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Weekly Contest 369 Q1
Bit Manipulation
Array

2917. Find the K-or of an Array

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Description

You are given an integer array nums, and an integer k. Let's introduce K-or operation by extending the standard bitwise OR. In K-or, a bit position in the result is set to 1 if at least k numbers in nums have a 1 in that position.

Return the K-or of nums.

 

Example 1:

Input: nums = [7,12,9,8,9,15], k = 4

Output: 9

Explanation:

Represent numbers in binary:

Number Bit 3 Bit 2 Bit 1 Bit 0
7 0 1 1 1
12 1 1 0 0
9 1 0 0 1
8 1 0 0 0
9 1 0 0 1
15 1 1 1 1
Result = 9 1 0 0 1

Bit 0 is set in 7, 9, 9, and 15. Bit 3 is set in 12, 9, 8, 9, and 15.
Only bits 0 and 3 qualify. The result is (1001)2 = 9.

Example 2:

Input: nums = [2,12,1,11,4,5], k = 6

Output: 0

Explanation: No bit appears as 1 in all six array numbers, as required for K-or with k = 6. Thus, the result is 0.

Example 3:

Input: nums = [10,8,5,9,11,6,8], k = 1

Output: 15

Explanation: Since k == 1, the 1-or of the array is equal to the bitwise OR of all its elements. Hence, the answer is 10 OR 8 OR 5 OR 9 OR 11 OR 6 OR 8 = 15.

 

Constraints:

  • 1 <= nums.length <= 50
  • 0 <= nums[i] < 231
  • 1 <= k <= nums.length

Solutions

Solution 1: Enumeration

We can enumerate each bit $i$ in the range $[0, 32)$, and count the number of numbers in the array $nums$ whose $i$-th bit is $1$, denoted as $cnt$. If $cnt \ge k$, we add $2^i$ to the answer.

After the enumeration, we return the answer.

The time complexity is $O(n \times \log M)$, where $n$ and $M$ are the length of the array $nums$ and the maximum value in $nums$, respectively. The space complexity is $O(1)$.

Python3

class Solution:
    def findKOr(self, nums: List[int], k: int) -> int:
        ans = 0
        for i in range(32):
            cnt = sum(x >> i & 1 for x in nums)
            if cnt >= k:
                ans |= 1 << i
        return ans

Java

class Solution {
    public int findKOr(int[] nums, int k) {
        int ans = 0;
        for (int i = 0; i < 32; ++i) {
            int cnt = 0;
            for (int x : nums) {
                cnt += (x >> i & 1);
            }
            if (cnt >= k) {
                ans |= 1 << i;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int findKOr(vector<int>& nums, int k) {
        int ans = 0;
        for (int i = 0; i < 32; ++i) {
            int cnt = 0;
            for (int x : nums) {
                cnt += (x >> i & 1);
            }
            if (cnt >= k) {
                ans |= 1 << i;
            }
        }
        return ans;
    }
};

Go

func findKOr(nums []int, k int) (ans int) {
	for i := 0; i < 32; i++ {
		cnt := 0
		for _, x := range nums {
			cnt += (x >> i & 1)
		}
		if cnt >= k {
			ans |= 1 << i
		}
	}
	return
}

TypeScript

function findKOr(nums: number[], k: number): number {
    let ans = 0;
    for (let i = 0; i < 32; ++i) {
        let cnt = 0;
        for (const x of nums) {
            cnt += (x >> i) & 1;
        }
        if (cnt >= k) {
            ans |= 1 << i;
        }
    }
    return ans;
}

C#

public class Solution {
    public int FindKOr(int[] nums, int k) {
        int ans = 0;
        for (int i = 0; i < 32; ++i) {
            int cnt = 0;
            foreach (int x in nums) {
                cnt += (x >> i & 1);
            }
            if (cnt >= k) {
                ans |= 1 << i;
            }
        }
        return ans;
    }
}