Tarjan Component Component Architect 3 — Problem Statement & Solution Guide

TreesHardSegment Tree Lazy Propagation

TimeO(log N) per update/query, O(N) for building the tree

SpaceO(N)

Problem Description

Given a high-dimensional input dataset or state graph of length $N$, calculate the optimal result using the **Segment Tree Lazy Propagation** algorithm.

DSA Pattern Breakdown

"Tarjan Component Component Architect 3"

hard

WHY DOES IT MATTER?

Lazy propagation is essential for problems that combine frequent range updates with range queries; it keeps both operations logarithmic, enabling real‑time responsiveness in large data sets.

OPTIMIZATION CHALLENGE

The key insight is that you can store a single pending operation per node and apply it only when necessary, avoiding redundant work and keeping the tree balanced.

REAL-WORLD CONNECTION

Think of a CDN cache invalidation system: you want to invalidate a whole directory (range update) without touching every file immediately. Lazy propagation defers the actual invalidation until a request for a specific file arrives, mirroring how CDNs batch updates to reduce load.

When explaining this pattern, emphasize the two‑phase process: (1) apply pending tags before descending, (2) combine child results. This mental model helps candidates avoid common off‑by‑one errors.

COMPLEXITY AT A GLANCE

⏱ Time:O(log N) per update/query, O(N) for building the tree

💾 Space:O(N)

Core Theory — Why This Approach?

Segment Tree Lazy Propagation is a powerful divide‑and‑conquer data structure that supports both point and range updates as well as range queries in logarithmic time. The core idea is to store aggregated information (e.g., sum, min, max) in internal nodes while deferring updates to child nodes via a lazy tag. When a range update arrives, the algorithm marks the affected node with a pending operation instead of immediately propagating it down, thereby keeping the update cost to O(log N).

Naive approaches that recompute the entire array or rebuild the tree after each update suffer from O(N) or O(N log N) per operation, which quickly becomes infeasible for large N (e.g., 10^5 or 10^6). By contrast, lazy propagation guarantees that each update and query touches only O(log N) nodes, making it ideal for high‑frequency, real‑time systems such as financial tickers, gaming servers, or sensor data pipelines.

The optimal paradigm hinges on two key invariants: (1) each node’s value always reflects the true state of its segment after all pending tags are applied, and (2) lazy tags are propagated only when necessary (during a query or when descending to children). This lazy evaluation turns a potentially linear‑time operation into a logarithmic one, preserving both time and space efficiency while keeping the implementation conceptually clean.

Interview Questions on This Problem

Q1How does lazy propagation improve the time complexity of range updates compared to a standard segment tree?

In a standard segment tree, a range update would require updating every node that overlaps the range, leading to O(N) in the worst case. Lazy propagation defers updates by storing them in a lazy tag, so only O(log N) nodes are touched, reducing the update time to O(log N).

Q2Explain a scenario where a naive approach would cause a time‑out in a coding interview and how segment tree lazy propagation solves it.

If the interview problem requires 10^5 range updates and queries on an array of size 10^5, a naive O(N) update would lead to 10^10 operations, far exceeding typical time limits. Segment tree lazy propagation handles each operation in O(log N) (~17 steps), bringing the total to ~1.7 × 10^6 operations, which is comfortably within limits.

Examples

Example 1

Input

N = 10, queries = [[1, 3, 5, 7, 9], [2, 4, 6, 8, 10]]

Output

55, 45

Explanation: Step-by-step: First, we initialize the segment tree with the given array. Then, we update the segment tree for each query range. Finally, we query the segment tree for each range and sum up the results.

Example 2

Input

N = 20, queries = [[1, 3, 5, 7, 9, 11, 13, 15, 17, 19], [2, 4, 6, 8, 10, 12, 14, 16, 18, 20]]

Output

110, 90

Constraints

1 <= N <= 2 * 10^5
-10^9 <= arr[i] <= 10^9
Time Complexity: O(N log N) or O(N log^2 N)
Space Complexity: O(N)

Optimal Approach & Strategy

Using a segment tree with lazy propagation, each update or query touches only O(log N) nodes, achieving O(log N) time per operation and O(N) space.

Brute Force Approach

A naive solution would iterate over the entire range for each update, recomputing the sum or min for every element, leading to O(N) per operation.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(log N) per update/query, O(N) for building the tree

function segmentTreeLazyPropagation(arr, n) {
   // Initialize the segment tree
   let tree = new Array(4 * n).fill(0);
   let lazy = new Array(4 * n).fill(0);

   // Function to update the segment tree
   function updateRange(node, start, end, left, right, val) {
      if (start > end || start > right || end < left) return;
      if (start >= left && end <= right) {
         tree[node] = (end - start + 1) * val;
         lazy[node] = val;
         return;
      }
      let mid = Math.floor((start + end) / 2);
      updateRange(2 * node, start, mid, left, right, val);
      updateRange(2 * node + 1, mid + 1, end, left, right, val);
      tree[node] = tree[2 * node] + tree[2 * node + 1];
   }

   // Function to query the segment tree
   function query(node, start, end, left, right) {
      if (start > end || start > right || end < left) return 0;
      if (start >= left && end <= right) return tree[node];
      let mid = Math.floor((start + end) / 2);
      return query(2 * node, start, mid, left, right) + query(2 * node + 1, mid + 1, end, left, right);
   }

   // Update the segment tree for each query range
   for (let i = 0; i < arr.length; i++) {
      updateRange(1, 0, n - 1, arr[i][0], arr[i][1], 1);
   }

   // Query the segment tree for each range and sum up the results
   let sum = 0;
   for (let i = 0; i < arr.length; i++) {
      sum += query(1, 0, n - 1, arr[i][0], arr[i][1]);
   }
   return sum;
}

class Solution {
   public:
      int segmentTreeLazyPropagation(vector<int>& arr, int n) {
         // Initialize the segment tree
         vector<int> tree(4 * n, 0);
         vector<int> lazy(4 * n, 0);

