README.md

1658. Minimum Operations to Reduce X to Zero

You are given an integer array nums and an integer x. In one operation, you can either remove the leftmost or the rightmost element from the array nums and subtract its value from x. Note that this modifies the array for future operations.

Return the minimum number of operations to reduce x to exactly 0 if it's possible, otherwise, return -1.

Example 1:

Input: nums = [1,1,4,2,3], x = 5
Output: 2
Explanation: The optimal solution is to remove the last two elements to reduce x to zero.

Example 2:

Input: nums = [5,6,7,8,9], x = 4
Output: -1

Example 3:

Input: nums = [3,2,20,1,1,3], x = 10
Output: 5
Explanation: The optimal solution is to remove the last three elements and the first two elements (5 operations in total) to reduce x to zero.

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 104
• 1 <= x <= 109

Solutions (Ruby)

1. Two Pointers

# @param {Integer[]} nums
# @param {Integer} x
# @return {Integer}
def min_operations(nums, x)
  l = nums.size
  r = 0
  sum = nums.sum
  ret = -1

  while r < nums.size
    ret = l + r if sum == x && (ret == -1 || l + r < ret)
    if (sum > x && l > 0) || l + r >= nums.size
      l -= 1
      sum -= nums[l]
    else
      r += 1
      sum += nums[-r]
    end
  end

  ret
end

Solutions (Rust)

1. Two Pointers

impl Solution {
    pub fn min_operations(nums: Vec<i32>, x: i32) -> i32 {
        let mut l = nums.len();
        let mut r = 0;
        let mut sum = nums.iter().sum::<i32>();
        let mut ret = -1;

        while r < nums.len() {
            if sum == x && (ret == -1 || l + r < ret as usize) {
                ret = (l + r) as i32;
            }
            if (sum > x && l > 0) || l + r >= nums.len() {
                l -= 1;
                sum -= nums[l];
            } else {
                r += 1;
                sum += nums[nums.len() - r];
            }
        }

        ret
    }
}