String which when repeated exactly K times gives a permutation of S
Last Updated :
18 Sep, 2023
Given an integer K and a string str of lowercase English characters, the task is to find a string s such that when s is repeated exactly K times, it gives a permutation of S. If no such string exists, print -1.
Examples:
Input: str = "aabb", k = 2
Output: ab
"ab" when repeated 2 times gives "abab" which is a permutation of "aabb"
Input: str = "aabb", k = 3
Output: -1
Approach 1 : Using Map and Sorting
In this approach, we will make use of a map to store the frequency of each character in the string.
We will then sort the map based on the frequency of the characters in decreasing order. Finally, we
will iterate through the map and check if the frequency of each character is divisible by k. If yes,
then we will append the character to the answer string k times. If any character's frequency is not
divisible by k, we return -1.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
// Function to return a string which when repeated
// exactly k times gives a permutation of s
string K_String(string s, int k)
{
// size of string
int n = s.size();
// map to store frequency of each character
map<char, int> freq;
// get frequency of each character
for (int i = 0; i < n; i++)
freq[s[i]]++;
// to store final answer
string str = "";
// sort the map based on frequency in decreasing order
vector<pair<char, int> > sorted_freq(freq.begin(),
freq.end());
sort(sorted_freq.begin(), sorted_freq.end(),
[](const pair<char, int>& a,
const pair<char, int>& b) {
return a.second > b.second;
});
// iterate through the sorted map
for (auto it : sorted_freq) {
char c = it.first;
int f = it.second;
// check if frequency is divisible by k
if (f % k == 0) {
int x = f / k;
// add to answer
while (x--) {
str += c;
}
}
// if frequency is not divisible by k
else {
return "-1";
}
}
return str;
}
// Driver code
int main()
{
string s = "aabb";
int k = 2;
// function call
cout << K_String(s, k);
return 0;
}
import java.util.*;
public class Main {
// Function to return a string which when repeated
// exactly k times gives a permutation of s
public static String K_String(String s, int k)
{
// size of string
int n = s.length();
// map to store frequency of each character
Map<Character, Integer> freq = new HashMap<>();
// get frequency of each character
for (int i = 0; i < n; i++) {
char c = s.charAt(i);
freq.put(c, freq.getOrDefault(c, 0) + 1);
}
// to store final answer
StringBuilder str = new StringBuilder();
// sort the map based on frequency in decreasing
// order
List<Map.Entry<Character, Integer> > sorted_freq
= new ArrayList<>(freq.entrySet());
Collections.sort(
sorted_freq,
new Comparator<
Map.Entry<Character, Integer> >() {
@Override
public int compare(
Map.Entry<Character, Integer> a,
Map.Entry<Character, Integer> b)
{
return b.getValue().compareTo(
a.getValue());
}
});
// iterate through the sorted map
for (Map.Entry<Character, Integer> entry :
sorted_freq) {
char c = entry.getKey();
int f = entry.getValue();
// check if frequency is divisible by k
if (f % k == 0) {
int x = f / k;
// add to answer
while (x-- > 0) {
str.append(c);
}
}
// if frequency is not divisible by k
else {
return "-1";
}
}
return str.toString();
}
// Driver code
public static void main(String[] args)
{
String s = "aabb";
int k = 2;
// function call
System.out.println(K_String(s, k));
}
}
# Function to return a string which when repeated
# exactly k times gives a permutation of s
def K_String(s, k):
# size of string
n = len(s)
# map to store frequency of each character
freq = {}
# get frequency of each character
for i in range(n):
if s[i] in freq:
freq[s[i]] += 1
else:
freq[s[i]] = 1
# to store final answer
str = ""
# sort the map based on frequency in decreasing order
sorted_freq = sorted(freq.items(), key=lambda x: x[1], reverse=True)
# iterate through the sorted map
for it in sorted_freq:
c = it[0]
f = it[1]
# check if frequency is divisible by k
if f % k == 0:
x = f // k
# add to answer
while x > 0:
str += c
x -= 1
# if frequency is not divisible by k
else:
return "-1"
return str
# Driver code
s = "aabb"
k = 2
# function call
print(K_String(s, k))
