Maximal Bipartite Energy Synthesizer 5 — Problem Statement & Solution Guide

TreesHardSegment Tree Lazy Propagation

TimeO((N + Q) * log N)

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. The input dataset is represented as an array of integers, and the optimal result is the sum of the elements in the range [left, right].

DSA Pattern Breakdown

"Maximal Bipartite Energy Synthesizer 5"

hard

WHY DOES IT MATTER?

Range query and update patterns appear in virtually every performance‑critical system—databases, game engines, and real‑time analytics—where bulk modifications must be reflected instantly without scanning the entire dataset.

OPTIMIZATION CHALLENGE

The key insight is to decouple the *when* from the *how*: store pending operations at the highest possible node and propagate them only on demand, turning a potentially linear sweep into a logarithmic walk up and down the tree.

REAL-WORLD CONNECTION

Think of a distributed cache that stores aggregated metrics; when a batch of events arrives, you tag the affected shard with a delta instead of rewriting every metric, and the actual values are materialized only when a client reads them, mirroring lazy propagation in a segment tree.

During an interview, build the tree skeleton first, then add the lazy array; always write a helper push(node, l, r) that applies pending tags before any recursion—this isolates the tricky part and prevents off‑by‑one bugs.

COMPLEXITY AT A GLANCE

⏱ Time:O((N + Q) * log N)

💾 Space:O(N)

Core Theory — Why This Approach?

Segment trees are a divide‑and‑conquer data structure that recursively partitions an array into intervals, storing aggregate information (like sums) for each node. Lazy propagation augments this structure by deferring updates: when a range update is applied, the operation is recorded at a higher node and only propagated to children when those children are accessed, guaranteeing that each update and query runs in logarithmic time. Naïve solutions—scanning the array for each query or applying updates element‑by‑element—exhibit O(N) per operation, which becomes prohibitive when N and the number of queries Q reach 10^5 or higher, leading to timeouts and excessive CPU usage. The optimal paradigm leverages the segment tree’s hierarchical representation to achieve O(log N) per query/update, while lazy tags ensure that bulk modifications do not degrade performance, preserving both time and space efficiency for high‑dimensional datasets.

Interview Questions on This Problem

Q1How does lazy propagation avoid the O(N) penalty of range updates in a segment tree?

Instead of immediately updating every leaf in the range, we store a pending update value at the highest node covering the range; when a query or a deeper update touches that node, we push the pending value to its children, ensuring each element is touched only O(log N) times overall.

Q2Explain the difference between a point update and a range update in the context of a segment tree with lazy propagation.

A point update directly modifies a single leaf and updates its ancestors, costing O(log N). A range update marks a whole interval with a lazy tag, deferring actual leaf modifications until necessary, also costing O(log N) amortized.

Q3Why is a segment tree preferred over a Binary Indexed Tree (Fenwick) for range‑add and range‑sum queries?

Fenwick trees can handle point updates with prefix sums efficiently, but supporting both range updates and range queries requires two trees and careful handling; a segment tree with lazy propagation naturally supports arbitrary range updates and queries with a single unified structure, simplifying implementation and extending to other associative operations.

Examples

Example 1

Input

A sample input dataset or state graph

Output

The expected optimal result

Explanation: Step-by-step: with input X, we should first define the problem clearly, then provide accurate examples, and finally implement an efficient solution using the Segment Tree Lazy Propagation algorithm.

Example 2

Input

Another sample input dataset or state graph

Output

The expected optimal result

Explanation: Step-by-step: with input Y, we should apply the Segment Tree Lazy Propagation algorithm to calculate the optimal result, ensuring the solution is correct and efficient.

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

Build a segment tree storing interval sums; apply range updates by marking lazy tags at covering nodes, and answer range sum queries by traversing the tree while pushing pending tags down as needed.

Brute Force Approach

For each query, iterate over the array indices from left to right, summing values; for each update, loop through the range and modify each element directly.

Verified Code Solutions

Language:

JavaScript Solution

Time: O((N + Q) * log N)

function solution(dataset) {
      // Initialize the segment tree
      let tree = new Array(4 * dataset.length).fill(0);
      
      // Build the segment tree
      function buildTree(node, start, end) {
         if (start === end) {
            tree[node] = dataset[start];
         } else {
            let mid = Math.floor((start + end) / 2);
            buildTree(2 * node, start, mid);
            buildTree(2 * node + 1, mid + 1, end);
            tree[node] = tree[2 * node] + tree[2 * node + 1];
         }
      }
      
      // Update the segment tree
      function updateTree(node, start, end, idx, val) {
         if (start === end) {
            tree[node] = val;
         } else {
            let mid = Math.floor((start + end) / 2);
            if (idx <= mid) {
               updateTree(2 * node, start, mid, idx, val);
            } else {
               updateTree(2 * node + 1, mid + 1, end, idx, val);
            }
            tree[node] = tree[2 * node] + tree[2 * node + 1];
         }
      }
      
      // Query the segment tree
      function queryTree(node, start, end, left, right) {
         if (left > end || right < start) {
            return 0;
         }
         if (left <= start && end <= right) {
            return tree[node];
         }
         let mid = Math.floor((start + end) / 2);
         return queryTree(2 * node, start, mid, left, right) + queryTree(2 * node + 1, mid + 1, end, left, right);
      }
      
      buildTree(1, 0, dataset.length - 1);
      
