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2983. Palindrome Rearrangement Queries

中文文档

Description

You are given a 0-indexed string s having an even length n.

You are also given a 0-indexed 2D integer array, queries, where queries[i] = [ai, bi, ci, di].

For each query i, you are allowed to perform the following operations:

  • Rearrange the characters within the substring s[ai:bi], where 0 <= ai <= bi < n / 2.
  • Rearrange the characters within the substring s[ci:di], where n / 2 <= ci <= di < n.

For each query, your task is to determine whether it is possible to make s a palindrome by performing the operations.

Each query is answered independently of the others.

Return a 0-indexed array answer, where answer[i] == true if it is possible to make s a palindrome by performing operations specified by the ith query, and false otherwise.

  • A substring is a contiguous sequence of characters within a string.
  • s[x:y] represents the substring consisting of characters from the index x to index y in s, both inclusive.

 

Example 1:

Input: s = "abcabc", queries = [[1,1,3,5],[0,2,5,5]]
Output: [true,true]
Explanation: In this example, there are two queries:
In the first query:
- a0 = 1, b0 = 1, c0 = 3, d0 = 5.
- So, you are allowed to rearrange s[1:1] => abcabc and s[3:5] => abcabc.
- To make s a palindrome, s[3:5] can be rearranged to become => abccba.
- Now, s is a palindrome. So, answer[0] = true.
In the second query:
- a1 = 0, b1 = 2, c1 = 5, d1 = 5.
- So, you are allowed to rearrange s[0:2] => abcabc and s[5:5] => abcabc.
- To make s a palindrome, s[0:2] can be rearranged to become => cbaabc.
- Now, s is a palindrome. So, answer[1] = true.

Example 2:

Input: s = "abbcdecbba", queries = [[0,2,7,9]]
Output: [false]
Explanation: In this example, there is only one query.
a0 = 0, b0 = 2, c0 = 7, d0 = 9.
So, you are allowed to rearrange s[0:2] => abbcdecbba and s[7:9] => abbcdecbba.
It is not possible to make s a palindrome by rearranging these substrings because s[3:6] is not a palindrome.
So, answer[0] = false.

Example 3:

Input: s = "acbcab", queries = [[1,2,4,5]]
Output: [true]
Explanation: In this example, there is only one query.
a0 = 1, b0 = 2, c0 = 4, d0 = 5.
So, you are allowed to rearrange s[1:2] => acbcab and s[4:5] => acbcab.
To make s a palindrome s[1:2] can be rearranged to become abccab.
Then, s[4:5] can be rearranged to become abccba.
Now, s is a palindrome. So, answer[0] = true.

 

Constraints:

  • 2 <= n == s.length <= 105
  • 1 <= queries.length <= 105
  • queries[i].length == 4
  • ai == queries[i][0], bi == queries[i][1]
  • ci == queries[i][2], di == queries[i][3]
  • 0 <= ai <= bi < n / 2
  • n / 2 <= ci <= di < n
  • n is even.
  • s consists of only lowercase English letters.

Solutions

Solution 1: Prefix Sum + Case Discussion

Let's denote the length of string $s$ as $n$, then half of the length is $m = \frac{n}{2}$. Next, we divide string $s$ into two equal-length segments, where the second segment is reversed to get string $t$, and the first segment remains as $s$. For each query $[a_i, b_i, c_i, d_i]$, where $c_i$ and $d_i$ need to be transformed to $n - 1 - d_i$ and $n - 1 - c_i$. The problem is transformed into: for each query $[a_i, b_i, c_i, d_i]$, determine whether $s[a_i, b_i]$ and $t[c_i, d_i]$ can be rearranged to make strings $s$ and $t$ equal.

We preprocess the following information:

  1. The prefix sum array $pre_1$ of string $s$, where $pre_1[i][j]$ represents the quantity of character $j$ in the first $i$ characters of string $s$;
  2. The prefix sum array $pre_2$ of string $t$, where $pre_2[i][j]$ represents the quantity of character $j$ in the first $i$ characters of string $t$;
  3. The difference array $diff$ of strings $s$ and $t$, where $diff[i]$ represents the quantity of different characters in the first $i$ characters of strings $s$ and $t$.

