LeetCode 2113: Elements in Array After Removing and Replacing Elements Solution

Master LeetCode problem 2113 (Elements in Array After Removing and Replacing Elements), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2113. Elements in Array After Removing and Replacing Elements

Medium

Array

Problem Explanation

Explanation

To solve this problem, we can iterate through the given array and maintain a set to keep track of all the elements that have been seen so far. For each element in the array, we check if it is present in the set. If it is present, we increment the element by 1 and check again. We repeat this process until we find an element that is not present in the set. Then, we add this element to the set and move on to the next element. The final array will contain the elements after all replacements have been made.

Time Complexity

The time complexity of this solution is O(n) where n is the number of elements in the input array.

Space Complexity

The space complexity of this solution is O(n) as we are using a set to store the unique elements in the array.

Solution Code

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

class Solution {
    public int[] replaceElements(int[] arr) {
        Set<Integer> seen = new HashSet<>();
        for (int i = 0; i < arr.length; i++) {
            int num = arr[i];
            while (seen.contains(num)) {
                num++;
            }
            arr[i] = num;
            seen.add(num);
        }
        return arr;
    }
}

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Frequently Asked Questions

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What is the time complexity of LeetCode 2113 (Elements in Array After Removing and Replacing Elements)?

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