LeetCode-in-Java
2873. Maximum Value of an Ordered Triplet I
Easy
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 <= 1001 <= nums[i] <= 106
Solution
public class Solution {
public long maximumTripletValue(int[] nums) {
final int n = nums.length;
final int[] iNumMaxs = new int[n];
int prev = 0;
for (int i = 0; i < n; i++) {
if (nums[i] <= prev) {
iNumMaxs[i] = prev;
} else {
prev = iNumMaxs[i] = nums[i];
}
}
long result = 0;
int kNumMax = nums[n - 1];
for (int j = n - 2; j > 0; j--) {
result = Math.max(result, (long) (iNumMaxs[j - 1] - nums[j]) * kNumMax);
if (nums[j] > kNumMax) {
kNumMax = nums[j];
}
}
return result;
}
}