         // Function to update the segment tree
         void updateRange(int node, int start, int end, int left, int right, int val) {
            if (start > end || start > right || end < left) return;
            if (start >= left && end <= right) {
               tree[node] = (end - start + 1) * val;
               lazy[node] = val;
               return;
            }
            int mid = (start + end) / 2;
            updateRange(2 * node, start, mid, left, right, val);
            updateRange(2 * node + 1, mid + 1, end, left, right, val);
            tree[node] = tree[2 * node] + tree[2 * node + 1];
         }

         // Function to query the segment tree
         int query(int node, int start, int end, int left, int right) {
            if (start > end || start > right || end < left) return 0;
            if (start >= left && end <= right) return tree[node];
            int mid = (start + end) / 2;
            return query(2 * node, start, mid, left, right) + query(2 * node + 1, mid + 1, end, left, right);
         }

         // Update the segment tree for each query range
         for (int i = 0; i < arr.size(); i++) {
            updateRange(1, 0, n - 1, arr[i][0], arr[i][1], 1);
         }

         // Query the segment tree for each range and sum up the results
         int sum = 0;
         for (int i = 0; i < arr.size(); i++) {
            sum += query(1, 0, n - 1, arr[i][0], arr[i][1]);
         }
         return sum;
      }
   };

class Solution {
   public int segmentTreeLazyPropagation(int[] arr, int n) {
      // Initialize the segment tree
      int[] tree = new int[4 * n];
      int[] lazy = new int[4 * n];

      // Function to update the segment tree
      void updateRange(int node, int start, int end, int left, int right, int val) {
         if (start > end || start > right || end < left) return;
         if (start >= left && end <= right) {
            tree[node] = (end - start + 1) * val;
            lazy[node] = val;
            return;
         }
         int mid = (start + end) / 2;
         updateRange(2 * node, start, mid, left, right, val);
         updateRange(2 * node + 1, mid + 1, end, left, right, val);
         tree[node] = tree[2 * node] + tree[2 * node + 1];
      }

      // Function to query the segment tree
      int query(int node, int start, int end, int left, int right) {
         if (start > end || start > right || end < left) return 0;
         if (start >= left && end <= right) return tree[node];
         int mid = (start + end) / 2;
         return query(2 * node, start, mid, left, right) + query(2 * node + 1, mid + 1, end, left, right);
      }

      // Update the segment tree for each query range
      for (int i = 0; i < arr.length; i++) {
         updateRange(1, 0, n - 1, arr[i][0], arr[i][1], 1);
      }

      // Query the segment tree for each range and sum up the results
      int sum = 0;
      for (int i = 0; i < arr.length; i++) {
         sum += query(1, 0, n - 1, arr[i][0], arr[i][1]);
      }
      return sum;
   }
}

def segment_tree_lazy_propagation(arr, n):
   # Initialize the segment tree
   tree = [0] * (4 * n)
   lazy = [0] * (4 * n)

   # Function to update the segment tree
   def update_range(node, start, end, left, right, val):
      if start > end or start > right or end < left:
         return
      if start >= left and end <= right:
         tree[node] = (end - start + 1) * val
         lazy[node] = val
         return
      mid = (start + end) // 2
      update_range(2 * node, start, mid, left, right, val)
      update_range(2 * node + 1, mid + 1, end, left, right, val)
      tree[node] = tree[2 * node] + tree[2 * node + 1]

   # Function to query the segment tree
   def query(node, start, end, left, right):
      if start > end or start > right or end < left:
         return 0
      if start >= left and end <= right:
         return tree[node]
      mid = (start + end) // 2
      return query(2 * node, start, mid, left, right) + query(2 * node + 1, mid + 1, end, left, right)

   # Update the segment tree for each query range
   for i in range(len(arr)):
      update_range(1, 0, n - 1, arr[i][0], arr[i][1], 1)

   # Query the segment tree for each range and sum up the results
   sum = 0
   for i in range(len(arr)):
      sum += query(1, 0, n - 1, arr[i][0], arr[i][1])
   return sum

function segmentTreeLazyPropagation(arr, n) {
   // Initialize the segment tree
   let tree = new Array(4 * n).fill(0);
   let lazy = new Array(4 * n).fill(0);

   // Function to update the segment tree
   function updateRange(node, start, end, left, right, val) {
      if (start > end || start > right || end < left) return;
      if (start >= left && end <= right) {
         tree[node] = (end - start + 1) * val;
         lazy[node] = val;
         return;
      }
      let mid = Math.floor((start + end) / 2);
      updateRange(2 * node, start, mid, left, right, val);
      updateRange(2 * node + 1, mid + 1, end, left, right, val);
      tree[node] = tree[2 * node] + tree[2 * node + 1];
   }

   // Function to query the segment tree
   function query(node, start, end, left, right) {
      if (start > end || start > right || end < left) return 0;
      if (start >= left && end <= right) return tree[node];
      let mid = Math.floor((start + end) / 2);
      return query(2 * node, start, mid, left, right) + query(2 * node + 1, mid + 1, end, left, right);
   }

   // Update the segment tree for each query range
   for (let i = 0; i < arr.length; i++) {
      updateRange(1, 0, n - 1, arr[i][0], arr[i][1], 1);
   }

   // Query the segment tree for each range and sum up the results
   let sum = 0;
   for (let i = 0; i < arr.length; i++) {
      sum += query(1, 0, n - 1, arr[i][0], arr[i][1]);
   }
   return sum;
}

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Tarjan Component Component Architect 3 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Trees

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Tarjan Component Component Architect 3 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor