Given a string containing only lowercase English letters, determine the minimum length of a contiguous substring that can be removed such that the remaining characters can be rearranged to form a palindrome.
Hi Kuai Great question! This problem is a fun one because it combines string manipulation with palindrome properties, which are always interesting to think about. Here's how I'd approach it,
The goal is to remove the smallest possible contiguous substring so that the remaining string can be rearranged into a palindrome. For a string to be rearranged into a palindrome, there can be at most one character with an odd frequency (for odd-length strings). This is a key observation!
I’d start by checking if the entire string can already be rearranged into a palindrome. If it can, then no removal is needed, and the answer is 0. If not, the challenge becomes finding the smallest substring to remove that makes the rest of the string palindromic.
We can use a two-pointer technique here:
Start with two pointers: one at the beginning (left) and one at the end (right).
If the characters at both pointers are the same, great! Move both pointers inward.
If the characters are different, you have two options:
Remove the character at the left pointer and check if the remaining string can form a palindrome.
Or remove the character at the right pointer and check again.
If either of these works, we return 1 because we only need to remove one character. If neither works, then we might need to remove two characters, so we return 2.
Neat problem! I sense this solution could be made cleaner but here's what I came up with:
A string of lowercase letters can be rearranged into a palindrome if and only if there is at most one letter with an odd frequency. In other words, if we examine the frequency distribution of the 26 letters in the string, either all the frequencies are even or one of the frequencies is odd and the rest even.
Now, instead of thinking about the problem as removing a substring, think about constructing a new string using some prefix + some suffix of the string. We want this concatenation to meet the "at most one odd frequency" requirement. Since we only care about the parity of the frequencies, we can think of the frequency distribution of each prefix and suffix as a bit pattern of 26 bits where each bit represents the frequency of a letter (from a-z) modulo 2. Notice that for each prefix, a valid matching suffix to meet the "at most one odd frequency" requirement must have a bit pattern that differs by at most one bit from the bit pattern of the prefix. Using a hashmap and keeping track of the minimum "removed" substring window, we can solve the problem in O(n) time and O(n) space.
The algorithm is as follows:
1. Starting from the end of the string, construct a hashmap with the keys being bit patterns for each suffix (stored as int/string/tuple) and the values being a list of the starting indices for the suffixes that have that bit pattern. Since we start from the end, the lists of indices are in sorted decreasing order. Add an entry for the empty suffix with index len(string)
2. Store the bit pattern of the final suffix (i.e. the entire string) as a separate variable
3. Starting from the beginning of the string including the empty prefix (""), keep track of the current prefix's bit pattern. For each bit pattern that differs by at most one bit from the prefix's bit pattern, find the earliest valid index of a suffix with that bit pattern. This is as simple as looking at the end of the list in the hashmap entry for that bit pattern. Update the variable that stores the minimum length of the removed substring accordingly. After processing a prefix, calculate the corresponding suffix bit pattern for that prefix (i.e. the rest of the string) and remove its index from the hashmap. This is a constant time operation since we remove it from the end of the list.
4. Return the length of the minimum length substring
Hi Kuai, Really good question!
I am sharing the solution here
1) have a hashmap
2)for this hashmap,check for the odd
3)return the oddcount-1 if its more than 0 else return 0
function checkPalindrome(str){
let count=0
let map=new Map()
for(let i=0;i<str.length;i++)
{
map.set(str[i],(map.get(str[i])||0)+1)
}
let oddCount=0
for(let [key,value] of map)
{
if(value%2!=0)
{
oddCount++
}
}
return oddCount>0?oddCount-1:0
}
Thanks Sneha for sharing your solution! We need to remove a contiguous substring, not individual characters from anywhere in the string.
Hi Kuai Great question! This problem is a fun one because it combines string manipulation with palindrome properties, which are always interesting to think about. Here's how I'd approach it,
The goal is to remove the smallest possible contiguous substring so that the remaining string can be rearranged into a palindrome. For a string to be rearranged into a palindrome, there can be at most one character with an odd frequency (for odd-length strings). This is a key observation!
I’d start by checking if the entire string can already be rearranged into a palindrome. If it can, then no removal is needed, and the answer is 0. If not, the challenge becomes finding the smallest substring to remove that makes the rest of the string palindromic.
We can use a two-pointer technique here:
Start with two pointers: one at the beginning (left) and one at the end (right).
If the characters at both pointers are the same, great! Move both pointers inward.
If the characters are different, you have two options:
Remove the character at the left pointer and check if the remaining string can form a palindrome.
Or remove the character at the right pointer and check again.
If either of these works, we return 1 because we only need to remove one character. If neither works, then we might need to remove two characters, so we return 2.
Neat problem! I sense this solution could be made cleaner but here's what I came up with:
A string of lowercase letters can be rearranged into a palindrome if and only if there is at most one letter with an odd frequency. In other words, if we examine the frequency distribution of the 26 letters in the string, either all the frequencies are even or one of the frequencies is odd and the rest even.
Now, instead of thinking about the problem as removing a substring, think about constructing a new string using some prefix + some suffix of the string. We want this concatenation to meet the "at most one odd frequency" requirement. Since we only care about the parity of the frequencies, we can think of the frequency distribution of each prefix and suffix as a bit pattern of 26 bits where each bit represents the frequency of a letter (from a-z) modulo 2. Notice that for each prefix, a valid matching suffix to meet the "at most one odd frequency" requirement must have a bit pattern that differs by at most one bit from the bit pattern of the prefix. Using a hashmap and keeping track of the minimum "removed" substring window, we can solve the problem in O(n) time and O(n) space.
The algorithm is as follows:
1. Starting from the end of the string, construct a hashmap with the keys being bit patterns for each suffix (stored as int/string/tuple) and the values being a list of the starting indices for the suffixes that have that bit pattern. Since we start from the end, the lists of indices are in sorted decreasing order. Add an entry for the empty suffix with index len(string)
2. Store the bit pattern of the final suffix (i.e. the entire string) as a separate variable
3. Starting from the beginning of the string including the empty prefix (""), keep track of the current prefix's bit pattern. For each bit pattern that differs by at most one bit from the prefix's bit pattern, find the earliest valid index of a suffix with that bit pattern. This is as simple as looking at the end of the list in the hashmap entry for that bit pattern. Update the variable that stores the minimum length of the removed substring accordingly. After processing a prefix, calculate the corresponding suffix bit pattern for that prefix (i.e. the rest of the string) and remove its index from the hashmap. This is a constant time operation since we remove it from the end of the list.
4. Return the length of the minimum length substring