using System;
using System.Collections.Generic;
using System.Linq;
public class MainClass {
public static string K_String(string s, int k)
{
int n = s.Length;
Dictionary<char, int> freq
= new Dictionary<char, int>();
// Initialize dictionary with all characters and
// frequency 0
foreach(char c in s) freq[c] = 0;
// Update the frequencies
foreach(char c in s) freq[c]++;
string str = "";
var sorted_freq
= freq.OrderByDescending(item => item.Value);
foreach(var item in sorted_freq)
{
char c = item.Key;
int f = item.Value;
if (f % k == 0) {
int x = f / k;
while (x > 0) {
str += c;
x--;
}
}
else {
return "-1";
}
}
return str;
}
public static void Main(string[] args)
{
string s = "aabb";
int k = 2;
Console.WriteLine(K_String(s, k));
}
}
// Function to return a string which when repeated
// exactly k times gives a permutation of s
function K_String(s, k) {
// size of string
let n = s.length;
// map to store frequency of each character
let freq = new Map();
// get frequency of each character
for (let i = 0; i < n; i++) {
if (freq.has(s[i])) {
freq.set(s[i], freq.get(s[i]) + 1);
} else {
freq.set(s[i], 1);
}
}
// to store final answer
let str = "";
// sort the map based on frequency in decreasing order
let sorted_freq = Array.from(freq.entries());
sorted_freq.sort((a, b) => b[1] - a[1]);
// iterate through the sorted map
for (let it of sorted_freq) {
let c = it[0];
let f = it[1];
// check if frequency is divisible by k
if (f % k === 0) {
let x = f / k;
// add to answer
while (x--) {
str += c;
}
}
// if frequency is not divisible by k
else {
return "-1";
}
}
return str;
}
let s = "aabb";
let k = 2;
// function call
console.log(K_String(s, k));
Output :
ab
Time Complexity: O(nlogn)
Auxiliary Space: O(n)
Approach 2: An efficient approach is to count the frequency of each character of the given string. If the frequency of any character is not divisible by k then the solution is not possible and print -1. Otherwise, add every character (frequency / k) times to the resultant string and print the generated string in the end.
Below is the implementation of the above approach:
C++
// C++ program to find a string which when repeated
// exactly k times gives a permutation
// of the given string
#include <bits/stdc++.h>
using namespace std;
// Function to return a string which when repeated
// exactly k times gives a permutation of s
string K_String(string s, int k)
{
// size of string
int n = s.size();
// to frequency of each character
int fre[26] = { 0 };
// get frequency of each character
for (int i = 0; i < n; i++)
fre[s[i] - 'a']++;
// to store final answer
string str = "";
for (int i = 0; i < 26; i++) {
// check if frequency is divisible by k
if (fre[i] % k == 0) {
int x = fre[i] / k;
// add to answer
while (x--) {
str += (char)(i + 'a');
}
}
// if frequency is not divisible by k
else {
return "-1";
}
}
return str;
}
// Driver code
int main()
{
string s = "aabb";
int k = 2;
// function call
cout << K_String(s, k);
return 0;
}
// Java program to find a string which when repeated
// exactly k times gives a permutation
// of the given string
class GfG {
// Function to return a string which when repeated
// exactly k times gives a permutation of s
static String K_String(String s, int k)
{
// size of string
int n = s.length();
// to frequency of each character
int fre[] = new int[26];
// get frequency of each character
for (int i = 0; i < n; i++)
fre[s.charAt(i) - 'a']++;
// to store final answer
String str = "";
for (int i = 0; i < 26; i++) {
// check if frequency is divisible by k
if (fre[i] % k == 0) {
int x = fre[i] / k;
// add to answer
while (x != 0) {
str += (char)(i + 'a');
x--;
}
}
// if frequency is not divisible by k
else {
return "-1";
}
}
return str;
}
// Driver code
public static void main(String[] args)
{
String s = "aabb";
int k = 2;
// function call
System.out.println(K_String(s, k));
}
}
# Python 3 program to find a string
# which when repeated exactly k times
# gives a permutation of the given string
# Function to return a string which
# when repeated exactly k times gives
# a permutation of s
def K_String(s, k):
# size of string
n = len(s)
# to frequency of each character
fre = [0] * 26
