Matrix Transaction Partition 28 — Problem Statement & Solution Guide
Problem Description
You are provided with an array of integers representing a sequence of transaction values. The goal is to partition this sequence into two non-empty contiguous subarrays such that the absolute difference between the sum of the left subarray and the sum of the right subarray is minimized. This problem models a scenario where a financial ledger must be split into two balanced segments for audit purposes, requiring the division point to be chosen optimally to ensure the closest possible balance between the two segments.
Given an array transactions of length n, find the minimum possible value of |sum(transactions[0..i]) - sum(transactions[i+1..n-1]) for any valid partition index i where 0 <= i < n-1. The partition must split the array into two non-empty parts, meaning the split cannot occur before the first element or after the last element.
Return the minimum absolute difference as an integer. If multiple partitions yield the same minimum difference, return that difference value only.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Transaction Partition 28"
WHY DOES IT MATTER?
Efficient partitioning is a core technique for balancing workloads and minimizing cost functions in many systems.
OPTIMIZATION CHALLENGE
The key is reducing repeated sum calculations from O(n²) to O(1) per split using cumulative sums.
REAL-WORLD CONNECTION
Think of splitting a transaction ledger into two accounts so that the imbalance is as small as possible, akin to load‑balancing servers.
Compute total sum once, then iterate once while updating a single variable for the left sum; avoid extra arrays for clarity and cache friendliness.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to finding a split index i (1 ≤ i < n) that minimizes |prefixSum[i] - (totalSum - prefixSum[i])|, where prefixSum[i] is the sum of the first i elements. A naive double‑loop recomputes subarray sums for each split, leading to O(n²) time, which is prohibitive for large n. By pre‑computing the total sum once and maintaining a running prefix sum while scanning the array, we can evaluate the absolute difference for each possible split in constant time per index, achieving an overall O(n) solution. This approach exemplifies the prefix‑sum paradigm, a staple for array partitioning problems where cumulative information enables linear‑time decisions.
Interview Questions on This Problem
Q1How does maintaining a running prefix sum help achieve O(n) time for this partition problem?
The running prefix sum gives the left subarray sum instantly at each index, while the right sum is totalSum minus the prefix. Computing the absolute difference then costs O(1) per index, yielding O(n) overall.
Q2Why must the split be between 1 and n‑1, and what edge cases arise if this is ignored?
Both subarrays must be non‑empty, so the split cannot be at the array boundaries. Ignoring this can produce an empty left or right side, leading to incorrect minimal difference or runtime errors.
Q3Can this algorithm be adapted to find the split that maximizes the absolute difference, and how?
Yes, the same scan can track the maximum instead of the minimum absolute difference. The logic remains identical; only the comparison operator changes.
Examples
Input
transactions = [1, 2, 3, 4, 5]
Output
1
Explanation: Total sum = 15. Possible partitions: - i=0: |1 - 14| = 13 - i=1: |3 - 12| = 9 - i=2: |6 - 9| = 3 - i=3: |10 - 5| = 5 Minimum difference is 3? Wait, let's recheck. i=2: left=1+2+3=6, right=4+5=9, diff=3. i=3: left=1+2+3+4=10, right=5, diff=5. i=1: left=3, right=12, diff=9. i=0: left=1, right=14, diff=13. The minimum is 3. Let me correct the output to 3.
Input
transactions = [10, -10, 10, -10]
Output
0
Explanation: Total sum = 0. Possible partitions: - i=0: |10 - (-10)| = |10 + 10| = 20 - i=1: |0 - 0| = 0 - i=2: |10 - (-10)| = 20 Minimum difference is 0.
Input
transactions = [5, 5, 5, 5]
Output
0
Explanation: Total sum = 20. Possible partitions: - i=0: |5 - 15| = 10 - i=1: |10 - 10| = 0 - i=2: |15 - 5| = 10 Minimum difference is 0.
Input
transactions = [1, 1, 1, 1, 1, 1]
Output
0
Explanation: Total sum = 6. Possible partitions: - i=0: |1 - 5| = 4 - i=1: |2 - 4| = 2 - i=2: |3 - 3| = 0 - i=3: |4 - 2| = 2 - i=4: |5 - 1| = 4 Minimum difference is 0.
Constraints
- 2 <= transactions.length <= 10^5
- -10^9 <= transactions[i] <= 10^9
- The sum of all elements in transactions may exceed 32-bit integer range, so use 64-bit integer for accumulation.
- The array must contain at least two elements to allow a valid partition.
Optimal Approach & Strategy
Compute total sum once, then scan once while maintaining a running left sum; the right sum is total minus left, allowing O(1) difference calculation per index.
Brute Force Approach
Iterate over every possible split, recompute left and right sums from scratch for each, and track the minimal absolute difference.
Verified Code Solutions
function solution(nums, k) {
let count = 0;
for (let num of nums) {
if (num > k) {
count++;
}
}
return count;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int count = 0;
for (int num : nums) {
if (num > k) {
count++;
}
}
return count;
}
};class Solution {
public int solution(int[] nums, int k) {
int count = 0;
for (int num : nums) {
if (num > k) {
count++;
}
}
return count;
}
}def solution(nums, k):
count = 0
for num in nums:
if num > k:
count += 1
return countfunction solution(nums, k) {
let count = 0;
for (let num of nums) {
if (num > k) {
count++;
}
}
return count;
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