Balanced Cargo Partitioning — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Arrays and solve the Balanced Cargo Partitioning problem optimally.
O(n)O(1)Problem Description
You are given an array weights of length n, where each element represents the mass of a discrete cargo unit arranged in a linear sequence. Your task is to determine the optimal partition index k (where 1 <= k < n) that divides the array into two non-empty contiguous subarrays: a left segment weights[0...k-1] and a right segment weights[k...n-1]. The objective is to minimize the absolute difference between the sum of the left segment and the sum of the right segment. Return the minimum possible absolute difference achievable by any valid partition.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Balanced Cargo Partitioning"
WHY DOES IT MATTER?
This pattern exemplifies the "prefix sum / running total" technique, a cornerstone for many partitioning and range‑query problems where global information can be derived from local aggregates in constant time.
OPTIMIZATION CHALLENGE
The key insight is that the right‑hand sum can be expressed as totalSum – leftSum, eliminating the need to recompute the right side for each candidate split, thus collapsing a quadratic process into a linear one.
REAL-WORLD CONNECTION
Think of a conveyor belt loading containers onto two trucks; you continuously track the weight loaded onto the first truck and compare it to the remaining weight, adjusting the split point to keep the trucks as balanced as possible.
During an interview, compute totalSum first, then iterate once while updating leftSum; keep track of the minimal absolute difference and its index—no extra arrays, no nested loops.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The Balanced Cargo Partitioning problem asks for an index k that splits an array into two contiguous parts such that the absolute difference between the sum of the left part and the sum of the right part is minimized. A naive solution would recompute the sums for each possible split, leading to O(n^2) time, which quickly becomes infeasible for large n (e.g., n > 10^5). The optimal paradigm leverages prefix sums: by scanning the array once we can maintain the cumulative sum of the left side while the total sum of the entire array is known beforehand. At each index we compute the right‑side sum as total‑left, evaluate the absolute difference, and keep the best index. This reduces the problem to a single linear pass, achieving O(n) time and O(1) extra space.
Interview Questions on This Problem
Q1How would you modify the solution if the array could contain negative weights?
The same prefix‑sum approach works unchanged because the total sum and running left sum correctly account for negative values; the absolute difference formula remains valid.
Q2Can you extend the algorithm to return all indices that achieve the minimal difference, not just one?
Yes—during the linear scan, store the current minimal difference and maintain a list of indices; whenever a new smaller difference is found, reset the list, and when an equal difference is encountered, append the index.
Q3What would be the impact on time and space complexity if the array is streamed and cannot be stored entirely in memory?
You would need two passes: first to compute the total sum while streaming, second to recompute the minimal difference using the running left sum; this still yields O(n) time and O(1) auxiliary space, but requires the ability to rewind or store the stream.
Examples
Input
weights = [1, 2, 3, 4, 5]
Output
1
Explanation: Total sum is 15. Possible splits: k=1: Left=1, Right=14, Diff=|1-14|=13 k=2: Left=3, Right=12, Diff=|3-12|=9 k=3: Left=6, Right=9, Diff=|6-9|=3 k=4: Left=10, Right=5, Diff=|10-5|=5 The minimum difference is 1.
Input
weights = [10, 10, 10, 10]
Output
0
Explanation: Total sum is 40. Possible splits: k=1: Left=10, Right=30, Diff=20 k=2: Left=20, Right=20, Diff=0 k=3: Left=30, Right=10, Diff=20 The minimum difference is 0.
Input
weights = [5, 1, 1, 1, 5]
Output
4
Explanation: Total sum is 13. Possible splits: k=1: Left=5, Right=8, Diff=3 k=2: Left=6, Right=7, Diff=1 k=3: Left=7, Right=6, Diff=1 k=4: Left=8, Right=5, Diff=3 Wait, let me re-calculate. k=1: Left=5, Right=1+1+1+5=8, Diff=|5-8|=3 k=2: Left=5+1=6, Right=1+1+5=7, Diff=|6-7|=1 k=3: Left=5+1+1=7, Right=1+5=6, Diff=|7-6|=1 k=4: Left=5+1+1+1=8, Right=5, Diff=|8-5|=3 The minimum difference is 1. Let me adjust the example to have a different output for variety. Let's use weights = [3, 1, 2, 4, 5]. Total=15. k=1: L=3, R=12, D=9 k=2: L=4, R=11, D=7 k=3: L=6, R=9, D=3 k=4: L=10, R=5, D=5 Min is 3. Let's stick to the first calculation for the third example but correct the output. Actually, let's create a new example. weights = [2, 3, 4, 5, 6]. Total=20. k=1: L=2, R=18, D=16 k=2: L=5, R=15, D=10 k=3: L=9, R=11, D=2 k=4: L=14, R=6, D=8 Min is 2.
Constraints
- 2 <= weights.length <= 10^5
- 1 <= weights[i] <= 10^9
- The sum of all elements in weights will not exceed 10^14
Optimal Approach & Strategy
Compute the total sum once, then iterate once while maintaining a running left sum; derive the right sum as total‑left and update the minimal difference on the fly.
Brute Force Approach
For each possible split index, sum the left part and the right part separately and compute their absolute difference, keeping the minimum.
Code Solutions
function solution(nums, k) {
let minDiff = Infinity;
for (let i = 1; i < nums.length; i++) {
let leftSum = nums.slice(0, i).reduce((a, b) => a + b, 0);
let rightSum = nums.slice(i).reduce((a, b) => a + b, 0);
let diff = Math.abs(leftSum - rightSum);
if (diff % k === 0 && diff < minDiff) {
minDiff = diff;
}
}
return minDiff === Infinity ? -1 : minDiff;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int minDiff = INT_MAX;
for (int i = 1; i < nums.size(); i++) {
int leftSum = 0;
int rightSum = 0;
for (int j = 0; j < i; j++) {
leftSum += nums[j];
}
for (int j = i; j < nums.size(); j++) {
rightSum += nums[j];
}
int diff = abs(leftSum - rightSum);
if (diff % k == 0 && diff < minDiff) {
minDiff = diff;
}
}
return minDiff == INT_MAX ? -1 : minDiff;
}
};class Solution {
public int solution(int[] nums, int k) {
int minDiff = Integer.MAX_VALUE;
for (int i = 1; i < nums.length; i++) {
int leftSum = 0;
int rightSum = 0;
for (int j = 0; j < i; j++) {
leftSum += nums[j];
}
for (int j = i; j < nums.length; j++) {
rightSum += nums[j];
}
int diff = Math.abs(leftSum - rightSum);
if (diff % k == 0 && diff < minDiff) {
minDiff = diff;
}
}
return minDiff == Integer.MAX_VALUE ? -1 : minDiff;
}
}def solution(nums, k):
min_diff = float('inf')
for i in range(1, len(nums)):
left_sum = sum(nums[:i])
right_sum = sum(nums[i:])
diff = abs(left_sum - right_sum)
if diff % k == 0 and diff < min_diff:
min_diff = diff
return -1 if min_diff == float('inf') else min_difffunction solution(nums, k) {
let minDiff = Infinity;
for (let i = 1; i < nums.length; i++) {
let leftSum = nums.slice(0, i).reduce((a, b) => a + b, 0);
let rightSum = nums.slice(i).reduce((a, b) => a + b, 0);
let diff = Math.abs(leftSum - rightSum);
if (diff % k === 0 && diff < minDiff) {
minDiff = diff;
}
}
return minDiff === Infinity ? -1 : minDiff;
}Solve in Interative Editor
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