LeetCode #2736 — HARD

Maximum Sum Queries

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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] = [x_i, y_i].

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

Return an array answer where answer[i] is the answer to the i^th 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 x_i = 4 and y_i = 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 x_i = 1 and y_i = 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 x_i = 2 and y_i = 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 x_i = 3 and y_i = 3. For every index, j, either nums1[j] < x_i or nums2[j] < y_i. 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
x_i == queries[i][1]
y_i == queries[i][2]
1 <= x_i, y_i <= 10⁹

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Stack · Segment Tree

Example 1

[4,3,1,2]
[2,4,9,5]
[[4,1],[1,3],[2,5]]

Example 2

[3,2,5]
[2,3,4]
[[4,4],[3,2],[1,1]]

Example 3

[2,1]
[2,3]
[[3,3]]

Core Insight

What unlocks the optimal approach

Sort (x, y) tuples and queries by x-coordinate descending. Don’t forget to index queries before sorting so that you can answer them in the correct order.
Before answering a query (min_x, min_y), add all (x, y) pairs with x >= min_x to some data structure.
Use a monotone descending map to store (y, x + y) pairs. A monotone map has ascending keys and descending values. When inserting a pair (y, x + y), remove all pairs (y', x' + y') with y' < y and x' + y' <= x + y.
To find the insertion position use binary search (built-in in many languages).
When querying for max (x + y) over y >= y', use binary search to find the first pair (y, x + y) with y >= y'. It will have the maximum value of x + y because the map has monotone descending values.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2736: Maximum Sum 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;
    }
}

// Accepted solution for LeetCode #2736: Maximum Sum Queries
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
}

# Accepted solution for LeetCode #2736: Maximum Sum Queries
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

// Accepted solution for LeetCode #2736: Maximum Sum Queries
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2736: Maximum Sum 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;
//     }
// }

// Accepted solution for LeetCode #2736: Maximum Sum Queries
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O((n + m)

Space

O(n + m)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.