LeetCode 3097: Shortest Subarray With OR at Least K II Solution

Master LeetCode problem 3097 (Shortest Subarray With OR at Least K II), a medium 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.

3097. Shortest Subarray With OR at Least K II

Medium

Array Bit Manipulation Sliding Window

Problem Explanation

Explanation

To solve this problem, we can use a deque (double-ended queue) to maintain a monotonically increasing prefix sum array. We iterate through the input array and keep track of the prefix sum and its corresponding index in the deque. Whenever the current prefix sum is greater than or equal to k, we update the result with the minimum subarray length and remove elements from the front of the deque until the prefix sum is less than k. This helps in finding the shortest special subarray.

Time Complexity

The time complexity of this approach is O(n), where n is the number of elements in the input array.

Space Complexity

The space complexity of this approach is O(n) for the deque.

Solution Code

class Solution {
    public int shortestSubarray(int[] nums, int k) {
        Deque<Integer> deque = new LinkedList<>();
        int n = nums.length;
        int[] prefixSum = new int[n + 1];
        int result = Integer.MAX_VALUE;

        for (int i = 0; i < n; i++) {
            prefixSum[i + 1] = prefixSum[i] + nums[i];
        }

        for (int i = 0; i <= n; i++) {
            while (!deque.isEmpty() && prefixSum[i] - prefixSum[deque.peekFirst()] >= k) {
                result = Math.min(result, i - deque.pollFirst());
            }
            while (!deque.isEmpty() && prefixSum[i] <= prefixSum[deque.peekLast()]) {
                deque.pollLast();
            }
            deque.offerLast(i);
        }

        return result == Integer.MAX_VALUE ? -1 : result;
    }
}

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