text: /*
  Minimum Numbers of Function Calls to Make Target Array
  
  
  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.
  
  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).
  Example 4:
  
  Input: nums = [3,2,2,4]
  Output: 7
  Example 5:
  
  Input: nums = [2,4,8,16]
  Output: 8
   
  
  Constraints:
  
  1 <= nums.length <= 10^5
  0 <= nums[i] <= 10^9
  */
  
  class Solution {
  public:
      int minOperations(vector<int>& nums) {
          // to think backwards instead
          // if the num is odd, decrement it by 1 (operation 0)
          // else divide it by 2 (operation 1)
          int op0=0,op1=0,highest=0;
          for(int num:nums){
              int bits=0;
              while(num>0){
                  if(num&1) op0++;
                  bits++;
                  num>>=1;
              }
              // find the highest bit for op 1
              highest=max(highest,bits);
          }
          op1=highest-1;
          return op0+op1;
      }
  };