2599. Make the Prefix Sum Non-negative

Problem Description

This problem presents an optimization challenge with an integer array nums. The main objective is to transform this array so that its prefix sum array contains no negative integers. To clarify, a prefix sum array is one where each element at index i is the total sum of all elements from the start of the array to that index, inclusive. The transformation of the array nums can only be achieved through a series of operations, each of which involves selecting one element from the array and moving it to the end.

The task is to figure out the minimum number of such operations needed to ensure that all the sums in the prefix sum array are non-negative. It is confirmed that there is always a way to rearrange the elements of the initial array in order to achieve this non-negative prefix sum property.

Intuition

The solution hinges on the greedy algorithm and a priority queue (which is implemented as a min heap in Python). The intuition behind the greedy approach is that by continually relocating the most negative value from the array to the end, we decrease the sum in the most effective way possible, which in turn quickly leads to a non-negative prefix sum at each step.

We start by traversing the array and calculating the prefix sums using a running sum variable s. When we come across a negative number, we put it into our priority queue (min heap). If at any point our running sum s becomes negative, it means our current prefix sum array contains negative values. In order to fix this, we repeatedly remove the smallest (most negative) number from the heap, since removing this number will give us the largest positive change towards making the sum non-negative. The number of times we need to remove an element from the heap to make the running sum non-negative at each step is added to our answer total, ans.

The loop structure ensures that we only pop from the heap when necessary, i.e., when the running sum is negative. This combination of a greedy approach with a priority queue allows us to efficiently manage and adjust the most negative elements which are contributing to a negative prefix sum. The counting of heap removals gives us the minimum number of operations required to prevent any negatives in the prefix sum array.

Learn more about Greedy and Heap (Priority Queue) patterns.

Solution Approach

The provided solution uses a greedy approach in conjunction with a priority queue (min heap) data structure from Python's heapq library, which enables efficient retrieval of the smallest element.

Following are step-by-step implementation details:

1. Initialize an empty min heap h, an accumulator ans to count the number of moves, and a sum s to keep track of the prefix sum.

2. Iterate over each element x in the array nums:

   • Add the current element to the prefix sum s (= s + x).
   • If x is negative, push it onto the min heap h using heappush(h, x).

3. After each addition of an element to s, check if s is negative:

   • While s is negative, repeatedly:
     • Pop the smallest (most negative) element from the min heap h using heappop(h).
     • Subtract the popped element from s, which will make s less negative or non-negative.
     • Increment ans by 1 representing an operation.

4. Once all elements have been processed, return ans, which now contains the count of the minimum number of operations needed to ensure the prefix sum array is non-negative.

The algorithm efficiently rearranges the elements by virtually moving the most negative values to the end of the array, thus not needing to manipulate the array directly. Instead, we operate on the prefix sum and extract the negative impact whenever it threatens to pull the sum below zero.

As we traverse nums, we accumulate negative values into the min heap. Popping from the min heap gives us the smallest negative number quickly, which is the ideal candidate to "move" to the end (in a virtual sense) because its removal will have the biggest immediate positive (or least negative) impact on s.

The solution approach takes advantage of the properties of a min heap, where ensuring the heap structure after each insertion or deletion operation (i.e., heappush and heappop) takes O(log n) time, thus each operation on the heap is efficient even as nums grows in size.

In summary, the solution applies algorithms and data structures (greedy technique and priority queue) to cleverly and efficiently find the minimum set of adjustments to nums that guarantees a non-negative prefix sum array.

Example Walkthrough

Consider the following small example array nums: [3, -7, 4, -2, 1, 2]

1. We initialize the following:

   • Min heap h: []
   • Operation counter ans: 0
   • Prefix sum s: 0

2. Process the array nums:

   • For the first element (3):
     • s becomes 3 (0 + 3). s is non-negative, so no need for changes.
     • Min heap h remains empty.
   • For the second element (-7):
     • s becomes -4 (3 - 7). s is negative.
     • We push -7 onto the min heap h: [-7]
     • Since s is negative, we pop from h (-7), add it back to s (+7), and increment ans by 1.
     • s is now 3 (-4 + 7), and ans is 1.
     • Min heap h is now empty.
   • For the third element (4):
     • s becomes 7 (3 + 4). s is non-negative.
     • Min heap h remains empty.
   • For the fourth element (-2):
     • s becomes 5 (7 - 2). s is non-negative.
     • We push -2 onto the min heap h: [-2]
   • For the fifth element (1):
     • s becomes 6 (5 + 1). s is non-negative.
     • Min heap h remains as: [-2]
   • For the sixth element (2):
     • s becomes 8 (6 + 2). s is non-negative.
     • Min heap h remains as: [-2]

3. Since we've finished processing nums and s is non-negative, ans remains at 1.

4. Thus, the minimum number of operations needed to ensure the prefix sum array is non-negative for nums is 1.

In this example, only one operation was needed, which involved moving the -7 to the end of the array to ensure all prefix sums were non-negative. This step is conceptual as we actually just remove the negative influence of -7 from the running sum s.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(n * log n). This is because the function iterates through all n elements of the input list nums. For each negative number encountered, it performs a heappush operation which is O(log k), where k is the number of negative numbers encountered so far, leading to a maximum of O(log n) when all elements are negative. Additionally, when the sum s becomes negative, the code performs a heappop in a while loop until s is non-negative again. In the worst case, this could involve popping every number that was pushed, resulting in a sequence of heappop operations. Since each heappop operation is O(log k), and you could theoretically pop every element once, the total time for all the heap operations across the entire list will be O(n * log n).

The space complexity of the given code is O(n). This is due to the additional heap h that is used to store the negative numbers. In the worst-case scenario, all elements in the list are negative and will be added to the heap. Since the heap can contain all negative numbers of the list at once, the space required for the heap is proportional to the input size, hence the space complexity is O(n).

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