3354. Make Array Elements Equal to Zero

Description

You are given an integer array nums.

Start by selecting a starting position curr such that nums[curr] == 0, and choose a movement direction of either left or right.

After that, you repeat the following process:

• If curr is out of the range [0, n - 1], this process ends.
• If nums[curr] == 0, move in the current direction by incrementing curr if you are moving right, or decrementing curr if you are moving left.
• Else if nums[curr] > 0:
  • Decrement 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.

Return the number of possible valid selections.

 

Example 1:

Input: nums = [1,0,2,0,3]

Output: 2

Explanation:

The only possible valid selections are the following:

• Choose curr = 3, and a movement direction to the left.
  • [1,0,2,0,3] -> [1,0,2,0,3] -> [1,0,1,0,3] -> [1,0,1,0,3] -> [1,0,1,0,2] -> [1,0,1,0,2] -> [1,0,0,0,2] -> [1,0,0,0,2] -> [1,0,0,0,1] -> [1,0,0,0,1] -> [1,0,0,0,1] -> [1,0,0,0,1] -> [0,0,0,0,1] -> [0,0,0,0,1] -> [0,0,0,0,1] -> [0,0,0,0,1] -> [0,0,0,0,0].
• Choose curr = 3, and a movement direction to the right.
  • [1,0,2,0,3] -> [1,0,2,0,3] -> [1,0,2,0,2] -> [1,0,2,0,2] -> [1,0,1,0,2] -> [1,0,1,0,2] -> [1,0,1,0,1] -> [1,0,1,0,1] -> [1,0,0,0,1] -> [1,0,0,0,1] -> [1,0,0,0,0] -> [1,0,0,0,0] -> [1,0,0,0,0] -> [1,0,0,0,0] -> [0,0,0,0,0].

Example 2:

Input: nums = [2,3,4,0,4,1,0]

Output: 0

Explanation:

There are no possible valid selections.

 

Constraints:

• 1 <= nums.length <= 100
• 0 <= nums[i] <= 100
• There is at least one element i where nums[i] == 0.

 

Solutions

Solution: Prefix Sum

• Time complexity: O(n)
• Space complexity: O(1)

 

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
const countValidSelections = function (nums) {
  const n = nums.length;
  let prefixSum = 0;
  let suffixSum = nums.reduce((sum, num) => sum + num);
  let result = 0;

  for (let index = 0; index < n; index++) {
    const num = nums[index];

    prefixSum += num;
    suffixSum -= num;

    if (num) continue;
    if (prefixSum === suffixSum) {
      result += 2;
    }

    if (Math.abs(prefixSum - suffixSum) === 1) {
      result += 1;
    }
  }

  return result;
};