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Leetcode 2869. Minimum Operations to Collect Elements

Problem Statement

You are given an integer array nums, where each element in nums is within the range [1, nums.length]. You need to determine the minimum number of operations required to collect all the elements from the array. In one operation, you can remove any element from the array. You have to return the minimum number of operations required to make an array empty such that in each step, you remove an element while ensuring all unique elements are collected at least once.

Clarifying Questions

1. Will the elements always be within the range [1, nums.length]? Yes, as stated in the problem.

2. Are there any constraints on the size of the array? No specific constraints are provided, but typical constraints on LeetCode should apply.

3. What should be done if the same element appears multiple times? The goal is to ensure all unique elements are collected. After collecting an element once, additional appearances of that element can be safely ignored.

Strategy

1. Understand the Problem: The goal is to collect all unique elements with the minimum number of operations. An operation consists of removing any element, which means each element can be removed once.

2. Count Frequency: Use a HashSet to keep track of all unique elements since we just need to collect every unique element at least once.

3. Algorithm Outline:

   • Create a HashSet to store the unique elements of nums.
   • Traverse through the array nums and add each element to the HashSet.
   • The size of the HashSet represents the number of unique elements.
   • The minimum number of operations needed to collect all elements is equal to the size of the HashSet.

Time Complexity

• Time Complexity: (O(n)), where (n) is the length of the nums array because we are making one pass through the array.
• Space Complexity: (O(n)), in the worst case where all elements are unique, the HashSet will store n elements.

Code

import java.util.HashSet;
import java.util.Set;

public class Solution {
    public int minOperationsToCollectAllElements(int[] nums) {
        Set<Integer> uniqueElements = new HashSet<>();
        
        // Iterate through the array and add elements to the HashSet
        for (int num : nums) {
            uniqueElements.add(num);
        }
        
        // The size of the HashSet gives us the number of unique elements
        return uniqueElements.size();
    }
}

Explanation

1. Initialize HashSet: We use a HashSet to store unique elements.
2. Populate HashSet: Iterate through nums, adding each element to the HashSet. Duplicate elements will only be stored once.
3. Count Unique Elements: The size of the HashSet represents the count of unique elements in the array, which is the answer.

By following this strategy, we ensure that we collect all unique elements in the minimum number of operations, effectively removing duplicates and counting each element exactly once.

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