LeetCode 1703: Minimum Adjacent Swaps for K Consecutive Ones Solution

Master LeetCode problem 1703 (Minimum Adjacent Swaps for K Consecutive Ones), a hard 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.

1703. Minimum Adjacent Swaps for K Consecutive Ones

Hard

Array Greedy Sliding Window Prefix Sum

Problem Explanation

Explanation

To solve this problem, we can use a sliding window approach. We will maintain a window of size k, count the number of 1's within this window, and minimize the number of swaps required to make all elements within the window as 1's. We will keep track of the minimum number of swaps needed to achieve k consecutive 1's.

Here is the detailed algorithm:

Initialize variables: count = 0 (to count the number of 1's in the window), minSwaps = Integer.MAX_VALUE.
Initialize a window of size k and count the number of 1's in it.
Slide the window to the right by one position.
Update the count of 1's in the window:
- If the element going out of the window is 1, decrement count.
- If the new element coming into the window is 1, increment count.
Calculate the minimum swaps required to make all elements within the window as 1's.
Update the minSwaps with the minimum of current minSwaps and the swaps needed for the current window.
Repeat steps 3-6 until reaching the end of the array.
Return the minSwaps.

Time Complexity: O(n), where n is the length of the input array nums. We are iterating through the array once. Space Complexity: O(1), as we are using a constant amount of extra space.

Solution Code

class Solution {
    public int minMoves(int[] nums, int k) {
        int count = 0, minSwaps = Integer.MAX_VALUE;
        for (int i = 0; i < k; i++) {
            count += nums[i];
        }
        int ones = count;
        for (int i = k; i < nums.length; i++) {
            ones += nums[i] - nums[i - k];
            minSwaps = Math.min(minSwaps, k - ones);
        }
        return minSwaps;
    }
}

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