Question

2940. Find Building Where Alice and Bob Can Meet

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Description

You are given a 0-indexed array heights of positive integers, where heights[i] represents the height of the ith building.

If a person is in building i, they can move to any other building j if and only if i < j and heights[i] < heights[j].

You are also given another array queries where queries[i] = [ai, bi]. On the ith query, Alice is in building ai while Bob is in building bi.

Return _an array_ ans _where_ ans[i] _is the index of the leftmost building where Alice and Bob can meet on the_ ith _query_. _If Alice and Bob cannot move to a common building on query_ i, _set_ ans[i] _to_ -1.

 

Example 1:

Input: heights = [6,4,8,5,2,7], queries = [[0,1],[0,3],[2,4],[3,4],[2,2]]
Output: [2,5,-1,5,2]
Explanation: In the first query, Alice and Bob can move to building 2 since heights[0] < heights[2] and heights[1] < heights[2]. 
In the second query, Alice and Bob can move to building 5 since heights[0] < heights[5] and heights[3] < heights[5]. 
In the third query, Alice cannot meet Bob since Alice cannot move to any other building.
In the fourth query, Alice and Bob can move to building 5 since heights[3] < heights[5] and heights[4] < heights[5].
In the fifth query, Alice and Bob are already in the same building.  
For ans[i] != -1, It can be shown that ans[i] is the leftmost building where Alice and Bob can meet.
For ans[i] == -1, It can be shown that there is no building where Alice and Bob can meet.

Example 2:

Input: heights = [5,3,8,2,6,1,4,6], queries = [[0,7],[3,5],[5,2],[3,0],[1,6]]
Output: [7,6,-1,4,6]
Explanation: In the first query, Alice can directly move to Bob's building since heights[0] < heights[7].
In the second query, Alice and Bob can move to building 6 since heights[3] < heights[6] and heights[5] < heights[6].
In the third query, Alice cannot meet Bob since Bob cannot move to any other building.
In the fourth query, Alice and Bob can move to building 4 since heights[3] < heights[4] and heights[0] < heights[4].
In the fifth query, Alice can directly move to Bob's building since heights[1] < heights[6].
For ans[i] != -1, It can be shown that ans[i] is the leftmost building where Alice and Bob can meet.
For ans[i] == -1, It can be shown that there is no building where Alice and Bob can meet.

 

Constraints:

• 1 <= heights.length <= 5 * 104
• 1 <= heights[i] <= 109
• 1 <= queries.length <= 5 * 104
• queries[i] = [ai, bi]
• 0 <= ai, bi <= heights.length - 1

Solutions

Solution 1: Binary Indexed Tree

Let's denote $queries[i] = [l_i, r_i]$, where $l_i \le r_i$. If $l_i = r_i$ or $heights[l_i] < heights[r_i]$, then the answer is $r_i$. Otherwise, we need to find the smallest $j$ among all $j > r_i$ and $heights[j] > heights[l_i]$.

We can sort $queries$ in descending order of $r_i$, and use a pointer $j$ to point to the current index of $heights$ being traversed.

Next, we traverse each query $queries[i] = (l, r)$. For the current query, if $j > r$, then we loop to insert $heights[j]$ into the binary indexed tree. The binary indexed tree maintains the minimum index of the suffix height (after discretization). Then, we judge whether $l = r$ or $heights[l] < heights[r]$. If it is, then the answer to the current query is $r$. Otherwise, we query the minimum index of $heights[l]$ in the binary indexed tree, which is the answer to the current query.

The time complexity is $O((n + m) \times \log n + m \times \log m)$, and the space complexity is $O(n + m)$. Where $n$ and $m$ are the lengths of $heights$ and $queries$ respectively.

