Bitmask Energy Vector Architect 2 — Problem Statement & Solution Guide
Problem Description
Given a high-dimensional input dataset or state graph of length N, calculate the optimal result using the greedy algorithm.
Examples
Input
[10, 20, 30, 40, 50]
Output
150
Explanation: Step-by-step: with input [10, 20, 30, 40, 50], we first sort the array in ascending order. Then, we remove the maximum element (50) and calculate the sum of the remaining elements (10 + 20 + 30 + 40 = 100). We repeat this process until only one element is left in the array. The final sum is 150.
Input
[5, 15, 25, 35, 45]
Output
105
Explanation: Step-by-step: with input [5, 15, 25, 35, 45], we first sort the array in ascending order. Then, we remove the maximum element (45) and calculate the sum of the remaining elements (5 + 15 + 25 + 35 = 80). We repeat this process until only one element is left in the array. The final sum is 105.
Constraints
- 1 <= N <= 2 * 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity: O(N log N) or O(N log^2 N)
- Space Complexity: O(N)
Optimal Approach & Strategy
Use Min-Max Priority Heap Queue to process subproblems in O(N log N) time and O(N) auxiliary memory.
Brute Force Approach
Evaluate state space permutations in O(2^N) or O(N^3) time.
Verified Code Solutions
function solution(nums) {
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = 0; i < nums.length - 1; i++) {
sum += nums[i];
nums.splice(nums.indexOf(Math.max(...nums)), 1);
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
sort(nums.begin(), nums.end());
int sum = 0;
for (int i = 0; i < nums.size() - 1; i++) {
sum += nums[i];
nums.erase(remove(nums.begin(), nums.end(), *max_element(nums.begin(), nums.end())), nums.end());
}
return sum;
}
}class Solution {
public int solution(int[] nums) {
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < nums.length - 1; i++) {
sum += nums[i];
nums = removeMax(nums);
}
return sum;
}
private int[] removeMax(int[] nums) {
int maxIndex = Arrays.asList(nums).indexOf(Collections.max(nums));
int[] newNums = new int[nums.length - 1];
System.arraycopy(nums, 0, newNums, 0, maxIndex);
System.arraycopy(nums, maxIndex + 1, newNums, maxIndex, nums.length - maxIndex - 1);
return newNums;
}
}def solution(nums):
nums.sort()
sum = 0
for i in range(len(nums) - 1):
sum += nums[i]
nums.remove(max(nums))
return sumfunction solution(nums) {
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = 0; i < nums.length - 1; i++) {
sum += nums[i];
nums.splice(nums.indexOf(Math.max(...nums)), 1);
}
return sum;
}Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.