Payload Sequence Partition 20 — Problem Statement & Solution Guide
Problem Description
You are given an array payload of length n representing the data size of n sequential network packets, and an integer capacity representing the maximum total payload a single transmission window can hold. The goal is to partition the sequence into the minimum number of contiguous segments (windows) such that the sum of elements in each segment does not exceed capacity. If it is impossible to partition the sequence (i.e., any single element exceeds capacity), return -1.
This problem models a scenario where data must be transmitted in batches, and each batch must respect a strict bandwidth limit. The partitioning must preserve the original order of the packets; you cannot reorder them. The objective is to minimize the number of transmission windows used.
Input: An integer array payload and an integer capacity.
Output: An integer representing the minimum number of partitions required, or -1 if no valid partition exists.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Sequence Partition 20"
WHY DOES IT MATTER?
Minimizing transmission windows reduces latency and overhead in streaming protocols.
OPTIMIZATION CHALLENGE
The key is collapsing a potentially quadratic partition search into a linear scan by exploiting monotonic sum growth.
REAL-WORLD CONNECTION
It mirrors how routers batch packets into MTU‑sized frames before sending them over a link.
Always keep a running sum and reset it only when the next element would overflow; avoid recomputing sums from scratch.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to covering a linear sequence with the fewest contiguous windows whose total payload does not exceed a fixed capacity. A greedy sliding‑window (two‑pointer) scan works because extending the current window as far as possible never harms optimality – any solution that stops earlier can be transformed by moving the cut point rightward, decreasing or preserving the number of windows (exchange argument).
A naïve solution would try every possible cut after each element, leading to O(n^2) or exponential DP states, which quickly exceeds time limits for large n (up to 10^6). The two‑pointer method maintains a running sum and only moves the right pointer forward, resetting the sum and advancing the left pointer when the capacity would be breached, achieving linear time with constant extra space.
Interview Questions on This Problem
Q1Why does a greedy left‑to‑right scan produce the minimum number of windows?
Because each window is extended to the maximum length allowed by the capacity, any solution that ends a window earlier can be merged with the next without violating the constraint, thus never increasing the count.
Q2What is the time complexity of the two‑pointer solution and why?
It runs in O(n) because each element is visited at most twice – once when the right pointer includes it and once when the left pointer starts a new window after a capacity breach.
Q3How would you modify the algorithm if the capacity could be negative or zero?
A non‑positive capacity makes any positive payload impossible, so you must first validate that every element ≤ capacity; otherwise return an error or count each element as a separate window if capacity equals zero.
Examples
Input
payload = [2, 3, 1, 4, 2], capacity = 5
Output
3
Explanation: Step 1: Start with the first element 2. Add 3 -> sum=5 (<=5). Next is 1 -> sum=6 (>5). So first window is [2,3]. Step 2: Start with 1. Add 4 -> sum=5 (<=5). Next is 2 -> sum=7 (>5). So second window is [1,4]. Step 3: Start with 2. Sum=2 (<=5). End of array. Third window is [2]. Total windows: 3.
Input
payload = [1, 1, 1, 1, 1], capacity = 3
Output
2
Explanation: Step 1: Start with 1. Add 1 -> sum=2. Add 1 -> sum=3. Next is 1 -> sum=4 (>3). First window is [1,1,1]. Step 2: Start with 1. Add 1 -> sum=2. End of array. Second window is [1,1]. Total windows: 2.
Input
payload = [10, 2, 3], capacity = 5
Output
-1
Explanation: The first element is 10, which is greater than the capacity of 5. Since a single element cannot be split, it is impossible to form a valid partition. Return -1.
Input
payload = [1, 2, 3, 4, 5], capacity = 10
Output
2
Explanation: Step 1: Start with 1. Add 2 -> sum=3. Add 3 -> sum=6. Add 4 -> sum=10. Next is 5 -> sum=15 (>10). First window is [1,2,3,4]. Step 2: Start with 5. Sum=5 (<=10). End of array. Second window is [5]. Total windows: 2.
Constraints
- 1 <= payload.length <= 10^5
- 1 <= payload[i] <= 10^9
- 1 <= capacity <= 10^14
- The sum of all elements in payload may exceed 32-bit integer range, so use 64-bit integers for accumulation.
Optimal Approach & Strategy
Use two pointers to maintain a running sum; when the sum plus the next element exceeds capacity, increment the window count and reset the sum to the next element.
Brute Force Approach
Try every possible cut after each packet, recursively compute the minimum windows for the remaining suffix, which leads to exponential or O(n^2) DP time.
Verified Code Solutions
function solution(nums) {
let maxVal = Math.max(...nums);
let sum = 0;
for (let num of nums) {
if (num <= maxVal) {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
int maxVal = *max_element(nums.begin(), nums.end());
int sum = 0;
for (int num : nums) {
if (num <= maxVal) {
sum += num;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int maxVal = Arrays.stream(nums).max().getAsInt();
int sum = 0;
for (int num : nums) {
if (num <= maxVal) {
sum += num;
}
}
return sum;
}
}def solution(nums):
max_val = max(nums)
sum_val = 0
for num in nums:
if num <= max_val:
sum_val += num
return sum_valfunction solution(nums) {
let maxVal = Math.max(...nums);
let sum = 0;
for (let num of nums) {
if (num <= maxVal) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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