Question

2736. Maximum Sum Queries

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Description

You are given two 0-indexed integer arrays nums1 and nums2, each of length n, and a 1-indexed 2D array queries where queries[i] = [xi, yi].

For the ith query, find the maximum value of nums1[j] + nums2[j] among all indices j (0 <= j < n), where nums1[j] >= xi and nums2[j] >= yi, or -1 if there is no j satisfying the constraints.

Return _an array _answer_ where _answer[i]_ is the answer to the _ith_ query._

 

Example 1:

Input: nums1 = [4,3,1,2], nums2 = [2,4,9,5], queries = [[4,1],[1,3],[2,5]]
Output: [6,10,7]
Explanation: 
For the 1st query xi = 4 and yi = 1, we can select index j = 0 since nums1[j] >= 4 and nums2[j] >= 1. The sum nums1[j] + nums2[j] is 6, and we can show that 6 is the maximum we can obtain.

For the 2nd query xi = 1 and yi = 3, we can select index j = 2 since nums1[j] >= 1 and nums2[j] >= 3. The sum nums1[j] + nums2[j] is 10, and we can show that 10 is the maximum we can obtain. 

For the 3rd query xi = 2 and yi = 5, we can select index j = 3 since nums1[j] >= 2 and nums2[j] >= 5. The sum nums1[j] + nums2[j] is 7, and we can show that 7 is the maximum we can obtain.

Therefore, we return [6,10,7].

Example 2:

Input: nums1 = [3,2,5], nums2 = [2,3,4], queries = [[4,4],[3,2],[1,1]]
Output: [9,9,9]
Explanation: For this example, we can use index j = 2 for all the queries since it satisfies the constraints for each query.

Example 3:

Input: nums1 = [2,1], nums2 = [2,3], queries = [[3,3]]
Output: [-1]
Explanation: There is one query in this example with xi = 3 and yi = 3. For every index, j, either nums1[j] < xi or nums2[j] < yi. Hence, there is no solution. 

 

Constraints:

• nums1.length == nums2.length 
• n == nums1.length 
• 1 <= n <= 105
• 1 <= nums1[i], nums2[i] <= 109 
• 1 <= queries.length <= 105
• queries[i].length == 2
• xi == queries[i][1]
• yi == queries[i][2]
• 1 <= xi, yi <= 109

Solutions

Solution 1: Binary Indexed Tree

This problem belongs to the category of two-dimensional partial order problems.

A two-dimensional partial order problem is defined as follows: given several pairs of points $(a_1, b_1)$, $(a_2, b_2)$, …, $(a_n, b_n)$, and a defined partial order relation, now given a point $(a_i, b_i)$, we need to find the number/maximum value of point pairs $(a_j, b_j)$ that satisfy the partial order relation. That is:

$$ \left(a_{j}, b_{j}\right) \prec\left(a_{i}, b_{i}\right) \stackrel{\text { def }}{=} a_{j} \lesseqgtr a_{i} \text { and } b_{j} \lesseqgtr b_{i} $$

The general solution to two-dimensional partial order problems is to sort one dimension and use a data structure to handle the second dimension (this data structure is generally a binary indexed tree).

For this problem, we can create an array $nums$, where $nums[i]=(nums_1[i], nums_2[i])$, and then sort $nums$ in descending order according to $nums_1$. We also sort the queries $queries$ in descending order according to $x$.

Next, we iterate through each query $queries[i] = (x, y)$. For the current query, we loop to insert the value of $nums_2$ for all elements in $nums$ that are greater than or equal to $x$ into the binary indexed tree. The binary indexed tree maintains the maximum value of $nums_1 + nums_2$ in the discretized $nums_2$ interval. Therefore, we only need to query the maximum value corresponding to the interval greater than or equal to the discretized $y$ in the binary indexed tree. Note that since the binary indexed tree maintains the prefix maximum value, we can insert $nums_2$ in reverse order into the binary indexed tree in the implementation.

