862. Shortest Subarray with Sum at Least K

Problem Description

The problem asks us to find the length of the shortest contiguous subarray within an integer array nums such that the sum of its elements is at least a given integer k. If such a subarray does not exist, we are to return -1. The attention is on finding the minimum-length subarray that meets or exceeds the sum condition.

Intuition

To solve this efficiently, we utilize a monotonic queue and prefix sums technique. The intuition behind using prefix sums is that they allow us to quickly calculate the sum of any subarray in constant time. This makes the task of finding subarrays with a sum of at least k much faster, as opposed to calculating the sum over and over again for different subarrays.

A prefix sum array s is an array that holds the sum of all elements up to the current index. So for any index i, s[i] is the sum of nums[0] + nums[1] + ... + nums[i-1].

To understand the monotonic queue, which is a Double-Ended Queue (deque) in this case, let's look at its properties:

1. It is used to maintain a list of indices of prefix sums in increasing order.
2. When we add a new prefix sum, we remove all the larger sums at the end of the queue because the new sum and any future sums would always be better choices (smaller subarray) for finding a subarray of sum k.
3. We also continuously check if the current prefix sum minus the prefix sum at the start of the queue is at least k. If it is, we found a candidate subarray, and update ans with the subarray's length. Then we can pop that element off the queue since we've already considered this subarray and it won't be needed for future calculations.

In summary, the prefix sums help us quickly compute subarray sums, and the monotonic queue lets us store and traverse candidate subarray ranges efficiently, ensuring we always have the smallest length subarray that meets the sum condition, thus arriving at the optimal solution.

Learn more about Queue, Binary Search, Prefix Sum, Sliding Window, Monotonic Queue and Heap (Priority Queue) patterns.

Solution Approach

The solution makes use of prefix sums and a monotonic queue, specifically a deque, to achieve an efficient algorithm to find the shortest subarray summing to at least k. Let's explore the steps involved:

1. Prefix Sum Calculation: We initiate the computation by creating a list called s that contains the cumulative sum of the nums list, by using the accumulate function with an initial parameter set to 0. This denotes that the first element of s is 0 and is a requirement to consider subarrays starting at index 0.

2. Initialization: A deque q is initialized to maintain the monotonic queue of indices. A variable ans is initialized to inf (infinity) which will later store the smallest length of a valid subarray.

3. Iterate Over Prefix Sums: We iterate over each value v in the prefix sums array s and its index i.

4. Deque Front Comparison: While there are elements in q and the current prefix sum v minus the prefix sum at q[0] (the front of the deque) is greater than or equal to k, we've found a subarray that meets the requirement. We then compute its length i - q.popleft() and update ans with the minimum of ans and this length. The popleft() operation removes this index from the deque as it is no longer needed.

5. Deque Back Optimization: Before appending the current index i to q, we pop indices from the back of the deque if their corresponding prefix sums are greater than or equal to v because these are not conducive to the smallest length requirement and will not be optimal candidates for future comparisons.

6. Index Appending: Append the current index i to q. This index represents the right boundary for the potential subarray sums evaluated in future iterations.

After the loop, two cases may arise:

• If ans remains inf, it means no valid subarray summing to at least k was found, so we return -1.
• Otherwise, we return the value stored in ans as it represents the length of the shortest subarray that fulfills the condition.

Overall, the algorithm smartly maintains a set of candidate indices for the start of the subarray in a deque, ensuring that only those that can potentially give a smaller length subarray are considered. The key here is to understand how the prefix sum helps us quickly calculate the sum of a subarray and how the monotonically increasing property of the queue ensures we always get the shortest length possible. The use of these data structures makes the solution capable of working in O(n) time complexity.

Example Walkthrough

Let's illustrate the solution approach with an example. Suppose we have the following array and target sum k:

1nums = [2, 1, 5, 2, 3, 2]
2k = 7

We want to find the length of the smallest contiguous subarray with a sum greater than or equal to 7. Here's how we would apply the solution approach:

1. Prefix Sum Calculation: We compute the prefix sum array s:

   1s = [0, 2, 3, 8, 10, 13, 15]

   Notice that s[0] is 0 because we've added it artificially to account for subarrays starting at index 0.

2. Initialization: We create a deque q to maintain indices of prefix sums:

   1q = []

   And initialize ans to inf:

   1ans = inf

3. Iterate Over Prefix Sums: We iterate over s, looking for subarrays that sum up to at least k.

4. Deque Front Comparison: As we proceed, when we reach s[3] = 8 (consider nums[0..2]), we find it is greater than or equal to k. The q is empty, so we just move on.

5. Deque Back Optimization: Before appending index 3 to q, we don't remove anything from q because it's still empty. So we append 3 into q.

6. Index Appending: Our q is now [3].

   Continuing the iteration, we eventually come to s[5] = 13, which is the sum up to nums[0..4].

   We have a non-empty q, and s[5] - s[q[0]] = 13 - 8 = 5, which is not greater than or equal to k. So we can't pop anything from the queue yet.

7. Once we reach s[6] = 15, we notice that 15 - 8 = 7 is exactly our k. We then calculate the subarray length 6 - q.popleft() = 6 - 3 = 3. Now the ans becomes 3, the smallest subarray [5, 2, 3] found so far.

8. Deque Back Optimization is also done each time before we append a new index, which keeps the indices in the deque monotonically increasing in sums.

After considering all elements in s, our ans is 3, as no smaller subarray summing to at least 7 is found. So we return 3 as the length of the shortest subarray. If ans was still infinity, we would return -1 indicating no such subarray was found.

Solution Implementation

Time and Space Complexity

The given Python code is for finding the length of the shortest contiguous subarray whose sum is at least k. The code uses the concept of prefix sums and a monotonic queue to keep track of potential candidates for the shortest subarray.

Time Complexity

The time complexity of the code is O(N), where N is the length of the input array nums. This is because:

• The prefix sum array s is computed using itertools.accumulate, which is a single pass through the array, thus taking O(N) time.
• The deque q is maintained by iterating through each element of the array once. Each element is added and removed from the deque at most once. Since the operations of adding to and popping from a deque are O(1), the loop operations are also O(N) in total.
• Inside the loop, q.popleft() and q.pop() are each called at most once per iteration, and as a result, each element in nums contributes at most O(1) to the time complexity.

Space Complexity

The space complexity of the code is O(N) as well:

• The prefix sum array s requires O(N) space.
• The deque q potentially stores indices from the entire array in the worst-case scenario. In the worst case, it could hold all indices in nums, also requiring O(N) space.
• Apart from these two data structures, the code uses a constant amount of space for variables such as ans and v, which does not depend on the input size.

Learn more about how to find time and space complexity quickly using problem constraints.