Palindrome Permutation
A string can form a palindrome if at most one character occurs an odd number of times in it, and all other characters occur an even number of times. Here’s a simple approach to determining whether a given string could form a palindrome:
 Create a Frequency Counter: Create a dictionary to count the frequency of each character in the given string.
 Check Frequencies: Traverse the frequency counter and count how many characters occur an odd number of times.
 Decide: If there is more than one character that occurs an odd number of times, return false. Otherwise, return true.
Here’s a code snippet to implement this algorithm:


Key Takeaways
 A string can form a palindrome if at most one character occurs an odd number of times, and all others occur an even number of times.
 The algorithm has a time complexity of (O(n)) and a space complexity of (O(c)), where (n) is the length of the string, and (c) is the number of unique characters in the string.
Here is the code for solving the problem:


This code first initializes a dictionary counts
to store the counts of each character in the string s
. Then, it iterates over the string, incrementing the count of each character in the dictionary.
Next, the code initializes a variable odd_count
to 0. The code then iterates over the dictionary counts
, checking if the count of each character is even. If the count of a character is odd, then the code increments odd_count
by 1.
Finally, the code returns True
if odd_count <= 1
and False
otherwise.
I thought you have to generate a permutation and check if it is a palindrome. Is that approach wrong? Why?
Your approach is not wrong. It is also a valid way to solve the problem. However, my approach is more efficient, as it does not require generating all possible permutations of the string.
My approach works by counting the number of characters that appear an odd number of times in the string. If the number of characters that appear an odd number of times is 0 or 1, then the string can be permuted to form a palindrome. If the number of characters that appear an odd number of times is greater than 1, then the string cannot be permuted to form a palindrome.
Your approach, on the other hand, would require generating all possible permutations of the string. This would be a much more computationally expensive approach.
In general, it is always a good idea to think about the most efficient way to solve a problem before you start coding. This will help you to write more efficient and elegant code.
Is my approach a brute force?
Yes, your approach is a brute force approach. A brute force approach is an algorithm that solves a problem by trying all possible solutions. In your case, you would try all possible permutations of the string and check if each permutation is a palindrome.
Brute force approaches are often not very efficient, as they can take a long time to run. However, they are often simple to implement and can be used to solve a wide variety of problems.
In this case, there is a more efficient way to solve the problem, as I have shown. However, if you are just starting out with programming, it is a good idea to try to solve problems using brute force first. This will help you to understand the problem better and to develop your problemsolving skills.
As you become more experienced, you will learn to identify more efficient ways to solve problems.
How to recognize the inefficieny in this brute force approach for this problem?
The inefficiency of the brute force approach for this problem can be recognized by considering the following:
 The number of possible permutations of a string of length $n$ is $n!$, which is factorial of $n$. This means that the brute force approach would have to try $n!$ permutations in the worst case.
 The number of possible permutations of a string of length $n$ grows exponentially with $n$. This means that the brute force approach would become increasingly inefficient as the length of the string increases.
 There is a more efficient way to solve the problem, as I have shown. This means that the brute force approach is not the most efficient way to solve the problem.
To recognize the inefficiency in this brute force approach, you can consider the following questions:
 How many possible permutations are there of a string of a given length?
 How does the number of possible permutations grow with the length of the string?
 Is there a more efficient way to solve the problem?
If you can answer these questions, then you will be able to recognize the inefficiency of the brute force approach for this problem.
Here are some additional tips for recognizing the inefficiency of brute force approaches:
 Consider the time complexity of the brute force approach. Is the time complexity exponential? If so, then the brute force approach is likely to be inefficient.
 Consider the space complexity of the brute force approach. Does the brute force approach require a lot of memory? If so, then the brute force approach is likely to be inefficient.
 Look for patterns in the brute force approach. Can you identify any patterns that could be exploited to improve the efficiency of the approach?
If you can follow these tips, then you will be better able to recognize the inefficiency of brute force approaches and to identify more efficient ways to solve problems.
The first coding construct in this solution is a frequency counter. Am I right?
Yes, you are right. The first coding construct in the solution is a frequency counter. A frequency counter is a data structure that stores the number of times each element appears in a sequence. In this case, the sequence is the string s
. The frequency counter is used to count the number of times each character appears in the string.
The frequency counter is implemented using a dictionary. The dictionary is initialized with a default value of 0. As the code iterates over the string s
, it increments the count of each character in the dictionary.
The frequency counter is used to solve the problem by counting the number of characters that appear an odd number of times in the string. If the number of characters that appear an odd number of times is 0 or 1, then the string can be permuted to form a palindrome. If the number of characters that appear an odd number of times is greater than 1, then the string cannot be permuted to form a palindrome.
The frequency counter is a simple but effective way to solve this problem. It is also a versatile data structure that can be used to solve a variety of problems.
What other problems on leetcode can I solve using the frequency counter as the basic building block?
Here are some other problems on LeetCode that you can solve using the frequency counter as the basic building block:
Valid Anagram
 Given two strings, determine if they are anagrams. An anagram is a word or phrase formed by rearranging the letters of another word or phrase.
Counting Occurrences of Characters
 Given a string, count the number of occurrences of each character in the string.
Find All Anagrams in a String
 Given a string and a pattern, find all occurrences of the pattern in the string.
Minimum Unique Substring
 Given a string, find the minimum substring that contains all the unique characters in the string.
Maximum Frequent Character
 Given a string, find the character that appears the most number of times in the string.
These are just a few examples of problems that you can solve using the frequency counter. There are many other problems that can be solved using this technique.
The frequency counter is a powerful tool that can be used to solve a wide variety of problems. If you are familiar with how to use it, you will be able to solve many problems more easily.
What is the key insight that the solution is based upon?
The key insight that the solution is based upon is that a palindrome is a string that reads the same forward and backward. This means that the frequency of each character in a palindrome must be even. If the frequency of a character is odd, then the string cannot be a palindrome.
The solution works by counting the frequency of each character in the string. If the number of characters with odd frequencies is 0 or 1, then the string can be permuted to form a palindrome. If the number of characters with odd frequencies is greater than 1, then the string cannot be permuted to form a palindrome.
This insight is simple but powerful. It allows us to solve the problem in a relatively efficient way.
Here is an example of how the insight works. Consider the string "racecar"
. The frequency of each character in the string is as follows:
r
: 2a
: 2c
: 2
The number of characters with odd frequencies is 0, so the string can be permuted to form a palindrome. This is because the string "racecar"
is already a palindrome.
Another example is the string "code"
. The frequency of each character in the string is as follows:
c
: 1o
: 1d
: 1
The number of characters with odd frequencies is 1, so the string cannot be permuted to form a palindrome. This is because the string "code"
cannot be rearranged to form a palindrome.
As you can see, the insight that the solution is based upon is simple but powerful. It allows us to solve the problem in a relatively efficient way.