Your task is to form an integer array nums
from an initial array of zeros arr
that is the same size as nums
.
Return the minimum number of function calls to make nums
from arr
.
The answer is guaranteed to fit in a 32-bit signed integer.
Input: nums = [1,5] Output: 5 Explanation: Increment by 1 (second element): [0, 0] to get [0, 1] (1 operation). Double all the elements: [0, 1] -> [0, 2] -> [0, 4] (2 operations). Increment by 1 (both elements) [0, 4] -> [1, 4] -> [1, 5] (2 operations). Total of operations: 1 + 2 + 2 = 5.
Input: nums = [2,2] Output: 3 Explanation: Increment by 1 (both elements) [0, 0] -> [0, 1] -> [1, 1] (2 operations). Double all the elements: [1, 1] -> [2, 2] (1 operation). Total of operations: 2 + 1 = 3.
Input: nums = [4,2,5] Output: 6 Explanation: (initial)[0,0,0] -> [1,0,0] -> [1,0,1] -> [2,0,2] -> [2,1,2] -> [4,2,4] -> [4,2,5](nums).
Input: nums = [3,2,2,4] Output: 7
Input: nums = [2,4,8,16] Output: 8
1 <= nums.length <= 10^5
0 <= nums[i] <= 10^9
impl Solution {
pub fn min_operations(nums: Vec<i32>) -> i32 {
let mut nums = nums;
let mut op0 = 0;
let mut op1 = 0;
for mut num in nums {
let mut tmp = 0;
while num > 0 {
op0 += num % 2;
tmp += 1;
num /= 2;
}
op1 = op1.max(tmp - 1);
}
op0 + op1
}
}