Find the Suffix Array of given String with no repeating character

Last Updated : 24 Feb, 2022

Given a string str of size N, the task is to find the suffix array of the given string.

Note: A suffix array is a sorted array of all suffixes of a given string.

Examples:

Input: str = "prince"
Output: 4 5 2 3 0 1
Explanation: The suffixes are
0 prince 4 ce
1 rince Sort the suffixes 5 e
2 ince ----------------> 2 ince
3 nce alphabetically 3 nce
4 ce 0 prince
5 e 1 rince

Input: str = "abcd"
Output: 0 1 2 3

Approach: The methods of suffix array finding for any string are discussed here. In this article, the focus is on finding suffix array for strings with no repeating character. It is a simple implementation based problem. Follow the steps mentioned below to solve the problem:

Count occurrence of each character.
Find prefix sum of it.
Find start array by start[0] = 0, start[i+1] = prefix[i] for all i > 0 .
Find start array containing the index of the substring (suffix) with starting character of the respective column.

Follow the illustration below for better understanding.

Illustration:

Consider string "prince":
Given below is the suffix table

char a b c d e f g h i j k l m n o p q r s t u v w x y z
count 0 0 1 0 1 0 0 0 1 0 0 0 0 1 0 1 0 1 0 0 0 0 0 0 0 0
prefix 0 0 1 1 2 2 2 2 3 3 3 3 3 4 4 5 5 6 6 6 6 6 6 6 6 6
start 0 0 0 1 1 2 2 2 2 3 3 3 3 3 4 4 5 5 6 6 6 6 6 6 6 6

For char 'r' start value is 5 .
Implies substring (suffix) starting with char 'r' i.e. "rince" has rank 5 .
Rank is position in suffix array. ( 1 "rince" ) implies 5th position in suffix array ), refer first table.
Similarly, start value of char 'n' is 3 . Implies ( 3 "nce" ) 3rd position in suffix array .

Below is the implementation of the above approach

C++

// C++ code to implement above approach #include <bits/stdc++.h> using namespace std;  // Function to calculate the suffix array void suffixArray(string str, int N) {     // arr[] is array to count     // occurrence of each character     int arr[30] = { 0 };     for (int i = 0; i < N; i++) {         arr[str[i] - 'a']++;     }      // Finding prefix count of character     for (int i = 1; i < 30; i++) {         arr[i] = arr[i] + arr[i - 1];     }      int start[30];     start[0] = 0;     for (int i = 0; i < 29; i++) {         start[i + 1] = arr[i];     }      int ans[N] = { 0 };      // Iterating string in reverse order     for (int i = N - 1; i >= 0; i--) {          // Storing suffix array in ans[]         ans[start[str[i] - 'a']] = i;     }      for (int i = 0; i < N; i++)         cout << ans[i] << " "; }  // Driver code int main() {     string str = "prince";     int N = str.length();      suffixArray(str, N);     return 0; }

Java

// Java program for the above approach import java.io.*; import java.lang.*; import java.util.*;  class GFG {     // Function to calculate the suffix array   static void suffixArray(String str, int N)   {      // arr[] is array to count     // occurrence of each character     int arr[] = new int[30];     for (int i = 0; i < N; i++) {       arr[str.charAt(i) - 'a']++;     }      // Finding prefix count of character     for (int i = 1; i < 30; i++) {       arr[i] = arr[i] + arr[i - 1];     }      int start[] = new int[30];     start[0] = 0;     for (int i = 0; i < 29; i++) {       start[i + 1] = arr[i];     }      int ans[]  = new int[N];      // Iterating string in reverse order     for (int i = N - 1; i >= 0; i--) {        // Storing suffix array in ans[]       ans[start[str.charAt(i) - 'a']] = i;     }      for (int i = 0; i < N; i++)       System.out.print(ans[i] + " ");   }    // Driver code   public static void main (String[] args)   {     String str = "prince";     int N = str.length();      suffixArray(str, N);   } }  // This code is contributed by hrithikgarg03188

Python3

# Python code for the above approach  # Function to calculate the suffix array def suffixArray(str, N):        # arr[] is array to count     # occurrence of each character     arr = [0] * 30     for i in range(N):         arr[ord(str[i]) - ord('a')] += 1      # Finding prefix count of character     for i in range(1, 30):         arr[i] = arr[i] + arr[i - 1]      start = [0] * 30     start[0] = 0     for i in range(29):         start[i + 1] = arr[i]      ans = [0] * N      # Iterating string in reverse order     for i in range(N - 1, 0, -1):          # Storing suffix array in ans[]         ans[start[ord(str[i]) - ord('a')]] = i      for i in range(N):         print(ans[i], end=" ")  # Driver code str = "prince" N = len(str)  suffixArray(str, N)  # This code is contributed by gfgking

// C# code to implement above approach using System; class GFG {        // Function to calculate the suffix array     static void suffixArray(string str, int N)     {         // arr[] is array to count         // occurrence of each character         int[] arr = new int[30];         for (int i = 0; i < N; i++) {             arr[str[i] - 'a']++;         }          // Finding prefix count of character         for (int i = 1; i < 30; i++) {             arr[i] = arr[i] + arr[i - 1];         }          int[] start = new int[30];         start[0] = 0;         for (int i = 0; i < 29; i++) {             start[i + 1] = arr[i];         }          int[] ans = new int[N];          // Iterating string in reverse order         for (int i = N - 1; i >= 0; i--) {              // Storing suffix array in ans[]             ans[start[str[i] - 'a']] = i;         }          for (int i = 0; i < N; i++) {             Console.Write(ans[i]);             Console.Write(" ");         }     }      // Driver code     public static int Main()     {         string str = "prince";         int N = str.Length;          suffixArray(str, N);         return 0;     } }  // This code is contributed by Taranpreet

JavaScript

   <script>         // JavaScript code for the above approach            // Function to calculate the suffix array         function suffixArray(str, N) {             // arr[] is array to count             // occurrence of each character             let arr = new Array(30).fill(0);             for (let i = 0; i < N; i++) {                 arr[str[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;             }              // Finding prefix count of character             for (let i = 1; i < 30; i++) {                 arr[i] = arr[i] + arr[i - 1];             }              let start = new Array(30)             start[0] = 0;             for (let i = 0; i < 29; i++) {                 start[i + 1] = arr[i];             }              let ans = new Array(N).fill(0)              // Iterating string in reverse order             for (let i = N - 1; i >= 0; i--) {                  // Storing suffix array in ans[]                 ans[start[str[i].charCodeAt(0) - 'a'.charCodeAt(0)]] = i;             }              for (let i = 0; i < N; i++)                 document.write(ans[i] + " ")         }          // Driver code          let str = "prince";         let N = str.length;          suffixArray(str, N);            // This code is contributed by Potta Lokesh     </script>

Output

4 5 2 3 0 1

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

Count of all unique substrings with non-repeating characters

prince_patel27

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Find the Suffix Array of given String with no repeating character

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