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Make Array Elements Equal to Zero

Number: 3616

Difficulty: Easy

Paid? No

Companies: N/A

Problem Description

Given an integer array nums, the task is to simulate a process where you first select a starting position curr such that nums[curr] == 0 and choose an initial movement direction (left or right). During the process, if nums[curr] is 0, you simply move one step in the current direction; if nums[curr] > 0, you decrement it by 1, reverse your direction, and then take one step. The process terminates when curr goes out of bounds. A selection (starting index and direction) is valid if, by the time the process ends, every element in nums has been reduced to 0. The goal is to count the number of valid selections.

Key Insights

  • Simulate the process separately for every potential starting index that initially contains 0.
  • For each valid starting position, try both movement directions (left and right).
  • During simulation, use a copy of the original array since each simulation modifies the array.
  • The simulation continues until the current index goes out of bounds.
  • At the end of the simulation, check whether the entire array has been reduced to zero.

Space and Time Complexity

Time Complexity: O(n * simulation_steps) per candidate, where simulation_steps depend on the number of decrements performed. Given the constraints, simulation is efficient. Space Complexity: O(n) per simulation due to array copying.

Solution

The solution involves iterating over all indices of the array and considering those that initially have a value of 0 as candidate starting positions. For each candidate, we simulate the process twice—once moving left and once moving right. During the simulation, a deep copy of the array is used so that each simulation is independent. The simulation consists of a while loop that terminates when the current index goes out of bounds. Within the loop, if the value at the current index is 0, the pointer is moved in the current direction; if the value is greater than 0, it is decremented, the direction is reversed, and then the pointer is moved accordingly. After the simulation finishes, the resulting array is checked to ensure all elements are 0. If they are, the candidate selection is counted as valid.

Code Solutions

Below are implementations in Python, JavaScript, C++, and Java with detailed line-by-line comments.

def is_valid(nums, start, direction):
    # Create a copy of the array to simulate on without modifying the original
    arr = nums[:]
    curr = start
    # Set movement delta: 1 for right, -1 for left
    dir_delta = 1 if direction == "right" else -1
    
    # Process simulation until curr goes out of array bounds
    while 0 <= curr < len(arr):
        if arr[curr] == 0:
            # If current element is 0, move in the current direction
            curr += dir_delta
        else:
            # Decrement the value at current position
            arr[curr] -= 1
            # Reverse the movement direction
            dir_delta = -dir_delta
            # Move one step in the new direction
            curr += dir_delta
    # After simulation, verify all elements are reduced to 0
    return all(x == 0 for x in arr)

def countValidSelections(nums):
    count = 0
    # Iterate over all indices; candidate must start where element is 0
    for i in range(len(nums)):
        if nums[i] == 0:
            for direction in ["left", "right"]:
                if is_valid(nums, i, direction):
                    count += 1
    return count

# Example usage:
nums = [1, 0, 2, 0, 3]
print(countValidSelections(nums))
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