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++
// 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
// 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#
//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