# 2736. Maximum Sum Queries

## 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 $queries$.

• 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;
}
};

• 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


• 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;
}