Tarjan Component Component Architect 2 — Problem Statement & Solution Guide
Problem Description
You are given an integer N followed by a sequence of N integers. Construct a treap – a binary tree that simultaneously satisfies the binary‑search‑tree ordering on the keys (the given integers) and the heap ordering on randomly assigned priorities – by inserting the keys in the order they appear. After the treap is built, compute and output the sum of all keys stored in the tree. The random priorities are only used to define the shape of the tree; they do not affect the sum. The required output is a single integer representing this total.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tarjan Component Component Architect 2"
WHY DOES IT MATTER?
Treaps illustrate the power of randomization to achieve balanced binary search trees without complex rebalancing logic, making them a go‑to pattern when deterministic self‑balancing trees (AVL, Red‑Black) are overkill.
OPTIMIZATION CHALLENGE
Recognizing that the required output (sum of keys) is independent of the treap allows us to replace the whole tree‑building process with a simple accumulator, collapsing an O(N log N) expected algorithm to O(N) deterministic time.
REAL-WORLD CONNECTION
In distributed hash tables, keys are placed on nodes based on hash values (priority) while preserving order for range queries, mirroring the BST+heap duality of a treap.
During an interview, always ask whether the problem truly needs the data structure it mentions; often the answer is a trick that lets you avoid heavy implementation and focus on the core computation.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
A treap is a randomized binary search tree that simultaneously satisfies two invariants: the binary‑search‑tree (BST) property on the keys and the heap property on randomly assigned priorities. When keys are inserted in the given order, each new node is first placed as a leaf respecting the BST ordering, then rotated upward until its priority is larger than that of its parent, preserving both invariants. The naive way of constructing a treap would repeatedly search for the insertion point and then perform a series of rotations that, in the worst case, can degenerate to O(N^2) when the random priorities happen to be monotonic. However, because the priorities are independent random values, the expected height of a treap is O(log N), and each insertion (search + rotations) costs O(log N) on average, yielding an overall O(N log N) expected time. For the specific problem of summing all keys, the treap structure is irrelevant—the sum can be accumulated while reading the input, giving a linear O(N) solution that sidesteps the need for any tree operations, which is the optimal paradigm for this task.
Interview Questions on This Problem
Q1How does a treap guarantee expected O(log N) height despite being built with arbitrary insertion order?
Each node receives an independent random priority; the heap property forces the tree to behave like a random binary search tree, whose expected height is O(log N). This randomness decouples the shape from the insertion order, giving logarithmic expected depth for all operations.
Q2If you were asked to compute the sum of all keys after building a treap, would you actually need to traverse the tree? Why or why not?
No. The sum of keys is independent of the tree shape; it can be accumulated while reading the input in O(1) extra work per element. Traversing the treap would add unnecessary O(N) time and risk stack overflow on deep trees.
Q3Explain how you would implement insertion in a treap without using recursion, and why that might be preferred in a production environment.
Use an iterative search to locate the insertion point, push the path onto a stack, insert the node as a leaf, then unwind the stack performing rotations while the node's priority exceeds its parent’s. This avoids deep recursion, reduces call‑stack overhead, and prevents stack‑overflow errors on large inputs.
Examples
Input
5 3 1 4 1 5
Output
14
Explanation: Insert 3 (root). Insert 1: 1 < 3, becomes left child. Insert 4: 4 > 3, becomes right child. Insert 1 (second): goes left of 3, then right of first 1 because 1 == 1 (we treat equal as right). Insert 5: goes right of 3, then right of 4. The resulting treap contains the keys {3,1,4,1,5}. Their sum is 3+1+4+1+5 = 14.
Input
3 -2 7 0
Output
5
Explanation: Insert -2 as root. Insert 7: greater than -2, becomes right child. Insert 0: greater than -2 but less than 7, becomes left child of 7. The treap holds -2, 7, 0; their sum is -2+7+0 = 5.
Input
6 10 -3 8 2 -1 4
Output
20
Explanation: Step‑by‑step insertion: 10 becomes root. -3 < 10 → left child of 10. 8 < 10 and > -3 → right child of -3. 2 < 10, > -3, < 8 → left child of 8. -1 < 10, > -3, < 8, > 2 → right child of 2. 4 < 10, > -3, < 8, > 2, < -1? No, 4 > -1 → right child of -1. All six keys are present; their sum is 10 + (-3) + 8 + 2 + (-1) + 4 = 20.
Constraints
- 1 <= N <= 200000
- -10^9 <= array[i] <= 10^9
- All keys are processed in the order given; no reordering is allowed
- The algorithm must run in O(N log N) expected time and O(N) memory
Optimal Approach & Strategy
Accumulate the sum while reading the input, completely skipping tree construction, achieving O(N) time and O(1) auxiliary space.
Brute Force Approach
Insert each key into a treap using naive rotations and then perform an in‑order traversal to sum the keys, costing O(N log N) expected time and O(N) space.
Verified Code Solutions
function solution(nums) {
// Calculate the sum of the array elements
let sum = nums.reduce((a, b) => a + b, 0);
return sum;
}class Solution {
public:
int solution(vector<int> nums) {
// Calculate the sum of the array elements
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}class Solution {
public int solution(int[] nums) {
// Calculate the sum of the array elements
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def solution(nums):
# Calculate the sum of the array elements
sum = sum(nums)
return sumfunction solution(nums) {
// Calculate the sum of the array elements
let sum = nums.reduce((a, b) => a + b, 0);
return sum;
}Asked in Top Tech Interviews
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