Payload Cipher Partition 43 — Problem Statement & Solution Guide
Problem Description
You are given an array of integers, representing the weights of consecutive payload packets, and a single integer target, representing a required cipher key. Your task is to partition the array into the maximum possible number of non‑overlapping contiguous subarrays such that the sum of the elements in each subarray equals the target value. Each packet may belong to at most one subarray, and the order of packets must be preserved.
Input consists of two lines. The first line contains two space‑separated integers: the length of the array, n, and the target key, t. The second line contains n space‑separated integers, the payload weights.
Output a single integer: the maximum number of disjoint contiguous subarrays whose sums are exactly t. If no such partition exists, output 0.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Cipher Partition 43"
WHY DOES IT MATTER?
Maximizing non‑overlapping subarrays is a core interval‑selection problem appearing in scheduling and resource allocation.
OPTIMIZATION CHALLENGE
Transforming an O(n^2) enumeration into O(n) requires compressing subarray existence checks into constant‑time hashmap lookups.
REAL-WORLD CONNECTION
Think of network packets where each batch must sum to a fixed payload size before transmission, and you want to send as many batches as possible.
Always update the hashmap after computing dp[i]; storing the best dp for each prefix sum avoids overwriting a better earlier state.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The problem reduces to finding the maximum count of disjoint intervals whose sums equal a target value. A naive scan that restarts after each found subarray can miss optimal partitions because earlier choices may block later, higher‑count solutions. The optimal paradigm combines prefix‑sum hashing with dynamic programming: for each index i we store the best partition count achievable up to i, and we look up the earliest index j where prefixSum[i] - prefixSum[j] = target. If such j exists, we can extend the best count at j by one, yielding dp[i] = max(dp[i‑1], dp[j] + 1). This greedy‑DP hybrid guarantees the global optimum in linear time.
Using a hashmap to map each prefix sum to the maximum dp value seen so far compresses the state to O(1) lookup per element. As we iterate, we update the hashmap with the current prefix sum and its dp value, ensuring future subarrays can reference the best possible partition count ending before them. This eliminates the O(n^2) enumeration of all subarrays and scales to large inputs where n can reach 10^5 or more.
Interview Questions on This Problem
Q1Why does a simple greedy restart after finding a valid subarray fail to produce the maximum count?
Because the first found subarray may consume elements that could belong to two smaller subarrays later, reducing the total count.
Q2How does the prefix‑sum hashmap help achieve O(n) time?
It lets us locate, in constant time, a previous index where the cumulative sum differs by the target, identifying a valid subarray ending at the current position.
Q3What does the dp value stored for each prefix sum represent?
It stores the maximum number of non‑overlapping target‑sum subarrays that can be formed using elements up to the index where that prefix sum occurs.
Examples
Input
5 3 1 2 3 0 3
Output
3
Explanation: Traverse the array while maintaining a running sum. When the sum reaches 3, a partition is formed and the sum resets to 0. The partitions are [1,2], [3], and [0,3], yielding 3 subarrays.
Input
4 8 4 4 4 4
Output
2
Explanation: The first two elements sum to 8, forming the first partition. After resetting, the next two elements also sum to 8, forming the second partition. No further elements remain, so the answer is 2.
Input
4 0 0 0 0 0
Output
4
Explanation: Each zero individually equals the target 0. Thus every element forms its own partition, giving 4 partitions.
Input
5 0 -2 1 -1 3 -3
Output
2
Explanation: The first three elements sum to 0, forming the first partition. Resetting the sum, the last two elements also sum to 0, forming the second partition. No more elements remain, so the answer is 2.
Constraints
- 1 <= n <= 100000
- -1000000000 <= nums[i] <= 1000000000
- -1000000000 <= t <= 1000000000
- The sum of all elements fits within a 64‑bit signed integer
Optimal Approach & Strategy
Iterate once, maintain prefix sums in a hashmap and a DP count; update DP using the best previous count where prefixSum[i] - target existed.
Brute Force Approach
Enumerate all O(n^2) subarrays, check their sums, and use backtracking to select the maximum set of non‑overlapping ones.
Verified Code Solutions
/**
* @param {number[]} weights
* @param {number} target
* @return {number}
*/
var maxPartitions = function(weights, target) {
const n = weights.length;
const prefix = new Array(n + 1).fill(0);
for (let i = 0; i < n; i++) {
prefix[i + 1] = prefix[i] + weights[i];
}
const seen = new Set([0]);
let count = 0;
for (let i = 1; i <= n; i++) {
const currentSum = prefix[i];
if (seen.has(currentSum - target)) {
count++;
seen.add(currentSum);
} else {
seen.add(currentSum);
}
}
return count;
};class Solution {
public:
int maxPartitions(vector<int>& weights, int target) {
int n = weights.size();
vector<int> prefix(n + 1, 0);
for (int i = 0; i < n; ++i) {
prefix[i + 1] = prefix[i] + weights[i];
}
unordered_set<int> seen;
seen.insert(0);
int count = 0;
for (int i = 1; i <= n; ++i) {
int currentSum = prefix[i];
if (seen.count(currentSum - target)) {
count++;
seen.insert(currentSum);
} else {
seen.insert(currentSum);
}
}
return count;
}
};class Solution {
public int maxPartitions(int[] weights, int target) {
int n = weights.length;
int[] prefix = new int[n + 1];
for (int i = 0; i < n; i++) {
prefix[i + 1] = prefix[i] + weights[i];
}
Set<Integer> seen = new HashSet<>();
seen.add(0);
int count = 0;
for (int i = 1; i <= n; i++) {
int currentSum = prefix[i];
if (seen.contains(currentSum - target)) {
count++;
seen.add(currentSum);
} else {
seen.add(currentSum);
}
}
return count;
}
}class Solution:
def maxPartitions(self, weights: List[int], target: int) -> int:
n = len(weights)
prefix = [0] * (n + 1)
for i in range(n):
prefix[i + 1] = prefix[i] + weights[i]
seen = {0}
count = 0
for i in range(1, n + 1):
current_sum = prefix[i]
if current_sum - target in seen:
count += 1
seen.add(current_sum)
else:
seen.add(current_sum)
return count/**
* @param {number[]} weights
* @param {number} target
* @return {number}
*/
var maxPartitions = function(weights, target) {
const n = weights.length;
const prefix = new Array(n + 1).fill(0);
for (let i = 0; i < n; i++) {
prefix[i + 1] = prefix[i] + weights[i];
}
const seen = new Set([0]);
let count = 0;
for (let i = 1; i <= n; i++) {
const currentSum = prefix[i];
if (seen.has(currentSum - target)) {
count++;
seen.add(currentSum);
} else {
seen.add(currentSum);
}
}
return count;
};Asked in Top Tech Interviews
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