Payload Token Tracker 45 — Problem Statement & Solution Guide
Problem Description
You are tasked with implementing a monitoring routine for a distributed system that tracks payload sizes and token usage metrics. The system provides a 2D array of integers where each row represents a batch of measurements taken at a specific time interval. Your goal is to determine the peak metric value observed across all batches to trigger an alert if it exceeds a threshold.
Given a 2D array metrics where each sub-array contains integer values representing individual measurements, return the maximum integer value found in the entire structure. If the input array is empty or contains no elements, return 0. The solution must efficiently scan the data to identify the global maximum without unnecessary overhead.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Token Tracker 45"
WHY DOES IT MATTER?
Efficiently solving exponential‑size search spaces is critical for real‑time monitoring and resource allocation.
OPTIMIZATION CHALLENGE
The key is to replace redundant recursive calls with compact state representation, cutting the combinatorial blow‑up.
REAL-WORLD CONNECTION
Think of assigning tokens to payloads in a distributed system where each token can be used only once per interval.
Cache intermediate results using a bitmask and always update a global best to enable aggressive pruning.
COMPLEXITY AT A GLANCE
O(N·2^N)O(2^N)Core Theory — Why This Approach?
Backtracking systematically explores all configurations of a combinatorial search space by making a series of choices and recursively exploring the consequences. When a partial configuration cannot possibly lead to an optimal solution—detected via pruning conditions—it is abandoned, dramatically cutting the search tree. Naïve enumeration of every possible selection (e.g., picking a column for each row in an N×M matrix) incurs O(M^N) time, which explodes even for modest N, making it infeasible for hard‑level constraints. The optimal paradigm replaces pure backtracking with state‑compression DP (often a bitmask) that records the best achievable metric for each subset of rows or columns, turning exponential branching into O(N·2^N) time while preserving exactness.
Interview Questions on This Problem
Q1Why does plain recursion without pruning fail on large N for this problem?
It explores every possible assignment, leading to O(M^N) calls. Pruning eliminates branches that cannot improve the current best, reducing work dramatically.
Q2How does a bitmask DP improve over backtracking for the assignment variant?
It memoizes the optimal value for each subset of columns, avoiding recomputation of identical sub‑problems. This collapses the exponential tree to O(N·2^N) states.
Q3What is the role of the ‘upper bound’ heuristic in backtracking?
It estimates the maximum possible metric from the current partial solution. If the bound is ≤ the best found, the branch is pruned.
Examples
Input
metrics = [[12, 45, 7], [3, 99, 15], [22, 8, 60]]
Output
99
Explanation: Scan the first row [12, 45, 7], the maximum is 45. Scan the second row [3, 99, 15], the maximum is 99. Scan the third row [22, 8, 60], the maximum is 60. The global maximum among 45, 99, and 60 is 99.
Input
metrics = [[-5, -10], [-2, -8]]
Output
-2
Explanation: The first row [-5, -10] has a maximum of -5. The second row [-2, -8] has a maximum of -2. Comparing the row maxima, -2 is greater than -5, so the result is -2.
Input
metrics = []
Output
0
Explanation: The input array is empty. According to the problem specification, if the input is empty, the function must return 0.
Input
metrics = [[100]]
Output
100
Explanation: There is only one row with a single element 100. The maximum value in the entire array is 100.
Constraints
- 0 <= metrics.length <= 10^4
- 0 <= metrics[i].length <= 10^4
- -10^9 <= metrics[i][j] <= 10^9
- The total number of elements across all sub-arrays will not exceed 10^6
Optimal Approach & Strategy
Use a DP with a bitmask of used columns; for each row iterate over free columns, update the mask, and memoize the best sum for that mask.
Brute Force Approach
Recursively choose a column for each row, exploring all M^N possibilities and tracking the maximum sum.
Verified Code Solutions
function solution(matrix) {
if (matrix.length === 0) return 0;
let max = -Infinity;
for (let row of matrix) {
let rowMax = Math.max(...row);
max = Math.max(max, rowMax);
}
return max;
}class Solution {
public:
int solution(vector<vector<int>>& matrix) {
if (matrix.empty()) return 0;
int max = INT_MIN;
for (auto& row : matrix) {
int rowMax = *max_element(row.begin(), row.end());
max = max > rowMax ? max : rowMax;
}
return max;
}
}class Solution {
public int solution(int[][] matrix) {
if (matrix.length == 0) return 0;
int max = Integer.MIN_VALUE;
for (int[] row : matrix) {
int rowMax = Arrays.stream(row).max().getAsInt();
max = Math.max(max, rowMax);
}
return max;
}
}def solution(matrix):
if not matrix:
return 0
max_val = float('-inf')
for row in matrix:
row_max = max(row)
max_val = max(max_val, row_max)
return max_valfunction solution(matrix) {
if (matrix.length === 0) return 0;
let max = -Infinity;
for (let row of matrix) {
let rowMax = Math.max(...row);
max = Math.max(max, rowMax);
}
return max;
}Asked in Top Tech Interviews
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