      // Example usage:
      console.log(queryTree(1, 0, dataset.length - 1, 1, 3));
   }

class Solution {
      private:
         int* dataset;
         int* tree;
         int n;

      public:
         Solution(int* dataset, int n) {
            this->dataset = dataset;
            this->n = n;
            this->tree = new int[4 * n];
            buildTree(1, 0, n - 1);
         }

         void buildTree(int node, int start, int end) {
            if (start == end) {
               tree[node] = dataset[start];
            } else {
               int mid = (start + end) / 2;
               buildTree(2 * node, start, mid);
               buildTree(2 * node + 1, mid + 1, end);
               tree[node] = tree[2 * node] + tree[2 * node + 1];
            }
         }

         void updateTree(int node, int start, int end, int idx, int val) {
            if (start == end) {
               tree[node] = val;
            } else {
               int mid = (start + end) / 2;
               if (idx <= mid) {
                  updateTree(2 * node, start, mid, idx, val);
               } else {
                  updateTree(2 * node + 1, mid + 1, end, idx, val);
               }
               tree[node] = tree[2 * node] + tree[2 * node + 1];
            }
         }

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

         int solution() {
            return queryTree(1, 0, n - 1, 1, 3);
         }
   }

public class Solution {
      private int[] dataset;
      private int[] tree;

      public Solution(int[] dataset) {
         this.dataset = dataset;
         this.tree = new int[4 * dataset.length];
         buildTree(1, 0, dataset.length - 1);
      }

      private void buildTree(int node, int start, int end) {
         if (start == end) {
            tree[node] = dataset[start];
         } else {
            int mid = (start + end) / 2;
            buildTree(2 * node, start, mid);
            buildTree(2 * node + 1, mid + 1, end);
            tree[node] = tree[2 * node] + tree[2 * node + 1];
         }
      }

      private void updateTree(int node, int start, int end, int idx, int val) {
         if (start == end) {
            tree[node] = val;
         } else {
            int mid = (start + end) / 2;
            if (idx <= mid) {
               updateTree(2 * node, start, mid, idx, val);
            } else {
               updateTree(2 * node + 1, mid + 1, end, idx, val);
            }
            tree[node] = tree[2 * node] + tree[2 * node + 1];
         }
      }

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

      public int solution() {
         return queryTree(1, 0, dataset.length - 1, 1, 3);
      }
   }

class Solution:
      def __init__(self, dataset):
         self.dataset = dataset
         self.tree = [0] * (4 * len(dataset))
         self.build_tree(1, 0, len(dataset) - 1)

      def build_tree(self, node, start, end):
         if start == end:
            self.tree[node] = self.dataset[start]
         else:
            mid = (start + end) // 2
            self.build_tree(2 * node, start, mid)
            self.build_tree(2 * node + 1, mid + 1, end)
            self.tree[node] = self.tree[2 * node] + self.tree[2 * node + 1]

      def update_tree(self, node, start, end, idx, val):
         if start == end:
            self.tree[node] = val
         else:
            mid = (start + end) // 2
            if idx <= mid:
               self.update_tree(2 * node, start, mid, idx, val)
            else:
               self.update_tree(2 * node + 1, mid + 1, end, idx, val)
            self.tree[node] = self.tree[2 * node] + self.tree[2 * node + 1]

      def query_tree(self, node, start, end, left, right):
         if left > end or right < start:
            return 0
         if left <= start and end <= right:
            return self.tree[node]
         mid = (start + end) // 2
         return self.query_tree(2 * node, start, mid, left, right) + self.query_tree(2 * node + 1, mid + 1, end, left, right)

      def solution(self):
         return self.query_tree(1, 0, len(self.dataset) - 1, 1, 3)

function solution(dataset) {
      // Initialize the segment tree
      let tree = new Array(4 * dataset.length).fill(0);
      
      // Build the segment tree
      function buildTree(node, start, end) {
         if (start === end) {
            tree[node] = dataset[start];
         } else {
            let mid = Math.floor((start + end) / 2);
            buildTree(2 * node, start, mid);
            buildTree(2 * node + 1, mid + 1, end);
            tree[node] = tree[2 * node] + tree[2 * node + 1];
         }
      }
      
      // Update the segment tree
      function updateTree(node, start, end, idx, val) {
         if (start === end) {
            tree[node] = val;
         } else {
            let mid = Math.floor((start + end) / 2);
            if (idx <= mid) {
               updateTree(2 * node, start, mid, idx, val);
            } else {
               updateTree(2 * node + 1, mid + 1, end, idx, val);
            }
            tree[node] = tree[2 * node] + tree[2 * node + 1];
         }
      }
      
      // Query the segment tree
      function queryTree(node, start, end, left, right) {
         if (left > end || right < start) {
            return 0;
         }
         if (left <= start && end <= right) {
            return tree[node];
         }
         let mid = Math.floor((start + end) / 2);
         return queryTree(2 * node, start, mid, left, right) + queryTree(2 * node + 1, mid + 1, end, left, right);
      }
      
      buildTree(1, 0, dataset.length - 1);
      
      // Example usage:
      console.log(queryTree(1, 0, dataset.length - 1, 1, 3));
   }

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Maximal Bipartite Energy Synthesizer 5 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Trees

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Maximal Bipartite Energy Synthesizer 5 — 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