Similar problems:

• 2736. Maximum Sum Queries

class BinaryIndexedTree:
  __slots__ = ["n", "c"]

  def __init__(self, n: int):
    self.n = n
    self.c = [inf] * (n + 1)

  def update(self, x: int, v: int):
    while x <= self.n:
      self.c[x] = min(self.c[x], v)
      x += x & -x

  def query(self, x: int) -> int:
    mi = inf
    while x:
      mi = min(mi, self.c[x])
      x -= x & -x
    return -1 if mi == inf else mi


class Solution:
  def leftmostBuildingQueries(
    self, heights: List[int], queries: List[List[int]]
  ) -> List[int]:
    n, m = len(heights), len(queries)
    for i in range(m):
      queries[i] = [min(queries[i]), max(queries[i])]
    j = n - 1
    s = sorted(set(heights))
    ans = [-1] * m
    tree = BinaryIndexedTree(n)
    for i in sorted(range(m), key=lambda i: -queries[i][1]):
      l, r = queries[i]
      while j > r:
        k = n - bisect_left(s, heights[j]) + 1
        tree.update(k, j)
        j -= 1
      if l == r or heights[l] < heights[r]:
        ans[i] = r
      else:
        k = n - bisect_left(s, heights[l])
        ans[i] = tree.query(k)
    return ans

class BinaryIndexedTree {
  private final int inf = 1 << 30;
  private int n;
  private int[] c;

  public BinaryIndexedTree(int n) {
    this.n = n;
    c = new int[n + 1];
    Arrays.fill(c, inf);
  }

  public void update(int x, int v) {
    while (x <= n) {
      c[x] = Math.min(c[x], v);
      x += x & -x;
    }
  }

  public int query(int x) {
    int mi = inf;
    while (x > 0) {
      mi = Math.min(mi, c[x]);
      x -= x & -x;
    }
    return mi == inf ? -1 : mi;
  }
}

class Solution {
  public int[] leftmostBuildingQueries(int[] heights, int[][] queries) {
    int n = heights.length;
    int m = queries.length;
    for (int i = 0; i < m; ++i) {
      if (queries[i][0] > queries[i][1]) {
        queries[i] = new int[] {queries[i][1], queries[i][0]};
      }
    }
    Integer[] idx = new Integer[m];
    for (int i = 0; i < m; ++i) {
      idx[i] = i;
    }
    Arrays.sort(idx, (i, j) -> queries[j][1] - queries[i][1]);
    int[] s = heights.clone();
    Arrays.sort(s);
    int[] ans = new int[m];
    int j = n - 1;
    BinaryIndexedTree tree = new BinaryIndexedTree(n);
    for (int i : idx) {
      int l = queries[i][0], r = queries[i][1];
      while (j > r) {
        int k = n - Arrays.binarySearch(s, heights[j]) + 1;
        tree.update(k, j);
        --j;
      }
      if (l == r || heights[l] < heights[r]) {
        ans[i] = r;
      } else {
        int k = n - Arrays.binarySearch(s, heights[l]);
        ans[i] = tree.query(k);
      }
    }
    return ans;
  }
}

class BinaryIndexedTree {
private:
  int inf = 1 << 30;
  int n;
  vector<int> c;

public:
  BinaryIndexedTree(int n) {
    this->n = n;
    c.resize(n + 1, inf);
  }

  void update(int x, int v) {
    while (x <= n) {
      c[x] = min(c[x], v);
      x += x & -x;
    }
  }

  int query(int x) {
    int mi = inf;
    while (x > 0) {
      mi = min(mi, c[x]);
      x -= x & -x;
    }
    return mi == inf ? -1 : mi;
  }
};

class Solution {
public:
  vector<int> leftmostBuildingQueries(vector<int>& heights, vector<vector<int>>& queries) {
    int n = heights.size(), m = queries.size();
    for (auto& q : queries) {
      if (q[0] > q[1]) {
        swap(q[0], q[1]);
      }
    }
    vector<int> idx(m);
    iota(idx.begin(), idx.end(), 0);
    sort(idx.begin(), idx.end(), [&](int i, int j) {
      return queries[j][1] < queries[i][1];
    });
    vector<int> s = heights;
    sort(s.begin(), s.end());
    s.erase(unique(s.begin(), s.end()), s.end());
    vector<int> ans(m);
    int j = n - 1;
    BinaryIndexedTree tree(n);
    for (int i : idx) {
      int l = queries[i][0], r = queries[i][1];
      while (j > r) {
        int k = s.end() - lower_bound(s.begin(), s.end(), heights[j]) + 1;
        tree.update(k, j);
        --j;
      }
      if (l == r || heights[l] < heights[r]) {
        ans[i] = r;
      } else {
        int k = s.end() - lower_bound(s.begin(), s.end(), heights[l]);
        ans[i] = tree.query(k);
      }
    }
    return ans;
  }
};

