Is Minimum Swaps to Move Zeros to End easy or hard?

The problem is typically classified as Easy because the optimal solution requires a single pass and simple counting logic. The main challenge is recognizing that each zero must swap with every non‑zero element to its right, which converts the problem into a straightforward counting task.

Minimum Swaps to Move Zeros to End Python/Java solution

Both Python and Java implementations follow the same linear scan logic. Iterate from the end of the array, maintain a count of non‑zero values, and add that count whenever a zero appears. The algorithm runs in O(n) time and constant space in either language.

How to solve Minimum Swaps to Move Zeros to End in O(n)?

Traverse the array from right to left and track how many non‑zero values have been seen so far. When encountering a zero, add the current non‑zero count to the swap total because the zero must cross each of those elements. When encountering a non‑zero value, increment the counter. This single pass computes the exact number of required swaps.

What is the best approach for Minimum Swaps to Move Zeros to End?

The optimal approach scans the array from right to left while counting how many non‑zero elements have appeared. Every time a zero is found, it must cross all those non‑zero elements to reach the end, so the swap count increases by that number. This method runs in O(n) time and uses O(1) extra space.

Is Minimum Swaps to Move Zeros to End asked at Google/Amazon/Meta?

Variants of moving zeros or counting swaps in arrays appear frequently in coding interviews at companies like Amazon, Google, and Meta. The exact wording may differ, but the core idea—counting inversions or using two‑pointer array techniques—is common in interview question banks.

What data structure is used in Minimum Swaps to Move Zeros to End?

The problem primarily uses arrays with simple counters or two pointers. No advanced data structures are required. The key idea is recognizing the relationship between zeros and the non‑zero elements that appear after them.

What is the time complexity of Minimum Swaps to Move Zeros to End?

The optimal solution runs in O(n) time because the array is scanned once while maintaining a counter of non‑zero elements. Space complexity is O(1) since only a few integer variables are used regardless of input size.

3936. Minimum Swaps to Move Zeros to End - Practice Online

You are given an integer array nums.

In one operation, you can choose any two distinct indices i and j and swap nums[i] and nums[j].

Return an integer denoting the minimum number of operations required to move all 0s to the end of the array.

Example 1:

Input: nums = [0,1,0,3,12]

Output: 2

Explanation:

We perform the following swap operations:

Swap nums[0] and nums[3], giving nums = [3, 1, 0, 0, 12].
Swap nums[2] and nums[4], giving nums = [3, 1, 12, 0, 0].

Thus, the answer is 2.

Example 2:

Input: nums = [0,1,0,2]

Output: 1

Explanation:

We perform the following swap operations:

Swap nums[0] and nums[3], giving nums = [2, 1, 0, 0].

Thus, the answer is 1.

Example 3:

Input: nums = [1,2,0]

Output: 0

Explanation:

The array already satisfies the condition. Therefore, no swap operations are needed.

Constraints:

1 <= nums.length <= 100
0 <= nums[i] <= 100