For each query $[a_i, b_i, c_i, d_i]$, let's assume $a_i \le c_i$, then we need to discuss the following cases:

  1. The prefix substrings $s[..a_i-1]$ and $t[..a_i-1]$ of strings $s$ and $t$ must be equal, and the suffix substrings $s[\max(b_i, d_i)+1..]$ and $t[\max(b_i, d_i)..]$ must also be equal, otherwise, it is impossible to rearrange to make strings $s$ and $t$ equal;
  2. If $d_i \le b_i$, it means the interval $[a_i, b_i]$ contains the interval $[c_i, d_i]$. If the substrings $s[a_i, b_i]$ and $t[a_i, b_i]$ contain the same quantity of characters, then it is possible to rearrange to make strings $s$ and $t$ equal, otherwise, it is impossible;
  3. If $b_i &lt; c_i$, it means the intervals $[a_i, b_i]$ and $[c_i, d_i]$ do not intersect. Then the substrings $s[b_i+1, c_i-1]$ and $t[b_i+1, c_i-1]$ must be equal, and the substrings $s[a_i, b_i]$ and $t[a_i, b_i]$ must be equal, and the substrings $s[c_i, d_i]$ and $t[c_i, d_i]$ must be equal, otherwise, it is impossible to rearrange to make strings $s$ and $t$ equal.
  4. If $c_i \le b_i &lt; d_i$, it means the intervals $[a_i, b_i]$ and $[c_i, d_i]$ intersect. Then the characters contained in $s[a_i, b_i]$, minus the characters contained in $t[a_i, c_i-1]$, must be equal to the characters contained in $t[c_i, d_i]$, minus the characters contained in $s[b_i+1, d_i]$, otherwise, it is impossible to rearrange to make strings $s$ and $t$ equal.

Based on the above analysis, we iterate through each query $[a_i, b_i, c_i, d_i]$, and determine whether it satisfies the above conditions.

The time complexity is $O((n + q) \times |\Sigma|)$, and the space complexity is $O(n \times |\Sigma|)$. Where $n$ and $q$ are the lengths of string $s$ and the query array $queries$ respectively; and $|\Sigma|$ is the size of the character set. In this problem, the character set is lowercase English letters, so $|\Sigma| = 26$.

class Solution:
    def canMakePalindromeQueries(self, s: str, queries: List[List[int]]) -> List[bool]:
        def count(pre: List[List[int]], i: int, j: int) -> List[int]:
            return [x - y for x, y in zip(pre[j + 1], pre[i])]

        def sub(cnt1: List[int], cnt2: List[int]) -> List[int]:
            res = []
            for x, y in zip(cnt1, cnt2):
                if x - y < 0:
                    return []
                res.append(x - y)
            return res

        def check(
            pre1: List[List[int]], pre2: List[List[int]], a: int, b: int, c: int, d: int
        ) -> bool:
            if diff[a] > 0 or diff[m] - diff[max(b, d) + 1] > 0:
                return False
            if d <= b:
                return count(pre1, a, b) == count(pre2, a, b)
            if b < c:
                return (
                    diff[c] - diff[b + 1] == 0
                    and count(pre1, a, b) == count(pre2, a, b)
                    and count(pre1, c, d) == count(pre2, c, d)
                )
            cnt1 = sub(count(pre1, a, b), count(pre2, a, c - 1))
            cnt2 = sub(count(pre2, c, d), count(pre1, b + 1, d))
            return bool(cnt1) and bool(cnt2) and cnt1 == cnt2

