Count of distinct permutation of a String obtained by swapping only unequal characters
Last Updated :
12 Nov, 2023
Given a string find the number of unique permutations that can be obtained by swapping two indices such that the elements at these indices are distinct.
NOTE: Swapping is always performed in the original string.
Examples:
Input: str = "sstt"
Output: 5
Explanation:
Swap str[0] with str[2], string obtained "tsst" which is valid (str[0]!=str[2])
Swap str[0] with str[3]. string obtained "tsts"
Swap str[1] with str[2], string obtained "stst"
Swap str[1] with str[3], string obtained "stts"
Hence total 5 strings can be obtained including the given string ("sstt")
Input: str = "abcd"
Output: 7
Naive Approach:
- Create a set to store unique strings
- Loop through each pair of indices in the given string
- Check if the elements at the pair of indices are distinct
- If the elements are distinct, swap the elements at those indices and add the resulting string to the set
- Swap the elements back to their original positions
- Add 1 to count the original string
- Return the number of unique permutations, including the original string.
Below is the implementation of the above approach:
C++
#include <iostream>
#include <set>
#include <string>
using namespace std;
int countUniquePermutations(string str)
{
// create a set to store unique strings
set<string> uniqueStrings;
// loop through each pair of indices
// in the given string
for (int i = 0; i < str.length(); i++) {
for (int j = i + 1; j < str.length(); j++) {
// check if the elements at the
// pair of indices are distinct
if (str[i] != str[j]) {
swap(str[i], str[j]);
uniqueStrings.insert(str);
swap(str[i], str[j]);
}
}
}
// add 1 to count the original string
// and return number of unique permutations
return uniqueStrings.size() + 1;
}
int main()
{
string str = "sstt";
cout << countUniquePermutations(str);
return 0;
}
import java.util.HashSet;
import java.util.Set;
public class UniquePermutations {
public static int countUniquePermutations(String str)
{
// Create a set to store unique strings
Set<String> uniqueStrings = new HashSet<>();
// Loop through each pair of indices in the given
// string
for (int i = 0; i < str.length(); i++) {
for (int j = i + 1; j < str.length(); j++) {
// Check if the elements at the pair of
// indices are distinct
if (str.charAt(i) != str.charAt(j)) {
// Swap the characters at the indices
char[] strArray = str.toCharArray();
char temp = strArray[i];
strArray[i] = strArray[j];
strArray[j] = temp;
String swappedStr
= new String(strArray);
// Add the swapped string to the set
uniqueStrings.add(swappedStr);
// Swap back to the original string
temp = strArray[i];
strArray[i] = strArray[j];
strArray[j] = temp;
}
}
}
// Add 1 to count the original string and return the
// number of unique permutations
return uniqueStrings.size() + 1;
}
public static void main(String[] args)
{
String str = "sstt";
System.out.println(countUniquePermutations(str));
}
}
def count_unique_permutations(s):
# Create a set to store unique strings
unique_strings = set()
# Loop through each pair of indices in the given string
for i in range(len(s)):
for j in range(i + 1, len(s)):
# Check if the elements at the pair of indices are distinct
if s[i] != s[j]:
# Swap the elements at the indices to generate a new string
# and add it to the set
s_list = list(s)
s_list[i], s_list[j] = s_list[j], s_list[i]
unique_strings.add("".join(s_list))
# Add 1 to count the original string and return the number of unique permutations
return len(unique_strings) + 1
# Driver Code
if __name__ == "__main__":
input_str = "sstt"
print(count_unique_permutations(input_str))
using System;
using System.Collections.Generic;
class Program
{
static int CountUniquePermutations(string str)
{
// Create a HashSet to store unique strings
HashSet<string> uniqueStrings = new HashSet<string>();
// Loop through each pair of indices in the given string
for (int i = 0; i < str.Length; i++)
{
for (int j = i + 1; j < str.Length; j++)
{
// Check if the elements at the pair of indices are distinct
if (str[i] != str[j])
{
char[] chars = str.ToCharArray();
// Swap the elements at indices i and j
char temp = chars[i];
chars[i] = chars[j];
chars[j] = temp;
string newString = new string(chars);
// Insert the unique string into the HashSet
uniqueStrings.Add(newString);
// Swap back to restore the original string
temp = chars[i];
chars[i] = chars[j];
chars[j] = temp;
}
}
}
// Add 1 to count the original string and return the number of unique permutations
return uniqueStrings.Count + 1;
}
static void Main()