The time complexity is $O((n + m) \times \log n + m \times \log m)$, and the space complexity is $O(n + m)$. Here, $n$ is the length of the array $nums$, and $m$ is the length of the array $queries$.

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

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

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

  def query(self, x: int) -> int:
    mx = -1
    while x:
      mx = max(mx, self.c[x])
      x -= x & -x
    return mx


class Solution:
  def maximumSumQueries(
    self, nums1: List[int], nums2: List[int], queries: List[List[int]]
  ) -> List[int]:
    nums = sorted(zip(nums1, nums2), key=lambda x: -x[0])
    nums2.sort()
    n, m = len(nums1), len(queries)
    ans = [-1] * m
    j = 0
    tree = BinaryIndexedTree(n)
    for i in sorted(range(m), key=lambda i: -queries[i][0]):
      x, y = queries[i]
      while j < n and nums[j][0] >= x:
        k = n - bisect_left(nums2, nums[j][1])
        tree.update(k, nums[j][0] + nums[j][1])
        j += 1
      k = n - bisect_left(nums2, y)
      ans[i] = tree.query(k)
    return ans

class BinaryIndexedTree {
  private int n;
  private int[] c;

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

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

  public int query(int x) {
    int mx = -1;
    while (x > 0) {
      mx = Math.max(mx, c[x]);
      x -= x & -x;
    }
    return mx;
  }
}

class Solution {
  public int[] maximumSumQueries(int[] nums1, int[] nums2, int[][] queries) {
    int n = nums1.length;
    int[][] nums = new int[n][0];
    for (int i = 0; i < n; ++i) {
      nums[i] = new int[] {nums1[i], nums2[i]};
    }
    Arrays.sort(nums, (a, b) -> b[0] - a[0]);
    Arrays.sort(nums2);
    int m = queries.length;
    Integer[] idx = new Integer[m];
    for (int i = 0; i < m; ++i) {
      idx[i] = i;
    }
    Arrays.sort(idx, (i, j) -> queries[j][0] - queries[i][0]);
    int[] ans = new int[m];
    int j = 0;
    BinaryIndexedTree tree = new BinaryIndexedTree(n);
    for (int i : idx) {
      int x = queries[i][0], y = queries[i][1];
      for (; j < n && nums[j][0] >= x; ++j) {
        int k = n - Arrays.binarySearch(nums2, nums[j][1]);
        tree.update(k, nums[j][0] + nums[j][1]);
      }
      int p = Arrays.binarySearch(nums2, y);
      int k = p >= 0 ? n - p : n + p + 1;
      ans[i] = tree.query(k);
    }
    return ans;
  }
}

class BinaryIndexedTree {
private:
  int n;
  vector<int> c;

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

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

  int query(int x) {
    int mx = -1;
    while (x > 0) {
      mx = max(mx, c[x]);
      x -= x & -x;
    }
    return mx;
  }
};

class Solution {
public:
  vector<int> maximumSumQueries(vector<int>& nums1, vector<int>& nums2, vector<vector<int>>& queries) {
    vector<pair<int, int>> nums;
    int n = nums1.size(), m = queries.size();
    for (int i = 0; i < n; ++i) {
      nums.emplace_back(-nums1[i], nums2[i]);
    }
    sort(nums.begin(), nums.end());
    sort(nums2.begin(), nums2.end());
    vector<int> idx(m);
    iota(idx.begin(), idx.end(), 0);
    sort(idx.begin(), idx.end(), [&](int i, int j) { return queries[j][0] < queries[i][0]; });
    vector<int> ans(m);
    int j = 0;
    BinaryIndexedTree tree(n);
    for (int i : idx) {
      int x = queries[i][0], y = queries[i][1];
      for (; j < n && -nums[j].first >= x; ++j) {
        int k = nums2.end() - lower_bound(nums2.begin(), nums2.end(), nums[j].second);
        tree.update(k, -nums[j].first + nums[j].second);
      }
      int k = nums2.end() - lower_bound(nums2.begin(), nums2.end(), y);
      ans[i] = tree.query(k);
    }
    return ans;
  }
};

