diff options
Diffstat (limited to 'apply_operations_arr_max_sqr.py')
-rw-r--r-- | apply_operations_arr_max_sqr.py | 31 |
1 files changed, 31 insertions, 0 deletions
diff --git a/apply_operations_arr_max_sqr.py b/apply_operations_arr_max_sqr.py new file mode 100644 index 0000000..e3b5ade --- /dev/null +++ b/apply_operations_arr_max_sqr.py @@ -0,0 +1,31 @@ +# 2897. Apply Operations on Array to Maximize Sum of Squares +from collections import Counter + +""" +you are given a 0-indexed integer array 'nums' and a positive integer 'k'. +you can do the following operation on the array any number of times: choose +any two distinct indices 'i' and 'j' and simultaneously update the values of +'nums[i]' to '(nums[i] & nums[j])' and 'nums[j]' to '(nums[i] | nums[j])'. +you have to choose 'k' elements from the final array and calcluate the sum of +their squares. return the maximum sum of squares you can achieve. since the +answer can be very large, return it modulo 10^9+7. +""" + + +class Solution(object): + def maxSum(self, nums, k): + """ + :type nums: List[int] + :type k: int + :rtype: int + """ + cnt = Counter(i for j in nums for i in range(32) if j & (1 << i)) + return sum(sum((cnt[i] > j) << i for i in range(32)) ** 2 for j in range(k)) % ( + 10**9 + 7 + ) + + +if __name__ == "__main__": + obj = Solution() + print(obj.maxSum(nums=[2, 6, 5, 8], k=2)) + print(obj.maxSum(nums=[4, 5, 4, 7], k=3)) |