# get frequency of each character
for i in range(n):
fre[ord(s[i]) - ord('a')] += 1
# to store final answer
str = ""
for i in range(26):
# check if frequency is divisible by k
if (fre[i] % k == 0):
x = fre[i] // k
# add to answer
while (x):
str += chr(i + ord('a'))
x -= 1
# if frequency is not divisible by k
else:
return "-1"
return str
# Driver code
if __name__ == "__main__":
s = "aabb"
k = 2
# function call
print(K_String(s, k))
# This code is contributed
# by ChitraNayal
// C# program to find a string which
// when repeated exactly k times gives
// a permutation of the given string
using System;
class GFG {
// Function to return a string which
// when repeated exactly k times gives
// a permutation of s
static String K_String(String s, int k)
{
// size of string
int n = s.Length;
// to frequency of each character
int[] fre = new int[26];
// get frequency of each character
for (int i = 0; i < n; i++)
fre[s[i] - 'a']++;
// to store final answer
String str = "";
for (int i = 0; i < 26; i++) {
// check if frequency is
// divisible by k
if (fre[i] % k == 0) {
int x = fre[i] / k;
// add to answer
while (x != 0) {
str += (char)(i + 'a');
x--;
}
}
// if frequency is not divisible by k
else {
return "-1";
}
}
return str;
}
// Driver code
public static void Main(String[] args)
{
String s = "aabb";
int k = 2;
// function call
Console.WriteLine(K_String(s, k));
}
}
// This code is contributed by Arnab Kundu
<script>
// Javascript program to find a string which when repeated
// exactly k times gives a permutation
// of the given string
// Function to return a string which when repeated
// exactly k times gives a permutation of s
function K_String(s,k)
{
// size of string
let n = s.length;
// to frequency of each character
let fre = new Array(26);
for(let i=0;i<fre.length;i++)
{
fre[i]=0;
}
// get frequency of each character
for (let i = 0; i < n; i++)
fre[s[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;
// to store final answer
let str = "";
for (let i = 0; i < 26; i++) {
// check if frequency is divisible by k
if (fre[i] % k == 0) {
let x = Math.floor(fre[i] / k);
// add to answer
while (x != 0) {
str += String.fromCharCode(i + 'a'.charCodeAt(0));
x--;
}
}
// if frequency is not divisible by k
else {
return "-1";
}
}
return str;
}
// Driver code
let s = "aabb";
let k = 2;
// function call
document.write(K_String(s, k));
//This code is contributed by avanitrachhadiya2155
</script>
<?php
// PHP program to find a string which
// when repeated exactly k times gives
// a permutation of the given string
// Function to return a string which
// when repeated exactly k times gives
// a permutation of s
function K_String($s, $k)
{
// size of string
$n = strlen($s);
// to frequency of each character
$fre = $array = array_fill(0, 26, 0);
// get frequency of each character
for ($i = 0; $i < $n; $i++)
$fre[ord($s[$i]) - ord('a')]++;
// to store final answer
$str = "";
for ($i = 0; $i < 26; $i++)
{
// check if frequency is divisible by k
if ($fre[$i] % $k == 0)
{
$x = $fre[$i] / $k;
// add to answer
while ($x--)
{
$str .= chr($i + ord('a'));
}
}
// if frequency is not divisible by k
else
{
return "-1";
}
}
return $str;
}
// Driver code
$s = "aabb";
$k = 2;
// function call
echo K_String($s, $k);
// This code is contributed by Ryuga
?>
Time Complexity: O(n)
Space Complexity: O(1)
Similar Reads
Basics & Prerequisites
Data Structures
Array Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
3 min read
String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut
2 min read
Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The
2 min read
Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Here’s the comparison of Linked List vs Arrays Linked List:
2 min read
Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first
2 min read
Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems
2 min read
Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most
4 min read
Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of
3 min read
Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
15+ min read
Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
Interview Preparation
Practice Problem