3354 - Make Array Elements Equal to Zero
problem
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
curris out of the range[0, n - 1], this process ends. - If
nums[curr] == 0, move in the current direction by incrementingcurrif you are moving right, or decrementingcurrif 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.
- Decrement
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 <= 1000 <= nums[i] <= 100- There is at least one element
iwherenums[i] == 0.
submission
// simulation is annoying and unnecessary
// the real question here is the sum of both parts
// separated by the choosen position
// if the absolute difference between both sum is 0
// then there are 2 valid selections available
// if the absolute difference between both sum is 1
// then there are only 1 valid selections
// anyother case there are none
impl Solution {
pub fn count_valid_selections(nums: Vec<i32>) -> i32 {
let sum: i32 = nums.iter().sum();
nums.into_iter()
.scan((0i32, sum), |&mut (ref mut f, ref mut b), val| {
*f += val;
*b -= val;
Some(match (val == 0, (*f - *b).abs()) {
(true, 0) => 2,
(true, 1) => 1,
_ => 0,
})
})
.sum()
}
}