{
string str = "sstt";
Console.WriteLine(CountUniquePermutations(str));
}
}
function countUniquePermutations(str) {
// Create a Set to store unique strings
let uniqueStrings = new Set();
// Loop through each pair of indices in the given string
for (let i = 0; i < str.length; i++) {
for (let j = i + 1; j < str.length; j++) {
// Check if the elements at the pair of indices are distinct
if (str[i] !== str[j]) {
// Swap the characters at indices i and j
let strArray = str.split('');
[strArray[i], strArray[j]] = [strArray[j], strArray[i]];
let modifiedStr = strArray.join('');
uniqueStrings.add(modifiedStr);
// Swap the characters back to their original positions
[strArray[i], strArray[j]] = [strArray[j], strArray[i]];
}
}
}
// Add 1 to count the original string and return the number of unique permutations
return uniqueStrings.size + 1;
}
let str = "sstt";
console.log(countUniquePermutations(str));
Time Complexity: The time complexity of this algorithm is O(n2 * log n), where n is the length of the input string. This is because we have two nested loops, and within each loop, we perform a constant-time set insertion operation, which has a time complexity of O(log n) on average. Therefore, the overall time complexity is O(n^2 * log n).
Auxiliary Space: The space complexity of this algorithm is also O(n2). This is because we create a set to store the unique strings, and the size of the set can be as large as n^2 in the worst case, if all possible string permutations are unique. Additionally, we store the input string itself, which has a size of n. Therefore, the overall space complexity is O(n^2 + n), which simplifies to O(n^2).
Efficient Approach: The problem can be solved using HashMap by the following steps:
- Create a hashmap and store the frequency of every character of the given string.
- Create a variable count, to store the total number of characters in the given string, i.e. count=str.length() and a variable ans to store the number of unique permutations possible and initialize ans=0.
- Traverse the string and for each character:
- Find the number of different characters present in the right of the current index. This can be done by subtracting the frequency of that character by the total count.
- Now add this calculated value to ans, as this is the number of characters with which the current character can be swapped to create a unique permutation.
- Now, Reduce the frequency of the current character and count by 1, so that it cannot interfere with the calculations of the same elements present to the right of it.
- Return ans+1, because the given string is also a unique permutation.
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate total
// number of valid permutations
int validPermutations(string str)
{
unordered_map<char, int> m;
// Creating count which is equal to the
// Total number of characters present and
// ans that will store the number of unique
// permutations
int count = str.length(), ans = 0;
// Storing frequency of each character
// present in the string
for (int i = 0; i < str.length(); i++) {
m[str[i]]++;
}
for (int i = 0; i < str.length(); i++) {
// Adding count of characters by excluding
// characters equal to current char
ans += count - m[str[i]];
// Reduce the frequency of the current character
// and count by 1, so that it cannot interfere
// with the calculations of the same elements
// present to the right of it.
m[str[i]]--;
count--;
}
// Return ans+1 (Because the given string
// is also a unique permutation)
return ans + 1;
}
// Driver Code
int main()
{
string str = "sstt";
cout << validPermutations(str);
return 0;
}
// Java program for the above approach
// Importing HashMap class
import java.util.HashMap;
class GFG {
// Function to calculate total
// number of valid permutations
static int validPermutations(String str)
{
HashMap<Character, Integer> m
= new HashMap<Character, Integer>();
// Creating count which is equal to the
// Total number of characters present and
// ans that will store the number of unique
// permutations
int count = str.length(), ans = 0;
// Storing frequency of each character
// present in the string
for (int i = 0; i < str.length(); i++) {
m.put(str.charAt(i),
m.getOrDefault(str.charAt(i), 0) + 1);
}
for (int i = 0; i < str.length(); i++) {
// Adding count of characters by excluding
// characters equal to current char
ans += count - m.get(str.charAt(i));
// Reduce the frequency of the current character
// and count by 1, so that it cannot interfere
// with the calculations of the same elements
// present to the right of it.
