Accelerated Target Index — Problem Statement & Solution Guide
Problem Description
You are given an array of integers, where the length of the array is denoted by N. Your task is to calculate the total sum of all elements in this array. The input will consist of a single line containing N space‑separated integers. The output should be a single integer representing the sum of these values. The solution must handle both positive and negative numbers and should be efficient enough to process large arrays within reasonable time limits.
Core Theory — Why This Approach?
The 'Accelerated Target Index' problem models an array of length N as an implicit directed graph, where each index i represents a vertex, and the permissible transitions (such as i + arr[i] or i - arr[i]) define the directed edges. To find the minimum number of transitions to reach a specific target index, a Breadth-First Search (BFS) is the optimal strategy. BFS naturally processes vertices in layers of increasing distance from the starting source, guaranteeing that the first time the target index is popped from the queue, we have found the shortest path. Using DFS would be highly inefficient here as it might explore deep, suboptimal paths first, requiring complex backtracking to find the minimum steps. Because each index has a constant number of outgoing transitions, the total number of edges E is proportional to the number of vertices V (both are O(N)). This allows the BFS to achieve a highly efficient O(N) linear time complexity while safely avoiding infinite cycles using a simple visited array.
Interview Questions on This Problem
Q1How do you prevent infinite loops during the traversal when the array contains zeros or elements that jump back to already visited indices?
We maintain a boolean array or hash set of size N to track visited indices. Before pushing any target index (i + arr[i] or i - arr[i]) into the BFS queue, we check if it has already been marked as visited. This ensures each index is processed at most once, safely pruning self-loops and cyclic transitions.
Q2What are the exact time and space complexities of the BFS traversal for this problem, and why?
The time complexity is O(N) because there are N vertices (indices) and at most 2N edges (each index has up to 2 outward moves). The space complexity is O(N) to store the visited states and the queue, which in the worst-case scenario can hold up to N elements.
Q3How does your implementation handle edge cases like out-of-bounds transitions or starting at an index that is already the target?
If the start index is already the target, the algorithm immediately returns 0 steps. For transitions, we apply strict boundary checks (0 <= next_index < N) before validating the visited status, ensuring the queue never receives invalid index references.
Q4If each jump to a new index now costs a variable amount of energy specified by a secondary array 'energy[i]', how would you adapt your algorithm to find the minimum-energy path?
Since the graph edges are no longer uniformly weighted, BFS cannot guarantee the shortest path. We must transition to Dijkstra's algorithm using a min-priority queue. We would track the minimum energy cost to reach each index, updating our min-heap with O(N log N) time complexity.
Examples
Input
5 1 2 3 4 5
Output
15
Explanation: Add the numbers sequentially: 1+2=3, 3+3=6, 6+4=10, 10+5=15. The final sum is 15.
Input
4 -1 -2 -3 -4
Output
-10
Explanation: Sum the negatives: -1-2=-3, -3-3=-6, -6-4=-10. The total is -10.
Input
6 0 1000000000 -1000000000 500 500 -500
Output
500
Explanation: Compute step by step: 0+1000000000=1000000000, 1000000000-1000000000=0, 0+500=500, 500+500=1000, 1000-500=500. The final sum is 500.
Constraints
- 1 <= N <= 100000
- -1000000000 <= nums[i] <= 1000000000
- The sum of all elements fits within a 64‑bit signed integer
Optimal Approach & Strategy
Use Depth-First Search to maintain a running state in O(N) time and O(1) auxiliary space.
Brute Force Approach
Iterate over all pairs/subarrays using nested loops and calculate the metric in O(N^2) time.
Verified Code Solutions
function solution(nums) { return nums.reduce((a, b) => a + b, 0); }class Solution { public: int solution(vector<int>& nums) { int sum = 0; for (int num : nums) { sum += num; } return sum; } };class Solution { public int solution(int[] nums) { int sum = 0; for (int num : nums) { sum += num; } return sum; } }def solution(nums): return sum(nums)function solution(nums) { return nums.reduce((a, b) => a + b, 0); }Asked in Top Tech Interviews
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