Tome Signal Consolidator 13 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a stream of integer values representing signal amplitudes in a distributed tome network. The system requires you to compute the 'Consolidator Value', which is defined as the maximum sum of any contiguous subarray within the input sequence. This metric indicates the peak energy accumulation possible over a continuous segment of the signal stream.
Given an array of integers signals, determine the maximum subarray sum. If all elements are negative, the maximum subarray sum is the largest single element (the least negative value), as the subarray must contain at least one element.
Input: An array of integers signals.
Output: An integer representing the maximum sum of any contiguous subarray.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Signal Consolidator 13"
WHY DOES IT MATTER?
Maximum subarray is a classic DP/greedy pattern that appears in many optimization problems.
OPTIMIZATION CHALLENGE
The key is reducing O(n²) pair enumeration to O(n) by discarding sub‑arrays that cannot improve the answer.
REAL-WORLD CONNECTION
It mirrors finding the most profitable contiguous time window in stock price changes or signal peaks.
Always initialize both current and global sums with the first array element to correctly handle negative‑only inputs.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The maximum subarray problem asks for the largest possible sum of a contiguous segment of an integer array. A naïve O(n²) scan enumerates every start‑end pair, which quickly becomes infeasible for large streams because the number of pairs grows quadratically.
Kadane's algorithm solves the problem in linear time by maintaining a running sum that resets to zero whenever it becomes negative, effectively discarding sub‑arrays that would diminish future sums. This greedy dynamic‑programming approach guarantees optimality because any optimal subarray either starts at the current index or extends the previous optimal subarray, allowing a single pass to capture the global maximum.
Interview Questions on This Problem
Q1What is the core idea behind Kadane's algorithm for maximum subarray sum?
It keeps a running sum that resets when negative, ensuring only beneficial prefixes are kept. The global maximum is updated whenever the running sum exceeds it.
Q2How does Kadane's algorithm handle all‑negative arrays?
By initializing the maximum with the first element and allowing the running sum to drop below zero, it correctly returns the largest (least negative) element. This avoids the common pitfall of always returning zero.
Q3Can Kadane's algorithm be adapted to return the subarray indices?
Yes, track start indices when the running sum resets and update end indices when a new maximum is found. The stored pair yields the exact segment achieving the maximum sum.
Examples
Input
signals = [-2, 1, -3, 4, -1, 2, 1, -5, 4]
Output
6
Explanation: The contiguous subarray [4, -1, 2, 1] has the largest sum = 4 + (-1) + 2 + 1 = 6.
Input
signals = [5, 4, -1, 7, 8]
Output
23
Explanation: The entire array is the optimal subarray. Sum = 5 + 4 + (-1) + 7 + 8 = 23.
Input
signals = [-2, -3, -1, -5]
Output
-1
Explanation: All elements are negative. The maximum sum is the single largest element, which is -1.
Input
signals = [1]
Output
1
Explanation: The array contains a single element. The maximum subarray sum is 1.
Constraints
- 1 <= signals.length <= 10^5
- -10^9 <= signals[i] <= 10^9
- The answer is guaranteed to fit in a 64-bit integer.
Optimal Approach & Strategy
Use Kadane's algorithm: iterate once, updating a running sum and global maximum, achieving O(n) time and O(1) space.
Brute Force Approach
Check every possible start and end index, summing each subarray to track the maximum, which costs O(n²) time.
Verified Code Solutions
function solution(nums) {
let prefixSum = [nums[0]];
let maxSum = nums[0];
for (let i = 1; i < nums.length; i++) {
prefixSum[i] = prefixSum[i - 1] + nums[i];
maxSum = Math.max(maxSum, prefixSum[i]);
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums) {
vector<int> prefixSum(nums.size());
int maxSum = nums[0];
prefixSum[0] = nums[0];
for (int i = 1; i < nums.size(); i++) {
prefixSum[i] = prefixSum[i - 1] + nums[i];
maxSum = max(maxSum, prefixSum[i]);
}
return maxSum;
}
};class Solution {
public int solution(int[] nums) {
int[] prefixSum = new int[nums.length];
int maxSum = nums[0];
prefixSum[0] = nums[0];
for (int i = 1; i < nums.length; i++) {
prefixSum[i] = prefixSum[i - 1] + nums[i];
maxSum = Math.max(maxSum, prefixSum[i]);
}
return maxSum;
}
}def solution(nums):
prefix_sum = [nums[0]]
max_sum = nums[0]
for i in range(1, len(nums)):
prefix_sum.append(prefix_sum[i - 1] + nums[i])
max_sum = max(max_sum, prefix_sum[i])
return max_sumfunction solution(nums) {
let prefixSum = [nums[0]];
let maxSum = nums[0];
for (let i = 1; i < nums.length; i++) {
prefixSum[i] = prefixSum[i - 1] + nums[i];
maxSum = Math.max(maxSum, prefixSum[i]);
}
return maxSum;
}Asked in Top Tech Interviews
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