You are given a 0-indexed integer array nums.
Return the maximum value over all triplets of indices (i, j, k) such that i < j < k. If all such triplets have a negative value, return 0.
The value of a triplet of indices (i, j, k) is equal to (nums[i] - nums[j]) * nums[k].
Example 1:
Input: nums = [12,6,1,2,7] Output: 77 Explanation: The value of the triplet (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77. It can be shown that there are no ordered triplets of indices with a value greater than 77.
Example 2:
Input: nums = [1,10,3,4,19] Output: 133 Explanation: The value of the triplet (1, 2, 4) is (nums[1] - nums[2]) * nums[4] = 133. It can be shown that there are no ordered triplets of indices with a value greater than 133.
Example 3:
Input: nums = [1,2,3] Output: 0 Explanation: The only ordered triplet of indices (0, 1, 2) has a negative value of (nums[0] - nums[1]) * nums[2] = -3. Hence, the answer would be 0.
Constraints:
3 <= nums.length <= 1051 <= nums[i] <= 106class Solution {
public long maximumTripletValue(int[] nums) {
int n = nums.length;
int[] prefixMax = new int[n];
int[] suffixMax = new int[n];
prefixMax[0] = nums[0];
suffixMax[n - 1] = nums[n - 1];
for (int i = 1; i < n; i++) {
prefixMax[i] = Math.max(prefixMax[i - 1], nums[i]);
suffixMax[n - 1 - i] = Math.max(suffixMax[n - i], nums[n - 1 - i]);
}
long ans = 0;
for (int i = 1; i < n - 1; i++)
ans = Math.max(ans, (prefixMax[i - 1] - (long) nums[i]) * suffixMax[i + 1]);
return ans;
}
}class Solution {
public long maximumTripletValue(int[] nums) {
int n = nums.length;
int[] prefixMax = new int[n];
int[] suffixMax = new int[n];
prefixMax[0] = nums[0];
suffixMax[n - 1] = nums[n - 1];
for (int i = 1; i < n; i++) {
prefixMax[i] = Math.max(prefixMax[i - 1], nums[i]);
suffixMax[n - 1 - i] = Math.max(suffixMax[n - i], nums[n - 1 - i]);
}
long ans = 0;
for (int i = 1; i < n - 1; i++)
ans = Math.max(ans, (prefixMax[i - 1] - (long) nums[i]) * suffixMax[i + 1]);
return ans;
}
}