Problem Description

You are given an integer array nums. The task is to start by selecting a starting position curr such that nums[curr] == 0, and then choose a movement direction, either left or right. Following that, repeat the following process:

• If curr is out of the range [0, n - 1], the process ends.
• If nums[curr] == 0, move in the current direction by incrementing curr if moving right, or decrementing curr if moving left.
• If nums[curr] > 0:
  • Decrease nums[curr] by 1.
  • Reverse your movement direction (left becomes right and vice versa).
  • Take a step in your new direction.

A selection of the initial position curr and movement direction is considered valid if every element in nums becomes 0 by the end of the process. The goal is to return the number of possible valid selections.

Intuition

The problem involves navigating through an array with certain initial choices, and the goal is to identify configurations that will result in the array becoming all zeros. The key insight is to think in terms of balance. We can align this "balance" concept by looking at prefix sums:

1. Calculate a total sum s of the array.
2. For each zero-value element in the array, calculate the sum of all the elements to its left (l). The elements to the right will sum to s - l.
3. Check two scenarios:
   • If the sum of the left and right are equal (l = s - l), moving from the zero in either direction will eventually balance the whole array to zero. Such a case can be approached from both directions, hence contributes 2 valid solutions.
   • If the absolute difference between the two sums (|l - (s - l)|) is 1, start from the zero and choose the direction towards the larger side to balance the entire array to zeros. This scenario provides 1 valid solution.

This logic is implemented efficiently with a single pass through the array and uses the prefix calculation.

Learn more about Prefix Sum patterns.

Solution Approach

The solution leverages the concept of prefix sums to keep track of the potential valid starting points and directions. Here's how it works:

1. Summation: First, compute the total sum s of all elements in the array nums.

2. Initialization: Initialize two variables: ans (to store the number of valid selections) and l (to keep a running prefix sum of elements encountered thus far).

3. Iteration: Iterate through each element x in nums:

   • If x is not zero, accumulate it into l.
   • If x is zero:
     • Check if 2 * l == s. This condition implies the sum of elements on both sides of the current zero are equal, hence the array can be balanced by starting movement either left or right from this point. Increase ans by 2 for these two possible directions.
     • Otherwise, if |2 * l - s| == 1, it indicates that the absolute difference between the left and right sums is 1. Depending on which side is larger, the direction from this zero value can be selected. Increase ans by 1, as there's only one valid direction.

4. Return Result: Finally, return ans as the count of valid selections.

Using this approach, the problem is solved efficiently with a single pass through the array, making it a linear time complexity solution.

Example Walkthrough

Let's consider the array nums = [1, 0, 2, 1, 0, 2, 3, 0] to illustrate the solution approach.

Step 1: Summation

Compute the total sum s of the array:
[ s = 1 + 0 + 2 + 1 + 0 + 2 + 3 + 0 = 9 ]

Step 2: Initialization

Initialize ans = 0 to store the number of valid selections, and l = 0 for the running prefix sum.

Step 3: Iteration

Iterate through each element:

1. Index 0 (x = 1):
   Add x to l:
   [ l = l + 1 = 1 ]

2. Index 1 (x = 0):
   Check conditions:

   • ( 2 \times l = 2 \times 1 = 2 \neq 9 )
   • (|2 \times l - s| = |2 - 9| = 7 \neq 1)
     No valid starting positions from this zero.

3. Index 2 (x = 2):
   Add x to l:
   [ l = l + 2 = 3 ]

4. Index 3 (x = 1):
   Add x to l:
   [ l = l + 1 = 4 ]

5. Index 4 (x = 0):
   Check conditions:

   • ( 2 \times l = 2 \times 4 = 8 \neq 9 )
   • (|2 \times l - s| = |8 - 9| = 1)
     The absolute difference is 1, thus a selection starting from here can balance the array.
     Increment ans by 1:
     [ ans = 1 ]

6. Index 5 (x = 2):
   Add x to l:
   [ l = l + 2 = 6 ]

7. Index 6 (x = 3):
   Add x to l:
   [ l = l + 3 = 9 ]

8. Index 7 (x = 0):
   Check conditions:

   • ( 2 \times l = 2 \times 9 = 18 \neq 9 )
   • (|2 \times l - s| = |18 - 9| = 9 \neq 1)
     No valid selections from this zero.

Step 4: Return Result

The count of valid selections is ( \text{ans} = 1 ).

Thus, using the solution approach, we find that there is 1 valid way to select a starting position and direction so that all elements in nums would eventually become zero.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n), where n is the length of the array nums. This is because each element in the array is processed exactly once in a single pass through the array.

The space complexity of the code is O(1), as no additional data structures that grow with the input size n are used. The variables s, ans, and l consume constant space.

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