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Problem solution · Java

Palindrome Rearrangement Queries

Palindrome Rearrangement Queries: a Java solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
82 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sliding window or two pointers

For Palindrome Rearrangement Queries, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 82 lines of Java from the credited upstream file 2983.java.
  • The implementation visibly relies on sequence storage.
  • 4 loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codePalindrome Rearrangement Queries · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public boolean[] canMakePalindromeQueries(String s, int[][] queries) {    final int n = s.length();    // mirroredDiffs[i] := the number of different letters between the first i    // letters of s[0..n / 2) and the first i letters of s[n / 2..n)[::-1]    final int[] mirroredDiffs = getMirroredDiffs(s);    // counts[i] := the count of s[0..i)    final int[][] counts = getCounts(s);    boolean[] ans = new boolean[queries.length];     for (int i = 0; i < queries.length; i++) {      // Use left-closed, right-open intervals to facilitate the calculation.      //   ...... [a, b) ...|... [rb, ra) ......      //   .... [rd, rc) .....|..... [c, d) ....      int[] query = queries[i];      final int a = query[0];      final int b = query[1] + 1;      final int c = query[2];      final int d = query[3] + 1;      final int ra = n - a; // the reflected index of a in s[n / 2..n)      final int rb = n - b; // the reflected index of b in s[n / 2..n)      final int rc = n - c; // the reflected index of c in s[n / 2..n)      final int rd = n - d; // the reflected index of d in s[n / 2..n)      // No difference is allowed outside the query ranges.      if ((Math.min(a, rd) > 0 && mirroredDiffs[Math.min(a, rd)] > 0) ||          (n / 2 > Math.max(b, rc) && mirroredDiffs[n / 2] - mirroredDiffs[Math.max(b, rc)] > 0) ||          (rd > b && mirroredDiffs[rd] - mirroredDiffs[b] > 0) ||          (a > rc && mirroredDiffs[a] - mirroredDiffs[rc] > 0)) {        ans[i] = false;      } else {        // The `count` map of the intersection of [a, b) and [rd, rc) in        // s[0..n / 2) must equate to the `count` map of the intersection of        // [c, d) and [rb, ra) in s[n / 2..n).        int[] leftRangeCount = subtractArrays(counts[b], counts[a]);        int[] rightRangeCount = subtractArrays(counts[d], counts[c]);        if (a > rd)          rightRangeCount =              subtractArrays(rightRangeCount, subtractArrays(counts[Math.min(a, rc)], counts[rd]));        if (rc > b)          rightRangeCount =              subtractArrays(rightRangeCount, subtractArrays(counts[rc], counts[Math.max(b, rd)]));        if (c > rb)          leftRangeCount =              subtractArrays(leftRangeCount, subtractArrays(counts[Math.min(c, ra)], counts[rb]));        if (ra > d)          leftRangeCount =              subtractArrays(leftRangeCount, subtractArrays(counts[ra], counts[Math.max(d, rb)]));        ans[i] = Arrays.stream(leftRangeCount).allMatch(freq -> freq >= 0) &&                 Arrays.stream(rightRangeCount).allMatch(freq -> freq >= 0) &&                 Arrays.equals(leftRangeCount, rightRangeCount);      }    }     return ans;  }   private int[] getMirroredDiffs(final String s) {    int[] diffs = new int[s.length() / 2 + 1];    for (int i = 0, j = s.length() - 1; i < j; i++, j--) {      diffs[i + 1] = diffs[i] + (s.charAt(i) != s.charAt(j) ? 1 : 0);    }    return diffs;  }   private int[][] getCounts(final String s) {    int[][] counts = new int[s.length() + 1][26];    int[] count = new int[26];    for (int i = 0; i < s.length(); ++i) {      ++count[s.charAt(i) - 'a'];      System.arraycopy(count, 0, counts[i + 1], 0, 26);    }    return counts;  }   private int[] subtractArrays(int[] a, int[] b) {    int[] res = new int[a.length];    for (int i = 0; i < a.length; ++i)      res[i] = a[i] - b[i];    return res;  }}

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