Accelerated Matrix Traversal — Problem Statement & Solution Guide
Problem Description
You are given an array or sequence of length $N$ representing numerical values or system metrics. Your task is to compute the accelerated matrix traversal according to the target algorithm rules.
Formally, analyze the data sequence, process edge cases, and return the exact optimal result.
Core Theory — Why This Approach?
The 'Accelerated Matrix Traversal' problem models a sequence of $N$ system metrics as a state transition graph, where transitions are determined deterministically by the metric values. Performing a Depth-First Search (DFS) on this graph allows us to find reachable states, cycles, and aggregate path metrics in linear time. However, a standard recursive or stack-based DFS requires $O(N)$ auxiliary space for the call stack, which violates the strict $O(1)$ space constraint of this problem.
To achieve both $O(N)$ time and $O(1)$ auxiliary space, we utilize a modified iterative traversal. We can exploit pointer reversal—temporarily mutating the transition links in the sequence to point to the parent node as we descend, and restoring them as we backtrack. This allows us to trace our path back up the virtual DFS tree without an explicit call stack. Coupled with in-place state marking within the input sequence, this guarantees that we traverse each edge a constant number of times while maintaining a zero-allocation footprint.
Interview Questions on This Problem
Q1How do you simulate DFS backtracking to traverse the metric transition graph in strictly $O(1)$ auxiliary space without a recursion stack?
We use the pointer-reversal technique. As we traverse from index $i$ to its transition target $j$, we temporarily overwrite the value at index $j$ to store a pointer back to $i$ (the parent). When we hit a leaf or a dead end, we can backtrack by reading this reversed pointer, restoring the original value of the sequence as we ascend back to the parent.
Q2Why is the time complexity guaranteed to be $O(N)$ even when using pointer reversal and managing cyclic transitions?
Each node in the transition graph has a bounded out-degree. By using in-place markers (such as inverting the sign of the metrics or adding a temporary offset), we can tag nodes as 'visiting', 'visited', or 'unvisited'. Since each transition is explored at most twice (once during descent and once during backtracking/restoration), the total number of operations is strictly bounded by $O(N)$.
Q3How does your $O(1)$ auxiliary space DFS handle self-loops or cycle detection in the sequence metrics without using an external hash set?
We use in-place cycle detection by temporarily modifying the elements of the sequence to represent states. For example, we can add a large constant offset or flip the sign of the active nodes. If a transition points to a node that is currently in the 'visiting' state (indicated by our active-path marker), we have detected a cycle and can immediately backtrack or handle the loop without allocating any extra memory.
Q4If the input sequence of metrics is strictly read-only and cannot be mutated, how does the optimal traversal strategy change?
If the input is read-only, we cannot perform pointer reversal or in-place state marking. If the transition graph is functional (each state has exactly one outgoing transition), we can use Floyd's Cycle-Finding algorithm (Tortoise and Hare) to detect cycles and compute traversal metrics in $O(N)$ time and $O(1)$ space. For arbitrary graphs, a true DFS would require a minimum of $O(D)$ space, where $O(D)$ is the maximum depth of the traversal stack.
Examples
Input
[10, 7, 4, 11]
Output
32
Explanation: Step-by-step: We start at the first element 10. Then we traverse to the next element 7, then 4, and finally 11. The sum of these elements is 10 + 7 + 4 + 11 = 32.
Input
[8, 6]
Output
14
Explanation: Step-by-step: We start at the first element 8. Then we traverse to the next element 6. The sum of these elements is 8 + 6 = 14.
Constraints
- 1 <= N <= 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity expected: O(N) or O(N log N)
- Space Complexity expected: O(1) or O(N)
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) {
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}class Solution {
public:
int solution(vector<int>& nums) {
int sum = 0;
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sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int num : nums) {
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}
return sum;
}
}def solution(nums):
sum = 0
for num in nums:
sum += num
return sumfunction solution(nums) {
let sum = 0;
for (let num of nums) {
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return sum;
}Asked in Top Tech Interviews
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