Problem 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.
Solution Code
class Solution {
public:
int findKOr(vector& nums, int k) {
int answer = 0;
for (int i = 0; i < 32; i++) {
int c = 0;
for (int num : nums) {
if ((num & (1 << i)) != 0) {
c++;
}
}
if (c >= k) {
answer |= (1 << i);
}
}
return answer;
}
};
Solution Explanation
The problem involves calculating the K-or of an array where a bit position in the result is set to 1 if at least k numbers in the array have a 1 in that position. The solution follows these steps:
- Iterate Over Bit Positions: Check each bit position from 0 to 31.
- Count Bits: Count how many numbers have a 1 at each bit position.
- Check Condition: If the count of 1s at a bit position is greater than or equal to
k, set that bit in the result. - Compute Result: Accumulate the result by setting the appropriate bits based on the condition.