Minimum Adjacent Swaps for K Consecutive Ones

Minimum Adjacent Swaps for K Consecutive Ones

What is this problem about?

The Minimum Adjacent Swaps for K Consecutive Ones is a challenging problem involving a binary array (containing only 0s and 1s). You are given an integer k, and you need to find the minimum number of adjacent swaps (swapping two neighbors) required to bring any k ones together so they form a continuous block of 1s. This problem combines array indexing, distance calculations, and optimization.

Why is this asked in interviews?

This is a classic Minimum Adjacent Swaps for K Consecutive Ones interview question for senior roles at companies like Google and Microsoft. It requires the "Median Property" insight—the idea that to minimize the total distance to a meeting point, the meeting point should be the median. It tests whether a candidate can reduce a physical "swap" problem into a mathematical "distance" problem using Sliding Windows and Prefix Sums.

Algorithmic pattern used

The problem uses a combination of the Sliding Window interview pattern and the Median property.

1. Extract the indices of all 1s into a new array.
2. For any window of k indices, the best place to gather them is around the median index of that window.
3. The cost to move 1s to the median can be calculated efficiently using Prefix Sums of the indices array, avoiding a nested loop for each window.

Example explanation

Suppose the array is [1, 0, 0, 1, 0, 1] and k = 3.

1. Indices of ones: pos = [0, 3, 5].
2. Median index is pos[1] = 3.
3. To bring them together at 3, the ones should end up at positions [2, 3, 4].
4. Swaps needed:
   • One at 0 moves to 2: |0 - 2| = 2 swaps.
   • One at 3 stays at 3: 0 swaps.
   • One at 5 moves to 4: |5 - 4| = 1 swap.
5. Total = 3 swaps.

Common mistakes candidates make

• Simulating the swaps: Actually trying to perform swaps in the array. This is far too slow and complex.
• Not using the Median: Picking the first 1 or a random spot as the "target" for the cluster, which doesn't yield the minimum result.
• Ignoring Prefix Sums: Calculating the distance for each window in $O(K)$ time, resulting in an $O(N \cdot K)$ solution, whereas $O(N)$ is required.

Interview preparation tip

Master the "Cost of gathering elements" pattern. Whenever you need to bring scattered items together with minimum movement, the Median is almost always the optimal meeting point. Practicing this Prefix Sum interview pattern will help you solve many "Meeting Point" style problems.