Suppose we have a list of numbers called nums and another value k. We have to following operation k times: Select any number on the list. In the binary representation of that number, select a bit that is 0 and make it 1. Finally, we have to return the minimum possible sum of all the numbers after performing k operations. If the answer is too high, return result mode 10^9+7.
So, if the input is like nums = [4, 7, 3] k = 2, then the output will be 17, as the binary representation of 4 is 100, 3 is 011, and 7 is 111. Since we need to set 2 bits, we can set the bits of 4 to make it 111 (7). Then the total sum is then 7 + 7 + 3 = 17.
To solve this, we will follow these steps:
ans := 0, i := 0
while k is non-zero, do
for each n in nums, do
if (n / 2^i) is even, then
ans := ans + 2^i
k := k - 1
if k is same as 0, then
come out from the loop
i := i + 1
return (ans + sum of all elements of nums) mod m
Let us see the following implementation to get better understanding:
Example
class Solution: def solve(self, nums, k): m = (10 ** 9 + 7) ans = 0 i = 0 while k: for n in nums: if (n >> i) & 1 == 0: ans += 1 << i k -= 1 if k == 0: break i += 1 return (ans + sum(nums)) % m ob = Solution() nums = [4, 7, 3] k = 2 print(ob.solve(nums, k))
Input
[4, 7, 3], 2
Output
17