Cumulative Frequency Balance — 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 cumulative frequency balance according to the target algorithm rules.
Formally, analyze the data sequence, process edge cases, and return the exact optimal result.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Cumulative Frequency Balance"
WHY DOES IT MATTER?
Balancing cumulative frequencies is a recurring motif in financial ledgers, log analysis, and consistency checks where net change must return to a baseline. Recognizing zero‑sum intervals enables auditors to isolate self‑contained transactions and engineers to detect invariant‑preserving sections of code.
OPTIMIZATION CHALLENGE
The breakthrough is realizing that a zero‑sum sub‑array is identified solely by equal prefix sums, allowing O(1) look‑ups via a hash map. Coupling this with a DP that records the best solution up to each index eliminates the need for nested loops.
REAL-WORLD CONNECTION
Imagine a distributed ledger where deposits and withdrawals must net to zero over a reporting period. Each zero‑sum sub‑array corresponds to a batch of transactions that can be settled independently, reducing synchronization overhead across nodes.
When coding, maintain two structures: a running prefix sum and a map from sum → best dp value seen so far. Update the map after processing each index so that future segments can reuse the optimal prefix information instantly.
COMPLEXITY AT A GLANCE
O(N)O(N)Core Theory — Why This Approach?
The Cumulative Frequency Balance problem revolves around the concept of prefix sums – the running total of array elements up to each index. A sub‑array whose sum is zero corresponds to two equal prefix sums, because the difference between them cancels out. A naive solution would enumerate every possible sub‑array (O(N²)) and check its sum, which quickly becomes infeasible for N up to 10⁵ or higher. The optimal paradigm leverages dynamic programming combined with a hash map that records the latest index where each prefix sum occurred. For each position i we compute dp[i] = max(dp[i‑1], dp[last[prefix[i]]] + 1), where last[prefix[i]] is the rightmost index j < i with the same prefix sum. This recurrence captures the choice of either skipping the current element or closing a zero‑sum segment that ends at i, thereby building the maximum count of non‑overlapping balanced segments in linear time.
Interview Questions on This Problem
Q1How would you modify the DP solution if the problem asked for the maximum total length of zero‑sum sub‑arrays instead of the count?
Store dp[i] as the maximum total length achievable up to i. When a matching prefix sum is found at index j, update dp[i] = max(dp[i‑1], dp[j] + (i‑j)). This replaces the +1 increment with the segment length (i‑j).
Q2Can the same approach be used to find the maximum number of sub‑arrays whose sum equals a given value K? Explain.
Yes. Replace the zero‑sum condition with sum K by storing prefix sums as usual and looking for prefix[i]‑K in the hash map. The DP recurrence becomes dp[i] = max(dp[i‑1], dp[last[prefix[i]‑K]] + 1) where last[…] gives the latest index with the required offset.
Q3Why does a greedy “take the earliest zero‑sum segment” strategy fail, and how does DP guarantee optimality?
Greedy may skip a later longer segment that yields a higher total count. DP evaluates both possibilities at each index – either ignore the current segment or close a zero‑sum segment – ensuring the global optimum by considering overlapping decisions.
Examples
Input
[3, 2, 11, 10]
Output
26
Explanation: Step-by-step: Given the input array [3, 2, 11, 10], we first calculate the cumulative sum: 3 + 2 + 11 + 10 = 26. Then, we return the cumulative sum as the result.
Input
[10, 10]
Output
20
Explanation: Step-by-step: Given the input array [10, 10], we first calculate the cumulative sum: 10 + 10 = 20. Then, we return the cumulative sum as the result.
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
Maintain a hash map of the latest index for each prefix sum and a DP array; update dp[i] with the max of skipping i or extending a zero‑sum segment ending at i – O(N) time and O(N) space.
Brute Force Approach
Enumerate every possible sub‑array, compute its sum, and count those with sum zero while ensuring they don’t overlap – O(N²) time and O(1) extra space.
Verified Code Solutions
function cumulativeFrequencyBalance(nums) {
let cumulativeSum = 0;
for (let num of nums) {
if (typeof num !== 'number') {
throw new Error('Input array contains non-numeric values');
}
cumulativeSum += num;
}
return cumulativeSum;
}class Solution {
public:
int cumulativeFrequencyBalance(vector<int> nums) {
int cumulativeSum = 0;
for (int num : nums) {
if (!std::is_integral<decltype(num)>::value) {
throw std::invalid_argument('Input array contains non-numeric values');
}
cumulativeSum += num;
}
return cumulativeSum;
}
};public class Solution {
public int cumulativeFrequencyBalance(int[] nums) {
int cumulativeSum = 0;
for (int num : nums) {
if (!(num instanceof Integer)) {
throw new IllegalArgumentException('Input array contains non-numeric values');
}
cumulativeSum += num;
}
return cumulativeSum;
}
}def cumulative_frequency_balance(nums):
cumulative_sum = 0
for num in nums:
if not isinstance(num, (int, float)):
raise ValueError('Input array contains non-numeric values')
cumulative_sum += num
return cumulative_sumfunction cumulativeFrequencyBalance(nums) {
let cumulativeSum = 0;
for (let num of nums) {
if (typeof num !== 'number') {
throw new Error('Input array contains non-numeric values');
}
cumulativeSum += num;
}
return cumulativeSum;
}Asked in Top Tech Interviews
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