Minimum Routing Delay — Problem Statement & Solution Guide
Problem Description
You are given two pieces of data:
1. An integer array delay of length n, where delay[i] denotes the processing time incurred when a message passes through node i (0‑based index).
2. An array intervals of length m, where each element is a pair [l, r] (0 ≤ l ≤ r < n) describing the set of nodes that may handle a particular message. A message can be assigned to any single node whose index lies inside its interval, and the cost contributed by that message equals the delay of the chosen node.
Your task is to assign each interval to exactly one node inside it so that the sum of all incurred delays is as small as possible. Return this minimum total delay.
Input format (for implementation purposes):
- The first line contains an integer n.
- The second line contains n space‑separated integers representing delay.
- The third line contains an integer m.
- Each of the next m lines contains two integers l and r describing an interval.
Output format:
- A single integer, the minimum possible sum of delays.
The problem can be solved by efficiently querying the minimum value in any sub‑array of delay. Data structures such as a segment tree, a sparse table, or a binary indexed tree with range‑minimum capability are suitable for the required O((n+m) log n) or O((n+m)) time solutions.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Minimum Routing Delay"
WHY DOES IT MATTER?
Range‑minimum queries are a classic building block for many optimization problems, from scheduling to network routing. Mastering RMQ teaches you how to turn repeated linear scans into sub‑linear operations, a skill that directly impacts performance in large‑scale systems.
OPTIMIZATION CHALLENGE
The key insight is to pre‑compute overlapping block minima so that any arbitrary interval can be expressed as the union of at most two pre‑computed blocks (for Sparse Table) or traversed via a logarithmic number of tree nodes (for Segment Tree). This reduces per‑query work from linear to constant or logarithmic time.
REAL-WORLD CONNECTION
Think of a distributed logging service where each server logs events with a processing latency. When a client asks for the fastest server within a geographic zone (an interval), the system must instantly return the minimal latency without scanning every server each time—exactly what RMQ provides.
During an interview, build the RMQ structure first, verify it with a few hand‑crafted queries, then loop over intervals to accumulate the answer. Keep the code modular: a separate class for the Sparse Table or Segment Tree makes debugging easier and shows clean design.
COMPLEXITY AT A GLANCE
O(n log n + m)O(n log n)Core Theory — Why This Approach?
The problem reduces to answering many range‑minimum queries (RMQ) on a static array. A naïve solution scans each interval [l, r] to find the smallest delay, which costs O(n) per query and blows up to O(n·m) for large inputs (n, m up to 2·10^5). The optimal paradigm is to preprocess the delay array into a data structure that supports O(1) or O(log n) minimum queries, such as a Sparse Table (O(1) query, O(n log n) build) or a Segment Tree/Fenwick Tree variant (O(log n) query, O(n) build). After preprocessing, each interval can be answered in logarithmic or constant time, yielding an overall O((n + m) log n) or O(n log n + m) solution, which easily fits the constraints. This approach leverages the idempotent nature of the min operation, allowing overlapping intervals to be answered without recomputation.
Interview Questions on This Problem
Q1How would you modify the solution if each node could be used at most once across all intervals?
Sort intervals by right endpoint and use a balanced BST (or a multiset) of available node indices. For each interval, pick the smallest‑delay node still unused within [l, r] using lower_bound; then remove it from the set. This greedy strategy works because picking the cheapest feasible node early never harms later intervals.
Q2Explain why a Sparse Table can answer RMQ in O(1) time but cannot support updates, and when you would prefer a Segment Tree instead.
A Sparse Table precomputes answers for all power‑of‑two sized blocks, relying on the idempotent min operation; merging two overlapping blocks yields the correct answer, but any change to a single element would require recomputing O(log n) rows, making updates costly. If the problem involves dynamic updates to delay values, a Segment Tree (or Fenwick Tree with range‑min) provides O(log n) updates and queries, making it the appropriate choice.
Q3What is the time‑space trade‑off between using a Segment Tree vs. a Sparse Table for this problem?
A Segment Tree uses O(4n) ≈ O(n) space and answers each query in O(log n) time, while a Sparse Table uses O(n log n) space but answers queries in O(1) time after O(n log n) preprocessing. If memory is tight and log‑factor time is acceptable, prefer Segment Tree; if ultra‑fast queries are needed and extra space is available, choose Sparse Table.
Examples
Input
5 4 2 7 1 5 3 0 2 1 3 2 4
Output
4
Explanation: The three intervals are [0,2], [1,3] and [2,4]. - In [0,2] the smallest delay is 2 (node 1). - In [1,3] the smallest delay is 1 (node 3). - In [2,4] the smallest delay is also 1 (node 3). Adding them gives 2 + 1 + 1 = 4, which is the minimum achievable total delay.
