LeetCode #2569 — HARD

Handling Sum Queries After Update

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

Solve on LeetCode
The Problem

Problem Statement

You are given two 0-indexed arrays nums1 and nums2 and a 2D array queries of queries. There are three types of queries:

  1. For a query of type 1, queries[i] = [1, l, r]. Flip the values from 0 to 1 and from 1 to 0 in nums1 from index l to index r. Both l and r are 0-indexed.
  2. For a query of type 2, queries[i] = [2, p, 0]. For every index 0 <= i < n, set nums2[i] = nums2[i] + nums1[i] * p.
  3. For a query of type 3, queries[i] = [3, 0, 0]. Find the sum of the elements in nums2.

Return an array containing all the answers to the third type queries.

Example 1:

Input: nums1 = [1,0,1], nums2 = [0,0,0], queries = [[1,1,1],[2,1,0],[3,0,0]]
Output: [3]
Explanation: After the first query nums1 becomes [1,1,1]. After the second query, nums2 becomes [1,1,1], so the answer to the third query is 3. Thus, [3] is returned.

Example 2:

Input: nums1 = [1], nums2 = [5], queries = [[2,0,0],[3,0,0]]
Output: [5]
Explanation: After the first query, nums2 remains [5], so the answer to the second query is 5. Thus, [5] is returned.

Constraints:

  • 1 <= nums1.length,nums2.length <= 105
  • nums1.length = nums2.length
  • 1 <= queries.length <= 105
  • queries[i].length = 3
  • 0 <= l <= r <= nums1.length - 1
  • 0 <= p <= 106
  • 0 <= nums1[i] <= 1
  • 0 <= nums2[i] <= 109
Patterns Used
📐Segment Tree→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given two 0-indexed arrays nums1 and nums2 and a 2D array queries of queries. There are three types of queries: For a query of type 1, queries[i] = [1, l, r]. Flip the values from 0 to 1 and from 1 to 0 in nums1 from index l to index r. Both l and r are 0-indexed. For a query of type 2, queries[i] = [2, p, 0]. For every index 0 <= i < n, set nums2[i] = nums2[i] + nums1[i] * p. For a query of type 3, queries[i] = [3, 0, 0]. Find the sum of the elements in nums2. Return an array containing all the answers to the third type queries.

Baseline thinking

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

Pattern signal: Array · Segment Tree

Example 1

[1,0,1]
[0,0,0]
[[1,1,1],[2,1,0],[3,0,0]]

Example 2

[1]
[5]
[[2,0,0],[3,0,0]]
Step 02

Core Insight

What unlocks the optimal approach

  • Use the Lazy Segment Tree to process the queries quickly.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #2569: Handling Sum Queries After Update
class Node {
    int l, r;
    int s, lazy;
}

class SegmentTree {
    private Node[] tr;
    private int[] nums;

    public SegmentTree(int[] nums) {
        int n = nums.length;
        this.nums = nums;
        tr = new Node[n << 2];
        for (int i = 0; i < tr.length; ++i) {
            tr[i] = new Node();
        }
        build(1, 1, n);
    }

    private void build(int u, int l, int r) {
        tr[u].l = l;
        tr[u].r = r;
        if (l == r) {
            tr[u].s = nums[l - 1];
            return;
        }
        int mid = (l + r) >> 1;
        build(u << 1, l, mid);
        build(u << 1 | 1, mid + 1, r);
        pushup(u);
    }

    public void modify(int u, int l, int r) {
        if (tr[u].l >= l && tr[u].r <= r) {
            tr[u].lazy ^= 1;
            tr[u].s = tr[u].r - tr[u].l + 1 - tr[u].s;
            return;
        }
        pushdown(u);
        int mid = (tr[u].l + tr[u].r) >> 1;
        if (l <= mid) {
            modify(u << 1, l, r);
        }
        if (r > mid) {
            modify(u << 1 | 1, l, r);
        }
        pushup(u);
    }

    public int query(int u, int l, int r) {
        if (tr[u].l >= l && tr[u].r <= r) {
            return tr[u].s;
        }
        pushdown(u);
        int mid = (tr[u].l + tr[u].r) >> 1;
        int res = 0;
        if (l <= mid) {
            res += query(u << 1, l, r);
        }
        if (r > mid) {
            res += query(u << 1 | 1, l, r);
        }
        return res;
    }

    private void pushup(int u) {
        tr[u].s = tr[u << 1].s + tr[u << 1 | 1].s;
    }

    private void pushdown(int u) {
        if (tr[u].lazy == 1) {
            int mid = (tr[u].l + tr[u].r) >> 1;
            tr[u << 1].s = mid - tr[u].l + 1 - tr[u << 1].s;
            tr[u << 1].lazy ^= 1;
            tr[u << 1 | 1].s = tr[u].r - mid - tr[u << 1 | 1].s;
            tr[u << 1 | 1].lazy ^= 1;
            tr[u].lazy ^= 1;
        }
    }
}

class Solution {
    public long[] handleQuery(int[] nums1, int[] nums2, int[][] queries) {
        SegmentTree tree = new SegmentTree(nums1);
        long s = 0;
        for (int x : nums2) {
            s += x;
        }
        int m = 0;
        for (var q : queries) {
            if (q[0] == 3) {
                ++m;
            }
        }
        long[] ans = new long[m];
        int i = 0;
        for (var q : queries) {
            if (q[0] == 1) {
                tree.modify(1, q[1] + 1, q[2] + 1);
            } else if (q[0] == 2) {
                s += 1L * q[1] * tree.query(1, 1, nums2.length);
            } else {
                ans[i++] = s;
            }
        }
        return ans;
    }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n + m × log n)
Space
O(n)

Approach Breakdown

BRUTE FORCE
O(n × q) time
O(1) space

For each of q queries, scan the entire range to compute the aggregate (sum, min, max). Each range scan takes up to O(n) for a full-array query. With q queries: O(n × q) total. Point updates are O(1) but queries dominate.

SEGMENT TREE
O(n + q log n) time
O(n) space

Building the tree is O(n). Each query or update traverses O(log n) nodes (tree height). For q queries: O(n + q log n) total. Space is O(4n) ≈ O(n) for the tree array. Lazy propagation adds O(1) per node for deferred updates.

Shortcut: Build O(n), query/update O(log n) each. When you need both range queries AND point updates.
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.

Related Problems
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