        n = len(s)
        m = n // 2
        t = s[m:][::-1]
        s = s[:m]
        pre1 = [[0] * 26 for _ in range(m + 1)]
        pre2 = [[0] * 26 for _ in range(m + 1)]
        diff = [0] * (m + 1)
        for i, (c1, c2) in enumerate(zip(s, t), 1):
            pre1[i] = pre1[i - 1][:]
            pre2[i] = pre2[i - 1][:]
            pre1[i][ord(c1) - ord("a")] += 1
            pre2[i][ord(c2) - ord("a")] += 1
            diff[i] = diff[i - 1] + int(c1 != c2)
        ans = []
        for a, b, c, d in queries:
            c, d = n - 1 - d, n - 1 - c
            ok = (
                check(pre1, pre2, a, b, c, d)
                if a <= c
                else check(pre2, pre1, c, d, a, b)
            )
            ans.append(ok)
        return ans
class Solution {
    public boolean[] canMakePalindromeQueries(String s, int[][] queries) {
        int n = s.length();
        int m = n / 2;
        String t = new StringBuilder(s.substring(m)).reverse().toString();
        s = s.substring(0, m);
        int[][] pre1 = new int[m + 1][0];
        int[][] pre2 = new int[m + 1][0];
        int[] diff = new int[m + 1];
        pre1[0] = new int[26];
        pre2[0] = new int[26];
        for (int i = 1; i <= m; ++i) {
            pre1[i] = pre1[i - 1].clone();
            pre2[i] = pre2[i - 1].clone();
            ++pre1[i][s.charAt(i - 1) - 'a'];
            ++pre2[i][t.charAt(i - 1) - 'a'];
            diff[i] = diff[i - 1] + (s.charAt(i - 1) == t.charAt(i - 1) ? 0 : 1);
        }
        boolean[] ans = new boolean[queries.length];
        for (int i = 0; i < queries.length; ++i) {
            int[] q = queries[i];
            int a = q[0], b = q[1];
            int c = n - 1 - q[3], d = n - 1 - q[2];
            ans[i] = a <= c ? check(pre1, pre2, diff, a, b, c, d)
                            : check(pre2, pre1, diff, c, d, a, b);
        }
        return ans;
    }

    private boolean check(int[][] pre1, int[][] pre2, int[] diff, int a, int b, int c, int d) {
        if (diff[a] > 0 || diff[diff.length - 1] - diff[Math.max(b, d) + 1] > 0) {
            return false;
        }
        if (d <= b) {
            return Arrays.equals(count(pre1, a, b), count(pre2, a, b));
        }
        if (b < c) {
            return diff[c] - diff[b + 1] == 0 && Arrays.equals(count(pre1, a, b), count(pre2, a, b))
                && Arrays.equals(count(pre1, c, d), count(pre2, c, d));
        }
        int[] cnt1 = sub(count(pre1, a, b), count(pre2, a, c - 1));
        int[] cnt2 = sub(count(pre2, c, d), count(pre1, b + 1, d));
        return cnt1 != null && cnt2 != null && Arrays.equals(cnt1, cnt2);
    }

    private int[] count(int[][] pre, int i, int j) {
        int[] cnt = new int[26];
        for (int k = 0; k < 26; ++k) {
            cnt[k] = pre[j + 1][k] - pre[i][k];
        }
        return cnt;
    }

    private int[] sub(int[] cnt1, int[] cnt2) {
        int[] cnt = new int[26];
        for (int i = 0; i < 26; ++i) {
            cnt[i] = cnt1[i] - cnt2[i];
            if (cnt[i] < 0) {
                return null;
            }
        }
        return cnt;
    }
}
class Solution {
public:
    vector<bool> canMakePalindromeQueries(string s, vector<vector<int>>& queries) {
        int n = s.length();
        int m = n / 2;
        string t = string(s.begin() + m, s.end());
        reverse(t.begin(), t.end());
        s = string(s.begin(), s.begin() + m);

        vector<vector<int>> pre1(m + 1, vector<int>(26));
        vector<vector<int>> pre2(m + 1, vector<int>(26));
        vector<int> diff(m + 1, 0);
        for (int i = 1; i <= m; ++i) {
            pre1[i] = pre1[i - 1];
            pre2[i] = pre2[i - 1];
            ++pre1[i][s[i - 1] - 'a'];
            ++pre2[i][t[i - 1] - 'a'];
            diff[i] = diff[i - 1] + (s[i - 1] == t[i - 1] ? 0 : 1);
        }

