Optimal Array Partition ā Problem Statement & Solution Guide
Problem Description
Given an array of integers weights of length n, determine the optimal partition point i that minimizes the absolute difference between the sum of weights in the left partition (0 to i-1) and the sum of weights in the right partition (i to n-1), such that the total number of elements in the two partitions is equal.
Examples
Input
[1, 1, 3, 4, 5]
Output
2
Explanation: Step-by-step: with input [1, 1, 3, 4, 5], we first calculate the total sum of the array, which is 1 + 1 + 3 + 4 + 5 = 14. Then, we calculate the sum of the left partition (0 to 1) and the sum of the right partition (2 to 4). The sum of the left partition is 1 + 1 = 2, and the sum of the right partition is 3 + 4 + 5 = 12. The absolute difference between the sums of the two partitions is |2 - 12| = 10. We repeat this process for each possible partition point and find that the optimal partition point is indeed 2.
Input
[1, 1, 1, 1, 1]
Output
2
Explanation: Step-by-step: with input [1, 1, 1, 1, 1], we first calculate the total sum of the array, which is 1 + 1 + 1 + 1 + 1 = 5. Then, we calculate the sum of the left partition (0 to 2) and the sum of the right partition (3 to 4). The sum of the left partition is 1 + 1 + 1 = 3, and the sum of the right partition is 1 + 1 = 2. The absolute difference between the sums of the two partitions is |3 - 2| = 1. We repeat this process for each possible partition point and find that the optimal partition point is indeed 2.
Constraints
- 2 <= number of containers <= 100
- 1 <= weight of each container <= 1000
Optimal Approach & Strategy
The optimal approach involves using a two-pointer technique to find the optimal partition point in linear time, resulting in a time complexity of O(n). This approach iterates over the array of weights, maintaining a running sum of the weights on the left and right sides of the partition point.
Brute Force Approach
The brute-force approach involves iterating over all possible partition points and calculating the absolute difference in total weight for each point, resulting in a time complexity of O(nĀ²). This approach is inefficient for large inputs. The naive approach would also involve checking all possible subsets of containers to find the optimal partition.
Verified Code Solutions
function optimalPartition(weights) {
let n = weights.length;
if (n === 0) return 0;
let totalSum = weights.reduce((a, b) => a + b, 0);
let partitionPoint = Math.floor(n / 2);
let leftSum = weights.slice(0, partitionPoint).reduce((a, b) => a + b, 0);
let rightSum = totalSum - leftSum;
let minDiff = Math.abs(leftSum - rightSum);
for (let i = partitionPoint + 1; i < n; i++) {
leftSum = weights.slice(0, i).reduce((a, b) => a + b, 0);
rightSum = totalSum - leftSum;
let diff = Math.abs(leftSum - rightSum);
if (diff < minDiff) {
minDiff = diff;
partitionPoint = i;
}
}
return partitionPoint;
}function optimalPartition(weights) {
let n = weights.length;
if (n === 0) return 0;
let totalSum = weights.reduce((a, b) => a + b, 0);
let partitionPoint = Math.floor(n / 2);
let leftSum = weights.slice(0, partitionPoint).reduce((a, b) => a + b, 0);
let rightSum = totalSum - leftSum;
let minDiff = Math.abs(leftSum - rightSum);
for (let i = partitionPoint + 1; i < n; i++) {
leftSum = weights.slice(0, i).reduce((a, b) => a + b, 0);
rightSum = totalSum - leftSum;
let diff = Math.abs(leftSum - rightSum);
if (diff < minDiff) {
minDiff = diff;
partitionPoint = i;
}
}
return partitionPoint;
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