Find Lexicographically smallest String - Coding

Given a circular string s, find whether there exists a permutation of s which is a K-periodic circular string and if it exists then find the lexicographically smallest permutation of s which is a K-periodic circular string. Return an empty string if there does not exist a valid permutation of s which is a K-periodic circular string.

Note: You can rotate the string in any direction. A K-periodic circular string is a circular string that remains the same when it is rotated by K units.

Examples:

Input: s = “abba”, K = 2
Output: abab
Explanation: “abab” when rotated by 2 units remains the same.

Input: s = “abbbbbb”, K = 4
Output: -1
Explanation: No permutation of s can be a 4-periodic circular string.

Approach: This can be solved with the following idea:

By keeping in mind, for each of the character at ith index, i == (i + k) % n == (i + 2 * k) % n == (i + 3 * k) % n, … should be same.

Below are the steps involved:

Intialise a map mp
- Store frequency of each character in mp.
intialize a req by n/ gcd(n,k).
Iterate in map,
- For each frequency, we need to add that particular character in f upto frequency/req.
For req time we need to add f in ans, which will be our required string.

Below is the implementation of the code:

C++

// C++ code for the above approach:
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
 
// Function to find lexographically
// smallest string
string kPeriodic(string s, int k)
{
    map<char, int> mp;
 
    // Store the frequency of each element
    for (int i = 0; i < s.size(); i++)
        mp[s[i]]++;
 
    int n = s.size();
    int req = n / __gcd(n, k);
 
    // If the string cannot be formed
    for (auto t : mp)
        if (t.second % req)
            return "-1";
 
    string f = "";
 
    // Iterate in map
    for (auto t : mp) {
        int fre = t.second / req;
        while (fre--)
            f += t.first;
    }
 
    // String will be returned
    string ans = "";
    while (req--)
        ans += f;
 
    // Return the final string
    return ans;
}
 
// Driver code
int main()
{
 
    string s = "abababa";
    int k = 2;
 
    // Function call
    cout << kPeriodic(s, k);
    return 0;
}

Java

import java.util.HashMap;
import java.util.Map;
 
public class LexicographicallySmallestString {
    public static String kPeriodic(String s, int k) {
        Map<Character, Integer> mp = new HashMap<>();
 
        // Store the frequency of each element
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            mp.put(ch, mp.getOrDefault(ch, 0) + 1);
        }
 
        int n = s.length();
        int gcd = gcd(n, k);
        int req = n / gcd;
 
        // If the string cannot be formed
        for (Map.Entry<Character, Integer> entry : mp.entrySet()) {
            if (entry.getValue() % req != 0) {
                return "-1";
            }
        }
 
        StringBuilder f = new StringBuilder();
 
        // Iterate in map
        for (Map.Entry<Character, Integer> entry : mp.entrySet()) {
            int fre = entry.getValue() / req;
            for (int i = 0; i < fre; i++) {
                f.append(entry.getKey());
            }
        }
 
        // String will be returned
        StringBuilder ans = new StringBuilder();
        for (int i = 0; i < req; i++) {
            ans.append(f);
        }
 
        // Return the final string
        return ans.toString();
    }
 
    public static int gcd(int a, int b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }
 
    public static void main(String[] args) {
        String s = "abababa";
        int k = 2;
 
        // Function call
        System.out.println(kPeriodic(s, k));
    }
}

Python3

import math
 
# Function to find lexicographically
# smallest string
def k_periodic(s, k):
    mp = {}
 
    # Store the frequency of each element
    for i in range(len(s)):
        if s[i] in mp:
            mp[s[i]] += 1
        else:
            mp[s[i]] = 1
 
    n = len(s)
    req = n // math.gcd(n, k)
 
    # If the string cannot be formed
    for t in mp.items():
        if t[1] % req != 0:
            return "-1"
 
    f = ""
 
    # Iterate in map
    for t in mp.items():
        fre = t[1] // req
        f += t[0] * fre
 
    # String will be returned
    ans = ""
    for _ in range(req):
        ans += f
 
    # Return the final string
    return ans
 
# Driver code
if __name__ == "__main__":
    s = "abababa"
    k = 2
 
    # Function call
    print(k_periodic(s, k))

C#

using System;
using System.Collections.Generic;
 
class GFG
{
    // Function to find lexographically smallest string
    static string KPeriodic(string s, int k)
    {
        Dictionary<char, int> mp = new Dictionary<char, int>();
        // Store the frequency of the each element
        foreach (char c in s)
        {
            if (mp.ContainsKey(c))
                mp++;
            else
                mp = 1;
        }
        int n = s.Length;
        int req = n / GCD(n, k);
        // If the string cannot be formed
        foreach (var t in mp)
        {
            if (t.Value % req != 0)
                return "-1";
        }
        string f = "";
        // Iterate in the dictionary
        foreach (var t in mp)
        {
            int fre = t.Value / req;
            while (fre-- > 0)
                f += t.Key;
        }
        // String will be returned
        string ans = "";
        while (req-- > 0)
            ans += f;
        // Return the final string
        return ans;
    }
    // Function to find the Greatest Common Divisor (GCD)
    static int GCD(int a, int b)
    {
        while (b != 0)
        {
            int temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }
    // Driver code
    static void Main()
    {
        string s = "abababa";
        int k = 2;
        // Function call
        Console.WriteLine(KPeriodic(s, k));
    }
}

Javascript

function kPeriodic(s, k) {
    let mp = new Map();
 
    // Store the frequency of each element
    for (let i = 0; i < s.length; i++) {
        let ch = s.charAt(i);
        mp.set(ch, (mp.get(ch) || 0) + 1);
    }
 
    let n = s.length;
    let gcd = calculateGCD(n, k);
    let req = n / gcd;
 
    // If the string cannot be formed
    for (let [key, value] of mp.entries()) {
        if (value % req !== 0) {
            return "-1";
        }
    }
 
    let f = "";
 
    // Iterate in map
    for (let [key, value] of mp.entries()) {
        let fre = value / req;
        for (let i = 0; i < fre; i++) {
            f += key;
        }
    }
 
    // String will be returned
    let ans = "";
    for (let i = 0; i < req; i++) {
        ans += f;
    }
 
    // Return the final string
    return ans;
}
 
function calculateGCD(a, b) {
    if (b === 0) {
        return a;
    }
    return calculateGCD(b, a % b);
}
 
// Test
let s = "abababa";
let k = 2;
 
// Function call
console.log(kPeriodic(s, k));

Output

-1

Time Complexity: O(n*log n)
Auxilairy Space: O(n)

Reffered: https://www.geeksforgeeks.org

DSA

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Category:	Coding
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