        vector<bool> ans(queries.size(), false);
        for (int i = 0; i < queries.size(); ++i) {
            vector<int> q = queries[i];
            int a = q[0], b = q[1];
            int c = n - 1 - q[3], d = n - 1 - q[2];
            ans[i] = (a <= c) ? check(pre1, pre2, diff, a, b, c, d) : check(pre2, pre1, diff, c, d, a, b);
        }
        return ans;
    }

private:
    bool check(const vector<vector<int>>& pre1, const vector<vector<int>>& pre2, const vector<int>& diff, int a, int b, int c, int d) {
        if (diff[a] > 0 || diff[diff.size() - 1] - diff[max(b, d) + 1] > 0) {
            return false;
        }

        if (d <= b) {
            return count(pre1, a, b) == count(pre2, a, b);
        }

        if (b < c) {
            return diff[c] - diff[b + 1] == 0 && count(pre1, a, b) == count(pre2, a, b) && count(pre1, c, d) == count(pre2, c, d);
        }

        vector<int> cnt1 = sub(count(pre1, a, b), count(pre2, a, c - 1));
        vector<int> cnt2 = sub(count(pre2, c, d), count(pre1, b + 1, d));

        return cnt1 != vector<int>() && cnt2 != vector<int>() && cnt1 == cnt2;
    }

    vector<int> count(const vector<vector<int>>& pre, int i, int j) {
        vector<int> cnt(26);
        for (int k = 0; k < 26; ++k) {
            cnt[k] = pre[j + 1][k] - pre[i][k];
        }
        return cnt;
    }

    vector<int> sub(const vector<int>& cnt1, const vector<int>& cnt2) {
        vector<int> cnt(26);
        for (int i = 0; i < 26; ++i) {
            cnt[i] = cnt1[i] - cnt2[i];
            if (cnt[i] < 0) {
                return vector<int>();
            }
        }
        return cnt;
    }
};
func canMakePalindromeQueries(s string, queries [][]int) (ans []bool) {
	n := len(s)
	m := n / 2
	t := reverse(s[m:])
	s = s[:m]

	pre1 := make([][]int, m+1)
	pre2 := make([][]int, m+1)
	diff := make([]int, m+1)
	pre1[0] = make([]int, 26)
	pre2[0] = make([]int, 26)

	for i := 1; i <= m; i++ {
		pre1[i] = slices.Clone(pre1[i-1])
		pre2[i] = slices.Clone(pre2[i-1])
		pre1[i][int(s[i-1]-'a')]++
		pre2[i][int(t[i-1]-'a')]++
		diff[i] = diff[i-1]
		if s[i-1] != t[i-1] {
			diff[i]++
		}
	}
	for _, q := range queries {
		a, b := q[0], q[1]
		c, d := n-1-q[3], n-1-q[2]
		if a <= c {
			ans = append(ans, check(pre1, pre2, diff, a, b, c, d))
		} else {
			ans = append(ans, check(pre2, pre1, diff, c, d, a, b))
		}
	}
	return
}

func check(pre1, pre2 [][]int, diff []int, a, b, c, d int) bool {
	if diff[a] > 0 || diff[len(diff)-1]-diff[max(b, d)+1] > 0 {
		return false
	}

	if d <= b {
		return slices.Equal(count(pre1, a, b), count(pre2, a, b))
	}

	if b < c {
		return diff[c]-diff[b+1] == 0 && slices.Equal(count(pre1, a, b), count(pre2, a, b)) && slices.Equal(count(pre1, c, d), count(pre2, c, d))
	}

	cnt1 := sub(count(pre1, a, b), count(pre2, a, c-1))
	cnt2 := sub(count(pre2, c, d), count(pre1, b+1, d))

