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2505. Bitwise OR of All Subsequence Sums
Description
Given an integer array nums
, return the value of the bitwise OR of the sum of all possible subsequences in the array.
A subsequence is a sequence that can be derived from another sequence by removing zero or more elements without changing the order of the remaining elements.
Example 1:
Input: nums = [2,1,0,3] Output: 7 Explanation: All possible subsequence sums that we can have are: 0, 1, 2, 3, 4, 5, 6. And we have 0 OR 1 OR 2 OR 3 OR 4 OR 5 OR 6 = 7, so we return 7.
Example 2:
Input: nums = [0,0,0] Output: 0 Explanation: 0 is the only possible subsequence sum we can have, so we return 0.
Constraints:
1 <= nums.length <= 105
0 <= nums[i] <= 109
Solutions
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class Solution { public long subsequenceSumOr(int[] nums) { long[] cnt = new long[64]; long ans = 0; for (int v : nums) { for (int i = 0; i < 31; ++i) { if (((v >> i) & 1) == 1) { ++cnt[i]; } } } for (int i = 0; i < 63; ++i) { if (cnt[i] > 0) { ans |= 1l << i; } cnt[i + 1] += cnt[i] / 2; } return ans; } }
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class Solution { public: long long subsequenceSumOr(vector<int>& nums) { vector<long long> cnt(64); long long ans = 0; for (int v : nums) { for (int i = 0; i < 31; ++i) { if (v >> i & 1) { ++cnt[i]; } } } for (int i = 0; i < 63; ++i) { if (cnt[i]) { ans |= 1ll << i; } cnt[i + 1] += cnt[i] / 2; } return ans; } };
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class Solution: def subsequenceSumOr(self, nums: List[int]) -> int: cnt = [0] * 64 ans = 0 for v in nums: for i in range(31): if (v >> i) & 1: cnt[i] += 1 for i in range(63): if cnt[i]: ans |= 1 << i cnt[i + 1] += cnt[i] // 2 return ans
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func subsequenceSumOr(nums []int) int64 { cnt := make([]int, 64) ans := 0 for _, v := range nums { for i := 0; i < 31; i++ { if v>>i&1 == 1 { cnt[i]++ } } } for i := 0; i < 63; i++ { if cnt[i] > 0 { ans |= 1 << i } cnt[i+1] += cnt[i] / 2 } return int64(ans) }