What is the time complexity of Replace Elements with Greatest Element on Right Side?

The optimal solution for Replace Elements with Greatest Element on Right Side runs in O(n) time and uses O(n) space. We only make a single pass through the array, leading to O(n) time complexity. The space complexity is O(n) due to the result array.

What is the brute force solution for Replace Elements with Greatest Element on Right Side?

The brute force approach for Replace Elements with Greatest Element on Right Side is Brute Force, with O(n²) time complexity and O(1) space complexity. For each element in the array, we look at all the elements to its right to find the maximum. This is straightforward but inefficient, especially for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Replace Elements with Greatest Element on Right Side?

Replace Elements with Greatest Element on Right Side uses the following concepts: Array. The recommended patterns to study are: Array, Two Pointers.

What are the interview tips for Replace Elements with Greatest Element on Right Side?

Always consider edge cases, like arrays with a single element. Explain your thought process clearly as you write code. Practice dry-running your solution with sample inputs.

What are common mistakes when solving Replace Elements with Greatest Element on Right Side?

Not initializing the maximum variable correctly. Forgetting to set the last element to -1.

#1299

Replace Elements with Greatest Element on Right Side

Easy

ArrayArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

Instead of checking all elements to the right for each element, we can keep track of the maximum element seen so far while iterating from the end of the array. This reduces the time complexity significantly.

⚙️

Algorithm

3 steps

1Step 1: Initialize a variable to keep track of the maximum element seen so far, starting with -1.
2Step 2: Iterate through the array from the last element to the first element.
3Step 3: For each element, store the current maximum in the result array, then update the maximum with the current element.

solution.py8 lines

1def replaceElements(arr):
2    n = len(arr)
3    result = [-1] * n
4    max_right = -1
5    for i in range(n - 1, -1, -1):
6        result[i] = max_right
7        max_right = max(max_right, arr[i])
8    return result

ℹ

Complexity note: We only make a single pass through the array, leading to O(n) time complexity. The space complexity is O(n) due to the result array.

1Iterating from the end allows us to keep track of the maximum efficiently.
2Using a single variable to store the maximum reduces the need for nested loops.

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