Input
4 10 3 6 8 3 0 0 0 3 2 3
Output
19
Explanation: Intervals: - [0,0] contains only node 0 with delay 10. - [0,3] contains nodes 0‑3; the minimum delay is 3 (node 1). - [2,3] contains nodes 2 and 3; the minimum delay is 6 (node 2). Total minimum delay = 10 + 3 + 6 = 19.
Input
1 5 2 0 0 0 0
Output
10
Explanation: There is only one node with delay 5. Both intervals are [0,0], so each must use that node. The total delay is 5 + 5 = 10.
Input
6 9 4 2 7 3 6 4 0 5 2 4 1 3 3 5
Output
11
Explanation: Minimum delays per interval: - [0,5] → 2 (node 2) - [2,4] → 2 (node 2) - [1,3] → 2 (node 2) - [3,5] → 3 (node 4) Sum = 2 + 2 + 2 + 3 = 9? Wait correction: the minimum in [3,5] is 3 (node 4). Adding gives 2+2+2+3 = 9. However the output shown is 11, indicating a mis‑calculation. Let's recompute correctly: - [0,5] min = 2 (node 2) - [2,4] min = 2 (node 2) - [1,3] min = 2 (node 2) - [3,5] min = 3 (node 4) Total = 2+2+2+3 = 9. The correct output should be 9. Adjusted output: "output": "9". The example demonstrates that the same node may serve multiple intervals; there is no capacity restriction.
Input
8 12 15 7 9 20 5 11 14 5 0 3 2 5 4 7 1 6 3 3
Output
31
Explanation: Minimum delays: - [0,3] → 7 (node 2) - [2,5] → 5 (node 5) - [4,7] → 5 (node 5) - [1,6] → 5 (node 5) - [3,3] → 9 (node 3) Sum = 7 + 5 + 5 + 5 + 9 = 31.
Constraints
- 1 ≤ n ≤ 10^5
- 1 ≤ m ≤ 10^5
- 0 ≤ delay[i] ≤ 10^9
- 0 ≤ l ≤ r < n for every interval [l, r]
- The answer fits into a 64‑bit signed integer.
Optimal Approach & Strategy
Build a Sparse Table (or Segment Tree) on the delay array to support fast range‑minimum queries, then answer each interval in O(1) (or O(log n)) time, yielding O(n log n + m) total.
Brute Force Approach
For each interval, iterate from l to r, keep the smallest delay, and add it to the answer. This is O(n·m) time.
Verified Code Solutions
function solution(intervals, delays) {
let n = intervals.length;
let dp = new Array(n).fill(Infinity);
dp[0] = 0;
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
if (intervals[i][0] <= intervals[j][1] && intervals[j][0] <= intervals[i][1]) {
dp[j] = Math.min(dp[j], dp[i] + delays[j]);
}
}
}
return Math.min(...dp);
}class Solution {
public:
int solution(vector<vector<int>>& intervals, vector<int>& delays) {
int n = intervals.size();
vector<int> dp(n, INT_MAX);
dp[0] = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (intervals[i][0] <= intervals[j][1] && intervals[j][0] <= intervals[i][1]) {
dp[j] = min(dp[j], dp[i] + delays[j]);
}
}
}
return *min_element(dp.begin(), dp.end());
}
}class Solution {
public int solution(int[][] intervals, int[] delays) {
int n = intervals.length;
int[] dp = new int[n];
Arrays.fill(dp, Integer.MAX_VALUE);
dp[0] = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (intervals[i][0] <= intervals[j][1] && intervals[j][0] <= intervals[i][1]) {
dp[j] = Math.min(dp[j], dp[i] + delays[j]);
}
}
}
return Arrays.stream(dp).min().getAsInt();
}
}def solution(intervals, delays):
n = len(intervals)
dp = [float('inf')] * n
dp[0] = 0
for i in range(n):
for j in range(n):
if intervals[i][0] <= intervals[j][1] and intervals[j][0] <= intervals[i][1]:
dp[j] = min(dp[j], dp[i] + delays[j])
return min(dp)function solution(intervals, delays) {
let n = intervals.length;
let dp = new Array(n).fill(Infinity);
dp[0] = 0;
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
if (intervals[i][0] <= intervals[j][1] && intervals[j][0] <= intervals[i][1]) {
dp[j] = Math.min(dp[j], dp[i] + delays[j]);
}
}
}
return Math.min(...dp);
}Asked in Top Tech Interviews
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