How do you solve Duplicate Zeros on LeetCode?

Duplicate Zeros (LeetCode #1089) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding in-place operations is crucial.

What is the brute force solution for Duplicate Zeros?

The brute force approach for Duplicate Zeros is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach involves creating a new array to accommodate the duplicated zeros while maintaining the order of the other elements. This is straightforward but inefficient as it requires extra space and multiple passes through the array. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Duplicate Zeros?

Duplicate Zeros uses the following concepts: Array, Two Pointers. The recommended patterns to study are: Two Pointers, Array.

What are the interview tips for Duplicate Zeros?

Clarify if in-place modification is required. Think about edge cases like arrays with no zeros. Practice explaining your thought process while coding.

What are common mistakes when solving Duplicate Zeros?

Not accounting for the length of the original array when duplicating zeros. Using extra space unnecessarily when an in-place solution is possible.

#1089

Duplicate Zeros

Easy

ArrayTwo PointersTwo PointersArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal solution uses a two-pass approach to count zeros and adjust the array in place without needing extra space. This method ensures we only traverse the array a couple of times, making it efficient.

⚙️

Algorithm

3 steps

1Step 1: Count the number of zeros in the array while determining the effective length of the new array.
2Step 2: Iterate backwards through the array, placing elements in their new positions based on the counted zeros.
3Step 3: If a zero is encountered, place two zeros in the new position.

solution.py13 lines

1# Full working Python code
2
3def duplicateZeros(arr):
4    n = len(arr)
5    zeros = arr.count(0)
6    for i in range(n - 1, -1, -1):
7        if i + zeros < n:
8            arr[i + zeros] = arr[i]
9        if arr[i] == 0:
10            zeros -= 1
11            if i + zeros < n:
12                arr[i + zeros] = 0
13

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array a couple of times. The space complexity is O(1) since we are modifying the array in place without using extra space.

1Understanding in-place operations is crucial.
2Counting elements before modifying helps avoid overwriting data.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.