type BinaryIndexedTree struct {
  n int
  c []int
}

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

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

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

func maximumSumQueries(nums1 []int, nums2 []int, queries [][]int) []int {
  n, m := len(nums1), len(queries)
  nums := make([][2]int, n)
  for i := range nums {
    nums[i] = [2]int{nums1[i], nums2[i]}
  }
  sort.Slice(nums, func(i, j int) bool { return nums[j][0] < nums[i][0] })
  sort.Ints(nums2)
  idx := make([]int, m)
  for i := range idx {
    idx[i] = i
  }
  sort.Slice(idx, func(i, j int) bool { return queries[idx[j]][0] < queries[idx[i]][0] })
  tree := NewBinaryIndexedTree(n)
  ans := make([]int, m)
  j := 0
  for _, i := range idx {
    x, y := queries[i][0], queries[i][1]
    for ; j < n && nums[j][0] >= x; j++ {
      k := n - sort.SearchInts(nums2, nums[j][1])
      tree.update(k, nums[j][0]+nums[j][1])
    }
    k := n - sort.SearchInts(nums2, y)
    ans[i] = tree.query(k)
  }
  return ans
}

class BinaryIndexedTree {
  private n: number;
  private c: number[];

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

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

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

function maximumSumQueries(nums1: number[], nums2: number[], queries: number[][]): number[] {
  const n = nums1.length;
  const m = queries.length;
  const nums: [number, number][] = [];
  for (let i = 0; i < n; ++i) {
    nums.push([nums1[i], nums2[i]]);
  }
  nums.sort((a, b) => b[0] - a[0]);
  nums2.sort((a, b) => a - b);
  const idx: number[] = Array(m)
    .fill(0)
    .map((_, i) => i);
  idx.sort((i, j) => queries[j][0] - queries[i][0]);
  const ans: number[] = Array(m).fill(0);
  let j = 0;
  const search = (x: number) => {
    let [l, r] = [0, n];
    while (l < r) {
      const mid = (l + r) >> 1;
      if (nums2[mid] >= x) {
        r = mid;
      } else {
        l = mid + 1;
      }
    }
    return l;
  };
  const tree = new BinaryIndexedTree(n);
  for (const i of idx) {
    const [x, y] = queries[i];
    for (; j < n && nums[j][0] >= x; ++j) {
      const k = n - search(nums[j][1]);
      tree.update(k, nums[j][0] + nums[j][1]);
    }
    const k = n - search(y);
    ans[i] = tree.query(k);
  }
  return ans;
}

Solution 2

class Solution {
  public int[] maximumSumQueries(int[] nums1, int[] nums2, int[][] q) {
    int n = nums1.length, m = q.length;
    int[][] a = new int[n][2];
    for (int i = 0; i < n; i++) {
      a[i][0] = nums1[i];
      a[i][1] = nums2[i];
    }
    int[][] b = new int[m][3];
    for (int i = 0; i < m; i++) {
      b[i][0] = q[i][0];
      b[i][1] = q[i][1];
      b[i][2] = i;
    }
    Arrays.sort(a, (o1, o2) -> o1[0] - o2[0]);
    Arrays.sort(b, (o1, o2) -> o1[0] - o2[0]);
    TreeMap<Integer, Integer> map = new TreeMap<>();
    int[] res = new int[m];
    int max = -1;
    for (int i = m - 1, j = n - 1; i >= 0; i--) {
      int x = b[i][0], y = b[i][1], idx = b[i][2];
      while (j >= 0 && a[j][0] >= x) {
        if (max < a[j][1]) {
          max = a[j][1];
          Integer key = map.floorKey(a[j][1]);
          while (key != null && map.get(key) <= a[j][0] + a[j][1]) {
            map.remove(key);
            key = map.floorKey(key);
          }
          map.put(max, a[j][0] + a[j][1]);
        }
        j--;
      }
      Integer key = map.ceilingKey(y);
      if (key == null)
        res[idx] = -1;
      else
        res[idx] = map.get(key);
    }
    return res;
  }
}