Monotonic Threshold Span Synthesizer 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 Divide and Conquer DP algorithm.
Examples
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: Given an array [1, 2, 3, 4, 5], we divide it into two halves [1, 2, 3] and [4, 5]. The maximum sum of the two halves is the maximum of the sum of the first half (9) and the sum of the second half (10), which is 10. However, we need to find the optimal result using the Divide and Conquer DP algorithm. This involves finding the maximum sum of two halves for each subarray and storing it in a table. The optimal result is then the maximum value in the table.
Input
[1, 2, 3, 4, 5, 6]
Output
18
Explanation: Step-by-step: Given an array [1, 2, 3, 4, 5, 6], we divide it into two halves [1, 2, 3] and [4, 5, 6]. The maximum sum of the two halves is the maximum of the sum of the first half (9) and the sum of the second half (10), which is 10. However, we need to find the optimal result using the Divide and Conquer DP algorithm. This involves finding the maximum sum of two halves for each subarray and storing it in a table. The optimal result is then the maximum value in the table.
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 Divide and Conquer 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).fill(0).map(() => new Array(n).fill(0));
for (let i = 0; i < n; i++) {
dp[i][i] = nums[i];
}
for (let len = 2; len <= n; len++) {
for (let i = 0; i <= n - len; i++) {
let j = i + len - 1;
dp[i][j] = Math.max(dp[i][j - 1], dp[i + 1][j]);
}
}
return dp[0][n - 1];
}class Solution {
public:
int solution(vector<int>& nums) {
int n = nums.size();
vector<vector<int>> dp(n, vector<int>(n, 0));
for (int i = 0; i < n; i++) {
dp[i][i] = nums[i];
}
for (int len = 2; len <= n; len++) {
for (int i = 0; i <= n - len; i++) {
int j = i + len - 1;
dp[i][j] = max(dp[i][j - 1], dp[i + 1][j]);
}
}
return dp[0][n - 1];
}
};class Solution {
public int solution(int[] nums) {
int n = nums.length;
int[][] dp = new int[n][n];
for (int i = 0; i < n; i++) {
dp[i][i] = nums[i];
}
for (int len = 2; len <= n; len++) {
for (int i = 0; i <= n - len; i++) {
int j = i + len - 1;
dp[i][j] = Math.max(dp[i][j - 1], dp[i + 1][j]);
}
}
return dp[0][n - 1];
}
}def solution(nums):
n = len(nums)
dp = [[0 for _ in range(n)] for _ in range(n)]
for i in range(n):
dp[i][i] = nums[i]
for len in range(2, n + 1):
for i in range(n - len + 1):
j = i + len - 1
dp[i][j] = max(dp[i][j - 1], dp[i + 1][j])
return dp[0][n - 1]function solution(nums) {
let n = nums.length;
let dp = new Array(n).fill(0).map(() => new Array(n).fill(0));
for (let i = 0; i < n; i++) {
dp[i][i] = nums[i];
}
for (let len = 2; len <= n; len++) {
for (let i = 0; i <= n - len; i++) {
let j = i + len - 1;
dp[i][j] = Math.max(dp[i][j - 1], dp[i + 1][j]);
}
}
return dp[0][n - 1];
}Asked in Top Tech Interviews
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