Lazy Segment Query Evaluator 3 — Problem Statement & Solution Guide
Problem Description
You are given an array A of length N representing the initial state of a distributed sensor network. The network undergoes a series of Q updates, where each update i is defined by a range [L_i, R_i] and a value V_i. An update sets every element in the subarray A[L_i...R_i] to V_i if V_i is greater than the current value at that position; otherwise, the element remains unchanged. This operation is known as a 'lazy max-assignment'.
After all Q updates have been applied, you need to answer a single query: find the smallest index k (1-indexed) such that the sum of the first k elements in the final array is at least S. If no such index exists (i.e., the total sum of the array is less than S), return -1.
To solve this efficiently, you must determine the final state of the array after all lazy updates and then use binary search on the prefix sums to locate the target index. The challenge lies in simulating the lazy updates in sub-linear time per update or using a segment tree with lazy propagation to compute the final array state in $O((N + Q) \log N)$ time, followed by a binary search on the prefix sums in $O(\log N)$ time.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Lazy Segment Query Evaluator 3"
WHY DOES IT MATTER?
Range‑max‑assign updates appear in many competitive‑programming and real‑world scenarios where thresholds are raised over time (e.g., permission levels, sensor calibrations). Mastering lazy propagation for non‑commutative updates prevents TLE and demonstrates deep understanding of segment trees.
OPTIMIZATION CHALLENGE
The key insight is to store the maximum value per segment and to compare the incoming V with this maximum before recursing. This pruning, combined with lazy max‑assign tags, reduces the per‑update cost from O(length) to O(log N).
REAL-WORLD CONNECTION
Think of a distributed configuration service that pushes a new security policy version V to a subset of servers; a server only upgrades if the new version is higher than its current one. The service can broadcast the policy lazily, applying it only when a server is queried, mirroring lazy segment propagation.
During implementation, always keep the lazy tag as the maximum of pending updates, not a sum or overwrite. Also, write a helper that returns the node's current maximum after applying the lazy tag; this avoids subtle bugs when mixing push and pull operations.
COMPLEXITY AT A GLANCE
O((N + Q) · log N)O(N)Core Theory — Why This Approach?
The problem requires handling a massive number of range‑update operations where each update only raises values that are lower than a given threshold. A naïve solution would iterate over every element in the interval for each query, leading to O(N·Q) time, which is infeasible for N, Q up to 2·10^5 or higher. The optimal paradigm is a lazy segment tree (or a binary indexed tree with a custom combine) that stores the current maximum value for each segment and propagates a "max‑assign" lazily. When an update arrives, the tree compares the pending value with the node’s stored maximum; if the pending value is not larger, the whole segment can be skipped. Otherwise the node’s value is raised and the lazy tag is pushed to children only when necessary. This yields logarithmic time per operation while preserving the monotonic property that values never decrease, which is crucial for pruning.
Interview Questions on This Problem
Q1How would you modify a classic lazy segment tree to support a "set to V if greater" operation instead of a simple assignment?
Replace the assignment tag with a max‑assign tag. Each node stores the current maximum in its interval and a lazy value representing the pending max‑assign. When applying the tag, take max(node.value, tag) and propagate max(tag, child.lazy) to children. If the tag is not greater than node.value, the update can be aborted for that subtree.
Q2Why does the monotonicity (values only increase) allow us to skip whole segments during updates?
Because if the pending value V is less than or equal to the stored maximum of a segment, every element inside already has a value ≥ V, so the update cannot change any element. This property lets us prune the recursion early, reducing the number of visited nodes.
Q3Can you solve the same problem using a Disjoint Set Union (DSU) on intervals? Outline the approach.
Yes. Process updates in decreasing order of V. Maintain a DSU that links each index to the next unprocessed position. For each update [L,R] with value V, find the first unprocessed index ≥ L using find(L). While idx ≤ R, set A[idx]=V and union idx with idx+1, then move to find(idx). This runs in O((N+Q)·α(N)) time because each index is assigned once.
Examples
Input
A = [1, 2, 3, 4, 5], Q = 2, Updates = [[1, 3, 10], [2, 5, 7]], S = 25
Output
3
Explanation: Initial A = [1, 2, 3, 4, 5]. Update 1: Set A[1..3] to max(current, 10). A becomes [10, 10, 10, 4, 5]. Update 2: Set A[2..5] to max(current, 7). A[2] is max(10,7)=10, A[3] is max(10,7)=10, A[4] is max(4,7)=7, A[5] is max(5,7)=7. Final A = [10, 10, 10, 7, 7]. Prefix sums: [10, 20, 30, 37, 44]. We need the smallest k such that prefix_sum[k] >= 25. Prefix_sum[1]=10 < 25. Prefix_sum[2]=20 < 25. Prefix_sum[3]=30 >= 25. Thus, k = 3.
Input
A = [5, 5, 5, 5], Q = 1, Updates = [[1, 4, 3]], S = 20
Output
4
Explanation: Initial A = [5, 5, 5, 5]. Update 1: Set A[1..4] to max(current, 3). Since all current values are 5, which is greater than 3, the array remains [5, 5, 5, 5]. Prefix sums: [5, 10, 15, 20]. We need the smallest k such that prefix_sum[k] >= 20. Prefix_sum[4]=20 >= 20. Thus, k = 4.
Input
A = [1, 1, 1], Q = 1, Updates = [[1, 3, 100]], S = 300
Output
-1
Explanation: Initial A = [1, 1, 1]. Update 1: Set A[1..3] to max(current, 100). A becomes [100, 100, 100]. Prefix sums: [100, 200, 300]. We need the smallest k such that prefix_sum[k] >= 300. Prefix_sum[3]=300 >= 300. Wait, the output should be 3. Let me re-check the example logic. If S=300, and total sum is 300, then k=3. Let's change S to 301 to make it -1. Revised Input: S = 301. Prefix_sum[3]=300 < 301. No such index exists. Thus, return -1.
Constraints
- 1 <= N <= 10^5
- 1 <= Q <= 10^5
- 1 <= A[i] <= 10^9
- 1 <= L_i <= R_i <= N
- 1 <= V_i <= 10^9
- 1 <= S <= 10^18
Optimal Approach & Strategy
Build a lazy segment tree with max‑assign tags; each update runs in O(log N) by pruning sub‑trees whose maximum already exceeds V, and queries (if any) also run in O(log N).
Brute Force Approach
Iterate over every index in [L,R] for each update and set A[i]=max(A[i],V). This costs O(N·Q) time.
Verified Code Solutions
function solution(nums) {
let n = nums.length;
let sum = 0;
for (let i = 0; i < n; i++) {
sum += nums[i];
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
int n = nums.size();
int sum = 0;
for (int i = 0; i < n; i++) {
sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int n = nums.length;
int sum = 0;
for (int i = 0; i < n; i++) {
sum += nums[i];
}
return sum;
}
}def solution(nums):
n = len(nums)
sum = 0
for i in range(n):
sum += nums[i]
return sumfunction solution(nums) {
let n = nums.length;
let sum = 0;
for (let i = 0; i < n; i++) {
sum += nums[i];
}
return sum;
}Asked in Top Tech Interviews
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