1177.can-make-palindrome-from-substring-cn

题目地址

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

题目描述

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

题中给出的字符串只包含小写字母（26个），让你判断给出的子串通过可重新排列和最多替换K个字母后，该子串是否可以组成回文字符串。

本题中，判断子串是否回文，有两个条件，

• 可重新排列，意思是只要子串中字母能组成回文即可不需要排好序的。 如：abb 可以组成回文字符串 bab。本身不是回文，但可以重新排列成回文字符串。

• 可最多替换K个字母，意思是本身或者通过从排列不能组成回文字符串，可以替换其中K个字母，也可以是回文字符串。 如：abcd, k = 2，可以替换cd -> ba, 组成abba。 或者是ab - dc, 组成dccd。 但如果abcd, k = 1，无论如何替换，都不可能组成回文字符串。

通过分析，我们可以发现规律，只要计算出子串中有多少个数为奇数的字母 oddCounts，如果oddCounts/2 >= K, 则我们可通过替换将子串变成回文。 那么这题就变为如何快速的统计出区间出现次数为奇数的字母个数，区间个数，我们可以用前缀和个数统计。

1. 为了快速统计区间个数，用前缀和数组prefixCount[i][j]记录前i个字母中，字母j出现的个数。

2. 统计区间[start, end]内出现的次数为奇数的字母个数 oddCounts += prefixCount[end+1][currChar] - prefixcount[start][currChar] 。

3. 如果oddCounts/2 >= K， 则可以通过替换将子串变成回文。

如下图构建的前缀和数组prefixCount[i][j], 由于这里字符串只包含26个小写字母，所以近似常数时间复杂度。

1177 can make palindrome

复杂度分析

• 时间复杂度: O(n + m * 26) - n is s length, m is queries length

• 空间复杂度: O(n * 26) - n is s length, n * 26 prefixcount matrix

Note, 可以用HashMap，这样不用每次都扫描26个字母，只要扫描出现过的字母即可。

判断字符串是否是回文：

回文字符串的特点：正读和反渎都是一样。如madam，noon, 12321， aaaa 等。

判断回文方法：

1. 第一种解法：直接reverse字符串，然后判断两字符串是否相同。return s == reverse(s)

2. 第二种解法：前后两个指针（left， right），left 从0开始，right从字符串最后一个字符开始，满足条件（left < right) 比较 s[left] 与s[right]是否相同,

   • 相同，则继续下一个，left往右移，left++, right往左移，right--

   • 不同，则直接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;
  }
}

关键点分析

• 前缀和数组计算前i个字母，j字母出现的次数

• 每一次query，扫描计算出次数为奇数的字母的个数，与k比较

代码 (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;
      }
      res.add(count / 2 <= k);
    }
    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