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Problem solution · Java

Path Existence Queries in a Graph II

Path Existence Queries in a Graph II: a Java solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Binary search
Source
walkccc LeetCode Solutions
Length
67 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

Binary search

For Path Existence Queries in a Graph II, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 67 lines of Java from the credited upstream file 3534.java.
  • The implementation visibly relies on sequence storage.
  • 8 loop blocks detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

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 codePath Existence Queries in a Graph II · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int[] pathExistenceQueries(int n, int[] nums, int maxDiff, int[][] queries) {    int[] ans = new int[queries.length];    int[] indexMap = new int[n];    int[] sortedNums = new int[n];    Pair<Integer, Integer>[] sortedNumAndIndexes = new Pair[n];     for (int i = 0; i < n; ++i)      sortedNumAndIndexes[i] = new Pair<>(nums[i], i);     Arrays.sort(sortedNumAndIndexes, Comparator.comparingInt(Pair::getKey));     for (int i = 0; i < n; ++i) {      final int num = sortedNumAndIndexes[i].getKey();      final int sortedIndex = sortedNumAndIndexes[i].getValue();      sortedNums[i] = num;      indexMap[sortedIndex] = i;    }     final int maxLevel = Integer.SIZE - Integer.numberOfLeadingZeros(n) + 1;    // jump[i][j] := the index of the j-th ancestor of i    int[][] jump = new int[n][maxLevel];     int right = 0;    for (int i = 0; i < n; ++i) {      while (right + 1 < n && sortedNums[right + 1] - sortedNums[i] <= maxDiff)        ++right;      jump[i][0] = right;    }     for (int level = 1; level < maxLevel; ++level)      for (int i = 0; i < n; ++i) {        final int prevJump = jump[i][level - 1];        jump[i][level] = jump[prevJump][level - 1];      }     for (int i = 0; i < queries.length; ++i) {      final int u = queries[i][0];      final int v = queries[i][1];      final int uIndex = indexMap[u];      final int vIndex = indexMap[v];      final int start = Math.min(uIndex, vIndex);      final int end = Math.max(uIndex, vIndex);      final int res = minJumps(jump, start, end, maxLevel - 1);      ans[i] = res == Integer.MAX_VALUE ? -1 : res;    }     return ans;  }   // Returns the minimum number of jumps from `start` to `end` using binary  // lifting.  private int minJumps(int[][] jump, int start, int end, int level) {    if (start == end)      return 0;    if (jump[start][0] >= end)      return 1;    if (jump[start][level] < end)      return Integer.MAX_VALUE;    int j = level;    for (; j >= 0; --j)      if (jump[start][j] < end)        break;    return (1 << j) + minJumps(jump, jump[start][j], end, j);  }}

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