m.put(str.charAt(i), m.get(str.charAt(i)) - 1);
count--;
}
// Return ans+1 (Because the given string
// is also a unique permutation)
return ans + 1;
}
public static void main(String[] args)
{
String str = "sstt";
System.out.println(validPermutations(str));
}
}
// This code is contributed by rajsanghavi9.
# Python 3 program for the above approach
# Function to calculate total
# number of valid permutations
def validPermutations(str):
m = {}
# Creating count which is equal to the
# Total number of characters present and
# ans that will store the number of unique
# permutations
count = len(str)
ans = 0
# Storing frequency of each character
# present in the string
for i in range(len(str)):
if(str[i] in m):
m[str[i]] += 1
else:
m[str[i]] = 1
for i in range(len(str)):
# Adding count of characters by excluding
# characters equal to current char
ans += count - m[str[i]]
# Reduce the frequency of the current character
# and count by 1, so that it cannot interfere
# with the calculations of the same elements
# present to the right of it.
m[str[i]] -= 1
count -= 1
# Return ans+1 (Because the given string
# is also a unique permutation)
return ans + 1
# Driver Code
if __name__ == '__main__':
str = "sstt"
print(validPermutations(str))
# This code is contributed by SURENDRA_GANGWAR.
// C# program for the above approach
// Importing Dictionary class
using System;
using System.Collections.Generic;
public class GFG {
// Function to calculate total
// number of valid permutations
static int validPermutations(String str)
{
Dictionary<char, int> m
= new Dictionary<char, int>();
// Creating count which is equal to the
// Total number of characters present and
// ans that will store the number of unique
// permutations
int count = str.Length, ans = 0;
// Storing frequency of each character
// present in the string
for (int i = 0; i < str.Length; i++) {
if(m.ContainsKey(str[i]))
m[str[i]]=m[str[i]]+1;
else
m.Add(str[i], 1);
}
for (int i = 0; i < str.Length; i++) {
// Adding count of characters by excluding
// characters equal to current char
ans += count - m[str[i]];
// Reduce the frequency of the current character
// and count by 1, so that it cannot interfere
// with the calculations of the same elements
// present to the right of it.
if(m.ContainsKey(str[i]))
m[str[i]]=m[str[i]]-1;
count--;
}
// Return ans+1 (Because the given string
// is also a unique permutation)
return ans + 1;
}
public static void Main(String[] args)
{
String str = "sstt";
Console.WriteLine(validPermutations(str));
}
}
// This code is contributed by shikhasingrajput
<script>
// JavaScript program for the above approach
// Function to calculate total
// number of valid permutations
function validPermutations(str) {
let m = new Map();
// Creating count which is equal to the
// Total number of characters present and
// ans that will store the number of unique
// permutations
let count = str.length,
ans = 0;
// Storing frequency of each character
// present in the string
for (let i = 0; i < str.length; i++) {
if (m.has(str[i])) {
m.set(str[i], m.get(str[i]) + 1);
} else {
m.set(str[i], 1);
}
}
for (let i = 0; i < str.length; i++)
{
// Adding count of characters by excluding
// characters equal to current char
ans += count - m.get(str[i]);
// Reduce the frequency of the current character
// and count by 1, so that it cannot interfere
// with the calculations of the same elements
// present to the right of it.
m.set(str[i], m.get(str[i]) - 1);
count--;
}
// Return ans+1 (Because the given string
// is also a unique permutation)
return ans + 1;
}
// Driver Code
let str = "sstt";
document.write(validPermutations(str));
</script>
Time Complexity: O(n)
Auxiliary Space: O(n)
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