algoadvance

Leetcode 2873. Maximum Value of an Ordered Triplet I

Problem Statement

Given an integer array nums, you need to find the maximum value of an ordered triplet (nums[i], nums[j], nums[k]) such that 0 <= i < j < k < n and nums[i] <= nums[j] <= nums[k]. Your task is to return the maximum value of such a triplet from the array.

Clarifying Questions

  1. Can the array have negative numbers?
    • Yes, the array can contain negative numbers.
  2. Can the triplet consist of equal values if the condition nums[i] <= nums[j] <= nums[k] is satisfied?
    • Yes, it is allowed as long as the indices i, j, and k satisfy 0 <= i < j < k < n.
  3. What should be returned if the array length is less than 3?
    • If the array length is less than 3, it’s impossible to form such a triplet, so we can return some indicative value or handle it via an exception.

Strategy

  1. Initialization:
    • Use three variables to store the candidate values for the triplet nums[i], nums[j], and nums[k].
  2. Traversal & Updates:
    • Traverse the array while keeping track of previous minimums and potential candidates for each of the triplet components.
  3. Data Structures:
    • We use arrays or variables to store minimums and maximums encountered so far.
  4. Edge Cases:
    • Handle cases where the array length is less than 3.

Code

public class MaxValueOrderedTriplet {
    
    public int maximumTripletValue(int[] nums) {
        int n = nums.length;
        // Early exit if the array length is less than 3, no triplet can be formed.
        if (n < 3) {
            throw new IllegalArgumentException("Array length must be at least 3");
        }
        
        // Variables to keep track of potential values of the triplet
        int max1 = Integer.MIN_VALUE, max2 = Integer.MIN_VALUE, max3 = Integer.MIN_VALUE;
        for (int i = 0; i < n - 2; i++) {
            for (int j = i + 1; j < n - 1; j++) {
                for (int k = j + 1; k < n; k++) {
                    if (nums[i] <= nums[j] && nums[j] <= nums[k]) {
                        int tripletValue = nums[i] + nums[j] + nums[k];
                        if (tripletValue > max3) {
                            max3 = tripletValue;
                        }
                    }
                }
            }
        }
        return max3;
    }

    public static void main(String[] args) {
        MaxValueOrderedTriplet solution = new MaxValueOrderedTriplet();
        int[] nums = {1, 2, 3, 4, 5};
        System.out.println(solution.maximumTripletValue(nums)); // Expected Output: 9
    }
}

Time Complexity

The time complexity of the above solution is O(n^3) due to the triple nested loops over the array.

However, this approach is not efficient for larger arrays. An optimal solution would involve improved traversal and potentially using more advanced data structures to maintain the current state efficiently in O(n log n) or O(n).

Notes:

  1. Efficiency Consideration:
    • Implementing data structures and dynamic programming to reduce the time complexity.
  2. Further Optimization:-
    • Exploring the usage of sorted data structures like TreeSet, or techniques such as prefix/suffix arrays to efficiently manage state updates.

Cut your prep time in half and DOMINATE your interview with AlgoAdvance AI