Algorithm


Problem Name: 1177. Can Make Palindrome from Substring

You are given a string s and array queries where queries[i] = [lefti, righti, ki]. We may rearrange the substring s[lefti...righti] for each query and then choose up to ki 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 a boolean array answer where answer[i] is the result of the ith query queries[i].

Note that each letter is counted individually for replacement, so if, for example s[lefti...righti] = "aaa", and ki = 2, we can only replace two of the letters. Also, note that no query modifies the initial string s.

 

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.

Example 2:

Input: s = "lyb", queries = [[0,1,0],[2,2,1]]
Output: [false,true]

 

Constraints:

  • 1 <= s.length, queries.length <= 105
  • 0 <= lefti <= righti < s.length
  • 0 <= ki <= s.length
  • s consists of lowercase English letters.
 

 

Code Examples

#1 Code Example with Java Programming

Code - Java Programming


class Solution {
  public List canMakePaliQueries(String s, int[][] queries) {
    int[][] count = new int[s.length() + 1][26];
    for (int i = 0; i  <  s.length(); i++) {
      count[i + 1] = count[i].clone();
      count[i + 1][s.charAt(i) - 'a']++;
    }
    List < Boolean> list = new ArrayList<>();
    for (int[] query : queries) {
      int sum = 0;
      for (int i = 0; i  <  26; i++) {
        sum += (count[query[1] + 1][i] - count[query[0]][i]) % 2;
      }
      list.add(sum / 2  < = query[2]);
    }
    return list;
  }
}
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Input

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s = "abcda", queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]]

Output

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[true,false,false,true,true]

#2 Code Example with Javascript Programming

Code - Javascript Programming


const canMakePaliQueries = function(s, queries) {
  const code = ch => ch.charCodeAt(0) - 'a'.charCodeAt(0)
  const preCount = [...s].reduce(
    (a, c) => {
      let nc = a[a.length - 1]
      nc ^= 1 << code(c) //NOT on one bit
      a.push(nc)
      return a
    },
    [0]
  )
  return queries.map(q => {
    let subCount = preCount[q[1] + 1] ^ preCount[q[0]]
    let oddChs = 0
    while (subCount > 0) {
      oddChs += subCount & 1
      subCount >>= 1
    }
    return Math.floor(oddChs / 2) <= q[2]
  })
}
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Input

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s = "abcda", queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]]

Output

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[true,false,false,true,true]

#3 Code Example with Python Programming

Code - Python Programming


class Solution:
    def canMakePaliQueries(self, s: str, queries: List[List[int]]) -> List[bool]:
        cnts = [{}]
        for i, c in enumerate(s):
            cnts.append(dict(cnts[-1]))
            cnts[-1][c] = cnts[-1].get(c, 0) + 1
        res = []
        for i, j, k in queries:
            res.append(sum((v - cnts[i].get(k, 0)) % 2 for k, v in cnts[j + 1].items()) - k * 2 <= 1)
        return res
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Input

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s = "lyb", queries = [[0,1,0],[2,2,1]]

Output

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[false,true]
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