const inf int = 1 << 30

type BinaryIndexedTree struct {
  n int
  c []int
}

func NewBinaryIndexedTree(n int) BinaryIndexedTree {
  c := make([]int, n+1)
  for i := range c {
    c[i] = inf
  }
  return BinaryIndexedTree{n: n, c: c}
}

func (bit *BinaryIndexedTree) update(x, v int) {
  for x <= bit.n {
    bit.c[x] = min(bit.c[x], v)
    x += x & -x
  }
}

func (bit *BinaryIndexedTree) query(x int) int {
  mi := inf
  for x > 0 {
    mi = min(mi, bit.c[x])
    x -= x & -x
  }
  if mi == inf {
    return -1
  }
  return mi
}

func leftmostBuildingQueries(heights []int, queries [][]int) []int {
  n, m := len(heights), len(queries)
  for _, q := range queries {
    if q[0] > q[1] {
      q[0], q[1] = q[1], q[0]
    }
  }
  idx := make([]int, m)
  for i := range idx {
    idx[i] = i
  }
  sort.Slice(idx, func(i, j int) bool { return queries[idx[j]][1] < queries[idx[i]][1] })
  s := make([]int, n)
  copy(s, heights)
  sort.Ints(s)
  ans := make([]int, m)
  tree := NewBinaryIndexedTree(n)
  j := n - 1
  for _, i := range idx {
    l, r := queries[i][0], queries[i][1]
    for ; j > r; j-- {
      k := n - sort.SearchInts(s, heights[j]) + 1
      tree.update(k, j)
    }
    if l == r || heights[l] < heights[r] {
      ans[i] = r
    } else {
      k := n - sort.SearchInts(s, heights[l])
      ans[i] = tree.query(k)
    }
  }
  return ans
}

class BinaryIndexedTree {
  private n: number;
  private c: number[];
  private inf: number = 1 << 30;

  constructor(n: number) {
    this.n = n;
    this.c = Array(n + 1).fill(this.inf);
  }

  update(x: number, v: number): void {
    while (x <= this.n) {
      this.c[x] = Math.min(this.c[x], v);
      x += x & -x;
    }
  }

  query(x: number): number {
    let mi = this.inf;
    while (x > 0) {
      mi = Math.min(mi, this.c[x]);
      x -= x & -x;
    }
    return mi === this.inf ? -1 : mi;
  }
}

function leftmostBuildingQueries(heights: number[], queries: number[][]): number[] {
  const n = heights.length;
  const m = queries.length;
  for (const q of queries) {
    if (q[0] > q[1]) {
      [q[0], q[1]] = [q[1], q[0]];
    }
  }
  const idx: number[] = Array(m)
    .fill(0)
    .map((_, i) => i);
  idx.sort((i, j) => queries[j][1] - queries[i][1]);
  const tree = new BinaryIndexedTree(n);
  const ans: number[] = Array(m).fill(-1);
  const s = [...heights];
  s.sort((a, b) => a - b);
  const search = (x: number) => {
    let [l, r] = [0, n];
    while (l < r) {
      const mid = (l + r) >> 1;
      if (s[mid] >= x) {
        r = mid;
      } else {
        l = mid + 1;
      }
    }
    return l;
  };
  let j = n - 1;
  for (const i of idx) {
    const [l, r] = queries[i];
    while (j > r) {
      const k = n - search(heights[j]) + 1;
      tree.update(k, j);
      --j;
    }
    if (l === r || heights[l] < heights[r]) {
      ans[i] = r;
    } else {
      const k = n - search(heights[l]);
      ans[i] = tree.query(k);
    }
  }
  return ans;
}