Maximized Matrix Traversal — Problem Statement & Solution Guide
Problem Description
You are given an integer array nums representing a set of weighted nodes in a linear graph. Your task is to reorder these nodes to maximize the cumulative traversal score. The traversal score is defined as the sum of all node weights after they have been arranged in strictly non-increasing order (descending).
To achieve this, you must sort the array such that the largest weight appears first and the smallest weight appears last. Once the array is sorted in this specific configuration, compute the total sum of all elements. Return this total as a single integer.
Note: The sorting operation is the core mechanism here. While the final result is a simple sum, the problem context emphasizes the reordering logic required to establish the 'maximized' state before aggregation. Ensure your solution handles negative values correctly, as the descending order must still place larger (less negative) numbers before smaller (more negative) ones.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Maximized Matrix Traversal"
WHY DOES IT MATTER?
Sorting underpins many optimization problems where ordering determines feasibility or performance; mastering it lets you convert unordered data into a structure that enables greedy or DP strategies.
OPTIMIZATION CHALLENGE
The key insight is recognizing that the problem's objective is invariant to permutation, so the only required work is to produce a descending order efficiently, which is achieved by a comparison‑based sort at the theoretical lower bound of O(n log n).
REAL-WORLD CONNECTION
Think of a load balancer that always routes traffic to the server with the highest remaining capacity—sorting server capacities descending each scheduling round mirrors this greedy allocation pattern.
In an interview, call out the built‑in sort immediately, justify its O(n log n) guarantee, and mention linear‑time alternatives (counting/radix) if the input domain is bounded.
COMPLEXITY AT A GLANCE
O(n log n)O(1) auxiliary (in‑place quicksort) or O(n) for stable mergesortCore Theory — Why This Approach?
Sorting is a fundamental algorithmic paradigm that rearranges elements into a defined order, enabling efficient computation of order‑dependent metrics. For the "Maximized Matrix Traversal" problem, the traversal score is simply the sum of the weights after they are placed in strictly non‑increasing order; because addition is commutative, the numeric sum does not change with ordering, but the problem statement forces a descending arrangement to model a greedy traversal that always picks the highest remaining weight. A naive approach that repeatedly scans for the maximum (selection sort) incurs O(n²) time, which quickly becomes prohibitive for large n (e.g., n > 10⁵). The optimal paradigm leverages comparison‑based sorting algorithms—such as quicksort, mergesort, or heapsort—that achieve the lower bound of Θ(n log n) comparisons, guaranteeing scalability while using only O(1)–O(n) auxiliary space depending on the implementation.
Modern interview environments expect candidates to recognize that the core of the problem reduces to a single sort operation. By invoking a built‑in, highly tuned sort (e.g., Arrays.sort with a custom comparator in Java or sort() in Python with reverse=True), you obtain a stable, O(n log n) solution with minimal code. Understanding the trade‑offs between in‑place (quick‑sort‑like) and stable (merge‑sort‑like) variants also demonstrates depth: in‑place sorts conserve memory, while stable sorts preserve relative order of equal weights, which can be relevant if additional metadata is attached to nodes.
Interview Questions on This Problem
Q1How would you modify your solution if the traversal score required the sum of prefix maxima rather than the total sum?
Compute the descending sort, then iterate once accumulating the running maximum; each prefix maximum is simply the current element because the array is sorted descending, so the score equals the sum of the sorted array. This insight reduces the problem to the same O(n log n) sort followed by O(n) accumulation.
Q2Can you achieve better than O(n log n) for this problem if the weight range is bounded?
Yes. If weights are integers within a known small range (e.g., 0 ≤ weight ≤ 10⁶), you can apply counting sort or radix sort to achieve O(n + k) time, where k is the range size, which is linear for practical bounds.
Q3Explain why a stable sort is unnecessary for this problem, but might be required in a variant where nodes carry timestamps.
The score depends only on weight values, not on original positions, so any order among equal weights yields the same sum. However, if each node also has a timestamp and the traversal must respect chronological order among equal weights, a stable sort preserves the original relative order, ensuring correctness.
Examples
Input
nums = [3, 1, 4, 1, 5, 9, 2, 6]
Output
31
Explanation: Step 1: Sort the array in descending order: [9, 6, 5, 4, 3, 2, 1, 1]. Step 2: Calculate the cumulative sum: 9 + 6 + 5 + 4 + 3 + 2 + 1 + 1 = 31. Step 3: Return 31.
Input
nums = [-5, -1, -3, -2, -4]
Output
-15
Explanation: Step 1: Sort the array in descending order (largest to smallest): [-1, -2, -3, -4, -5]. Step 2: Calculate the cumulative sum: -1 + (-2) + (-3) + (-4) + (-5) = -15. Step 3: Return -15.
Input
nums = [10, 10, 10, 10]
Output
40
Explanation: Step 1: Sort the array in descending order: [10, 10, 10, 10]. Step 2: Calculate the cumulative sum: 10 + 10 + 10 + 10 = 40. Step 3: Return 40.
Input
nums = [0, -1, 2, -3, 4]
Output
2
Explanation: Step 1: Sort the array in descending order: [4, 2, 0, -1, -3]. Step 2: Calculate the cumulative sum: 4 + 2 + 0 + (-1) + (-3) = 2. Step 3: Return 2.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The sum of all elements may exceed the range of a 32-bit integer, so use a 64-bit integer for accumulation.
Optimal Approach & Strategy
Apply a built‑in comparison sort with a descending comparator, then compute the sum in a single linear pass.
Brute Force Approach
Repeatedly scan the unsorted array to find the maximum, place it at the next position, and continue until all elements are ordered.
Verified Code Solutions
/**
* @param {number[]} nums
* @return {number}
*/
var maxMatrixTraversal = function(nums) {
nums.sort((a, b) => b - a);
return nums.reduce((acc, val) => acc + val, 0);
};class Solution {
public:
int maxMatrixTraversal(vector<int>& nums) {
sort(nums.begin(), nums.end(), greater<int>());
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};class Solution {
public int maxMatrixTraversal(int[] nums) {
Arrays.sort(nums);
int sum = 0;
for (int i = nums.length - 1; i >= 0; i--) {
sum += nums[i];
}
return sum;
}
}class Solution:
def maxMatrixTraversal(self, nums: List[int]) -> int:
nums.sort(reverse=True)
return sum(nums)/**
* @param {number[]} nums
* @return {number}
*/
var maxMatrixTraversal = function(nums) {
nums.sort((a, b) => b - a);
return nums.reduce((acc, val) => acc + val, 0);
};Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.