Print all distinct circular strings of length M in lexicographical order

Given a string and an integer M, print all distinct circular strings of length M in lexicographical order.

Examples:

Input: str = “baaaa”, M = 3
Output: aaa aab aba baa
All possible circular substrings of length 3 are “baa” “aaa” “aaa” “aab” “aba”
Out of the 6, 4 are distinct, and the lexicographical order is aaa aab aba baa

Input: str = “saurav”, M = 3
Output: aura avsa ravs saur urav vsau
All possible circular substrings of length 4 are saur aura urav ravs avsa vsau.
All the substrings are distinct, the lexicographical order is aura avsa ravs saur urav vsau.



Approach: The substr function is used to solve the problem. Append the string to itself at first. Iterate over the length of the string to generate all possible substrings of length M. Set is used in C++ to store all the distinct substrings of length 4, set by default stores all its element in lexicographical order. Once all the strings are generated, print the elements in the set from the beginning.

Below is the implementation of the above approach:

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// C++ program to print all
// distinct circular strings
// of length M in lexicographical order
#include <bits/stdc++.h>
using namespace std;
  
// Function to print all the distinct substrings
// in lexicographical order
void printStrings(string s, int l, int m)
{
    // stores all the distinct substrings
    set<string> c;
  
    // Append the string to self
    s = s + s;
  
    // Iterate over the length to generate
    // all substrings of length m
    for (int i = 0; i < l; i++) {
  
        // insert the substring of length m
        // in the set
        c.insert(s.substr(i, m));
    }
  
    // prints all the distinct circular
    // substrings  of length m
    while (!c.empty()) {
  
        // Prints the substring
        cout << *c.begin() << " ";
  
        // erases the beginning element after
        // printing
        c.erase(c.begin());
    }
}
  
// Driver code
int main()
{
    string str = "saurav";
    int N = str.length();
    int M = 4;
  
    printStrings(str, N, M);
  
    return 0;
}

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Output:

aura avsa ravs saur urav vsau

Time Complexity: O(N*M), where N is the length of the string.



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