	return !slices.Equal(cnt1, []int{}) && !slices.Equal(cnt2, []int{}) && slices.Equal(cnt1, cnt2)
}

func count(pre [][]int, i, j int) []int {
	cnt := make([]int, 26)
	for k := 0; k < 26; k++ {
		cnt[k] = pre[j+1][k] - pre[i][k]
	}
	return cnt
}

func sub(cnt1, cnt2 []int) []int {
	cnt := make([]int, 26)
	for i := 0; i < 26; i++ {
		cnt[i] = cnt1[i] - cnt2[i]
		if cnt[i] < 0 {
			return []int{}
		}
	}
	return cnt
}

func reverse(s string) string {
	runes := []rune(s)
	for i, j := 0, len(runes)-1; i < j; i, j = i+1, j-1 {
		runes[i], runes[j] = runes[j], runes[i]
	}
	return string(runes)
}
function canMakePalindromeQueries(s: string, queries: number[][]): boolean[] {
    const n: number = s.length;
    const m: number = n >> 1;
    const t: string = s.slice(m).split('').reverse().join('');
    s = s.slice(0, m);

    const pre1: number[][] = Array.from({ length: m + 1 }, () => Array(26).fill(0));
    const pre2: number[][] = Array.from({ length: m + 1 }, () => Array(26).fill(0));
    const diff: number[] = Array(m + 1).fill(0);
    for (let i = 1; i <= m; ++i) {
        pre1[i] = [...pre1[i - 1]];
        pre2[i] = [...pre2[i - 1]];
        ++pre1[i][s.charCodeAt(i - 1) - 'a'.charCodeAt(0)];
        ++pre2[i][t.charCodeAt(i - 1) - 'a'.charCodeAt(0)];
        diff[i] = diff[i - 1] + (s[i - 1] === t[i - 1] ? 0 : 1);
    }

    const ans: boolean[] = Array(queries.length).fill(false);
    for (let i = 0; i < queries.length; ++i) {
        const q: number[] = queries[i];
        const [a, b] = [q[0], q[1]];
        const [c, d] = [n - 1 - q[3], n - 1 - q[2]];
        ans[i] = a <= c ? check(pre1, pre2, diff, a, b, c, d) : check(pre2, pre1, diff, c, d, a, b);
    }
    return ans;
}

function check(
    pre1: number[][],
    pre2: number[][],
    diff: number[],
    a: number,
    b: number,
    c: number,
    d: number,
): boolean {
    if (diff[a] > 0 || diff[diff.length - 1] - diff[Math.max(b, d) + 1] > 0) {
        return false;
    }

    if (d <= b) {
        return arraysEqual(count(pre1, a, b), count(pre2, a, b));
    }

    if (b < c) {
        return (
            diff[c] - diff[b + 1] === 0 &&
            arraysEqual(count(pre1, a, b), count(pre2, a, b)) &&
            arraysEqual(count(pre1, c, d), count(pre2, c, d))
        );
    }

    const cnt1: number[] | null = sub(count(pre1, a, b), count(pre2, a, c - 1));
    const cnt2: number[] | null = sub(count(pre2, c, d), count(pre1, b + 1, d));

    return cnt1 !== null && cnt2 !== null && arraysEqual(cnt1, cnt2);
}

function count(pre: number[][], i: number, j: number): number[] {
    const cnt: number[] = new Array(26).fill(0);
    for (let k = 0; k < 26; ++k) {
        cnt[k] = pre[j + 1][k] - pre[i][k];
    }
    return cnt;
}

function sub(cnt1: number[], cnt2: number[]): number[] | null {
    const cnt: number[] = new Array(26).fill(0);
    for (let i = 0; i < 26; ++i) {
        cnt[i] = cnt1[i] - cnt2[i];
        if (cnt[i] < 0) {
            return null;
        }
    }
    return cnt;
}

function arraysEqual(arr1: number[], arr2: number[]): boolean {
    return JSON.stringify(arr1) === JSON.stringify(arr2);
}