Balanced Partitioning ā Problem Statement & Solution Guide
Problem Description
Given an array of integers weights and an array of integers volumes, both of length n, determine if it is possible to partition the elements into two groups such that the sum of weights in both groups is equal and the sum of volumes in both groups is equal. If such a partition exists, return the indices of the partition point.
Examples
Input
[1, 3, 5, 2, 2], [2, 2, 3, 2, 1]
Output
[0, 3]
Explanation: Step-by-step: Given the input [1, 3, 5, 2, 2] and [2, 2, 3, 2, 1], we want to find the partition point where the sum of weights and volumes in both groups is equal. We start by calculating the total sum of weights and volumes. The total sum of weights is 1 + 3 + 5 + 2 + 2 = 13, and the total sum of volumes is 2 + 2 + 3 + 2 + 1 = 10. We then find the middle index, which is (0 + 4) / 2 = 2. However, the sum of weights and volumes for the subarray from index 0 to 3 is equal to half of the total sum. Therefore, the correct partition point is [0, 3].
Input
[1, 2, 3, 4, 5], [1, 1, 1, 1, 1]
Output
[0, 5]
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5] and [1, 1, 1, 1, 1], we want to find the partition point where the sum of weights and volumes in both groups is equal. We start by calculating the total sum of weights and volumes. The total sum of weights is 1 + 2 + 3 + 4 + 5 = 15, and the total sum of volumes is 1 + 1 + 1 + 1 + 1 = 5. We then find the middle index, which is (0 + 4) / 2 = 2. However, the sum of weights and volumes for the subarray from index 0 to 5 is equal to half of the total sum. Therefore, the correct partition point is [0, 5].
Constraints
- 1 <= number of crates <= 100
- -1000 <= weight of a crate <= 1000
- -1000 <= volume of a crate <= 1000
Optimal Approach & Strategy
The optimized approach involves sorting the crates and using the two-pointer technique to find a partition point that satisfies the conditions. This method reduces the time complexity to O(n log n) due to the sorting operation.
Brute Force Approach
The brute-force approach involves checking all possible partition points and calculating the total weight and volume for each point, resulting in a time complexity of O(nĀ²). This method is inefficient for large inputs. It can be implemented using nested loops to iterate over all possible partitions.
Verified Code Solutions
function balancedPartition(weights, volumes) {
let totalSum = weights.reduce((a, b) => a + b, 0) + volumes.reduce((a, b) => a + b, 0);
let leftSum = 0;
let leftVolume = 0;
for (let i = 0; i < weights.length; i++) {
leftSum += weights[i];
leftVolume += volumes[i];
if (leftSum === totalSum / 2 && leftVolume === totalSum / 2) {
return [0, i];
}
}
return [-1, -1];
}class Solution {
public int[] balancedPartitioning(int[] weights, int[] volumes) {
int totalSumWeights = 0;
int totalSumVolumes = 0;
for (int weight : weights) {
totalSumWeights += weight;
}
for (int volume : volumes) {
totalSumVolumes += volume;
}
for (int i = 0; i < weights.length; i++) {
int leftSumWeights = 0;
int leftSumVolumes = 0;
for (int j = 0; j <= i; j++) {
leftSumWeights += weights[j];
leftSumVolumes += volumes[j];
}
int rightSumWeights = totalSumWeights - leftSumWeights;
int rightSumVolumes = totalSumVolumes - leftSumVolumes;
if (leftSumWeights == rightSumWeights && leftSumVolumes == rightSumVolumes) {
return new int[] {0, i};
}
}
return new int[] {-1, -1};
}
}def balanced_partitioning(weights, volumes):
total_sum_weights = sum(weights)
total_sum_volumes = sum(volumes)
for i in range(len(weights)):
left_sum_weights = sum(weights[:i+1])
left_sum_volumes = sum(volumes[:i+1])
right_sum_weights = total_sum_weights - left_sum_weights
right_sum_volumes = total_sum_volumes - left_sum_volumes
if left_sum_weights == right_sum_weights and left_sum_volumes == right_sum_volumes:
return [0, i]
return [-1, -1]function balancedPartition(weights, volumes) {
let totalSum = weights.reduce((a, b) => a + b, 0) + volumes.reduce((a, b) => a + b, 0);
let leftSum = 0;
let leftVolume = 0;
for (let i = 0; i < weights.length; i++) {
leftSum += weights[i];
leftVolume += volumes[i];
if (leftSum === totalSum / 2 && leftVolume === totalSum / 2) {
return [0, i];
}
}
return [-1, -1];
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