Question

1558. Minimum Numbers of Function Calls to Make Target Array

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Description

You are given an integer array nums. You have an integer array arr of the same length with all values set to 0 initially. You also have the following modify function:

You want to use the modify function to convert arr to nums using the minimum number of calls.

Return _the minimum number of function calls to make _nums_ from _arr.

The test cases are generated so that the answer fits in a 32-bit signed integer.

 

Example 1:

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.

Example 2:

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.

Example 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).

 

Constraints:

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

Solutions

Solution 1

class Solution:
  def minOperations(self, nums: List[int]) -> int:
    return sum(v.bit_count() for v in nums) + max(0, max(nums).bit_length() - 1)

class Solution {
  public int minOperations(int[] nums) {
    int ans = 0;
    int mx = 0;
    for (int v : nums) {
      mx = Math.max(mx, v);
      ans += Integer.bitCount(v);
    }
    ans += Integer.toBinaryString(mx).length() - 1;
    return ans;
  }
}

class Solution {
public:
  int minOperations(vector<int>& nums) {
    int ans = 0;
    int mx = 0;
    for (int v : nums) {
      mx = max(mx, v);
      ans += __builtin_popcount(v);
    }
    if (mx) ans += 31 - __builtin_clz(mx);
    return ans;
  }
};

func minOperations(nums []int) int {
  ans, mx := 0, 0
  for _, v := range nums {
    mx = max(mx, v)
    for v > 0 {
      ans += v & 1
      v >>= 1
    }
  }
  if mx > 0 {
    for mx > 0 {
      ans++
      mx >>= 1
    }
    ans--
  }
  return ans
}