarities of string with all of its suffixes - GeeksforGeeks

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: 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++
Java
Python3
C#
Javascript
// C++ implementation of the approach#include <iostream>#include <string>#include <vector>using namespace std; // Function to calculate the Z-array for the given stringvoid 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 sumint 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 codeint main(){    string s = "ababa";    int n = s.length();     cout << sumSimilarities(s, n);    return 0;}
// Java implementation of the above approach public class GFG{ // Function to calculate the Z-array for the given stringstatic 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 sumstatic 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 codepublic 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 implementation of the approachdef 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 Codes = "ababa" n = len(s) print(sumSimilarities(s, n)) # This code is contributed# by Mohit kumar 29
//C# implementation of the above approachusing System; public class GFG{    // Function to calculate the Z-array for the given stringstatic 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 sumstatic 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.}
<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>
9
 
Time Complexity: ON) Auxiliary Space: O(N) 
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