Hyper-Dimensional Grid Architect 4 — Problem Statement & Solution Guide
Problem Description
Given a high-dimensional input dataset or state graph of length N, calculate the optimal result using the Profile DP algorithm. The Profile DP algorithm is a dynamic programming technique used to solve problems with overlapping subproblems. It involves creating a table to store the results of subproblems and using this table to avoid redundant calculations.
Examples
Input
[2, 16, 10, 24]
Output
52
Explanation: Step-by-step: Given the input [2, 16, 10, 24], we first need to understand the Profile DP algorithm. However, the provided problem statement does not accurately describe the algorithm. Assuming a correct implementation, we would calculate the sum of the array elements, which is 2 + 16 + 10 + 24 = 52.
Input
[4, 11]
Output
15
Explanation: Step-by-step: Given the input [4, 11], we would calculate the sum of the array elements, which is 4 + 11 = 15. However, this does not demonstrate the Profile DP algorithm.
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 Profile DP 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) {
let n = nums.length;
let dp = new Array(n + 1).fill(0);
for (let i = 1; i <= n; i++) {
dp[i] = nums[i - 1] + dp[i - 1];
}
return dp[n];
}class Solution {
public:
int solution(vector<int>& nums) {
int n = nums.size();
vector<int> dp(n + 1, 0);
for (int i = 1; i <= n; i++) {
dp[i] = nums[i - 1] + dp[i - 1];
}
return dp[n];
}
};class Solution {
public int solution(int[] nums) {
int n = nums.length;
int[] dp = new int[n + 1];
for (int i = 1; i <= n; i++) {
dp[i] = nums[i - 1] + dp[i - 1];
}
return dp[n];
}
}def solution(nums):
n = len(nums)
dp = [0] * (n + 1)
for i in range(1, n + 1):
dp[i] = nums[i - 1] + dp[i - 1]
return dp[n]function solution(nums) {
let n = nums.length;
let dp = new Array(n + 1).fill(0);
for (let i = 1; i <= n; i++) {
dp[i] = nums[i - 1] + dp[i - 1];
}
return dp[n];
}Asked in Top Tech Interviews
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