# 1177.can-make-palindrome-from-substring-en

## Problem

https://leetcode.com/problems/can-make-palindrome-from-substring/

## Problem Description

``````Given a string s, we make queries on substrings of s.

For each query queries[i] = [left, right, k], we may rearrange the substring s[left], ..., s[right], and then choose up to k of them to replace with any lowercase English letter.

If the substring is possible to be a palindrome string after the operations above, the result of the query is true. Otherwise, the result is false.

Return an array answer[], where answer[i] is the result of the i-th query queries[i].

Note that: Each letter is counted individually for replacement so if for example s[left..right] = "aaa", and k = 2, we can only replace two of the letters.  (Also, note that the initial string s is never modified by any query.)

Example :

Input: s = "abcda", queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]]
Output: [true,false,false,true,true]
Explanation:
queries[0] : substring = "d", is palidrome.
queries[1] : substring = "bc", is not palidrome.
queries[2] : substring = "abcd", is not palidrome after replacing only 1 character.
queries[3] : substring = "abcd", could be changed to "abba" which is palidrome. Also this can be changed to "baab" first rearrange it "bacd" then replace "cd" with "ab".
queries[4] : substring = "abcda", could be changed to "abcba" which is palidrome.

Constraints:

1 <= s.length, queries.length <= 10^5
0 <= queries[i][0] <= queries[i][1] < s.length
0 <= queries[i][2] <= s.length
s only contains lowercase English letters.``````

## 思路

In this problem, there are 2 conditions for checking whether substring is palindrome:

• Rearrange, very useful here. e.g：`abb` can be rearranged to palindrome string `bab`, itself maybe not palindrome, but after rearrange, can be palindrome.

• Can replace with at most K letters, meaning itself or after rearrange itself still cannot become palindrome, it may be palindrome after replacing K letters. e.g: `abcd, k = 2`，can replace`cd -> ba`, to be palindrome`abba`; or replace`ab -> dc`, be palindrome `dccd`

e.g: `abcd, k = 1`，then no matter how to replace, it cannot become palindrome.

After analysis, found there is some pattern to form palindrome by rearranging and replace K letters.

If we calculate the number of letters, and track the odd number letters, as `oddCounts`, if `oddCounts/2 >= K` then we can through replace K letters to form a new palindrome string, otherwise false.

Then this problem becomes to calculate the number of the letter which appears odd times in a substring, queries give us range, so range prefix sum is useful in this case (so that for each given range, we can get the number in `~O(1)` time).

Steps:

1. Define prefix sum matrix: `prefixCount[i][j]` keep track of in string `s`, in range `[0, i]` aka, in `s[0,i]`, the number of letter `j`.

2. Count the number of letters which appears odd number times (`oddCounts`) in range `[start, end]`. `oddCounts += prefixCount[end+1][currChar] - prefixcount[start][currChar]`

3. If `oddCounts/2 >= K`, then substring can become palindrome after rearrange and replace `<=K`letters.

4. If `oddCounts/2 < K`, then substring cannot become palindrome, `return false;`.

As below pic, create a prefix sum matrix, `prefixCount[i][j]`, since string only contains 26 lowercase letters, it is `O(1)` for loop 26 letters.

### Complexity Analysis

• Time Complexity: `O(n + m * 26) - n is s length, m is queries length`

• Space Complexity: `O(n * 26) - n is s length, n * 26 prefixcount matrix`

Note, here we can also use HashMap，so that we don't need to loop 26 letters each time, only loop the number of letters in substring.

Validate a String is Palindrome

Palindrome String: forward and backward are exactly the same, e.g: `madam，noon, 12321， aaaa` etc.

Validate Palindrome Solution:

1. Solution #1: Reverse string s, then check whether original string equals to reversed string. `return s == reverse(s)`

2. Solution #2: use left and right two pointers, `（left， right）``left` start from index `0`, `right` from last index，when `（left < right)`, compare

-`s[left]` == `s[right]`, then continue next, `left move right, left++`, `right move left, right--`

• `s[left]` != `s[right]`，then `return false`

``````class ValidPalindrome {
public boolean isPalindrome(String s) {
if (s == null || s.length() < 2) return true;
int left = 0;
int right = s.length() - 1;
while (left < right) {
if (s.charAt(left) != s.charAt(right)) return false;
left++;
right--;
}
return true;
}
}``````

## Key Points

• prefix sum matrix count index from 0 to i substring, the number of times that `j` letter appears

• For each query, count the number of letters which appears odd times (`oddCounts`) in range `[start, end]`, then compare `oddCounts` and `k`

## Code (`Java/Python3`)

Java code

``````public class CanMakePalindromeFromSubstring {
public List<Boolean> canMakePaliQueries(String s, int[][] queries) {
int n = s.length();
int[][] prefixCount = new int[n + 1][26];
// prefixCount[i][j] matrix, calculate number of letter from range [0,i] for j letter.
for (int i = 0; i < n; i++) {
prefixCount[i + 1] = prefixCount[i].clone();
prefixCount[i + 1][s.charAt(i) - 'a']++;
}
List<Boolean> res = new ArrayList<>();
for (int[] q : queries) {
int start = q[0];
int end = q[1];
int k = q[2];
int count = 0;
// count odd letters number from range [start, end]
for (int i = 0; i < 26; i++) {
count += (prefixCount[end + 1][i] - prefixCount[start][i]) % 2;
}
}
return res;
}
}``````

Python3 code (slow :( )

``````class Solution:
def canMakePaliQueries(self, s: str, queries: List[List[int]]) -> List[bool]:
prefix_sum = [[0] * 26]
for i, char in enumerate(s):
prefix_sum.append(prefix_sum[i][:])
prefix_sum[i + 1][ord(char) - ord('a')] += 1
res = [True] * len(queries)
for i, (start, end, k) in enumerate(queries):
odd_count = 0
for j in range(26):
odd_count += (prefix_sum[end + 1][j] - prefix_sum[start][j]) % 2
res[i] = (odd_count // 2 <= k)
return res``````

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