3314. Construct the Minimum Bitwise Array I
Explanation:
To construct the array ans, we need to find the smallest values for each element such that the bitwise OR of an element and the next element is equal to the corresponding element in the nums array. We can achieve this by iterating through the nums array and finding the rightmost set bit in each element to determine the value of ans[i].
- Find the rightmost set bit in
nums[i]usingInteger.lowestOneBit(nums[i]). - Construct
ans[i]by setting all bits to the right of the rightmost set bit to 1. - Check if
ans[i] OR (ans[i] + 1)equalsnums[i]. If not, setans[i] = -1.
Time Complexity: O(n), where n is the length of the nums array.
Space Complexity: O(n) for storing the ans array.
:
class Solution {
public int[] solve(int[] nums) {
int n = nums.length;
int[] ans = new int[n];
for (int i = 0; i < n; i++) {
int rightmostSetBit = Integer.lowestOneBit(nums[i]);
int val = (1 << Integer.bitCount(rightmostSetBit) - 1) - 1;
ans[i] = val | (val + 1);
if ((ans[i] | (ans[i] + 1)) != nums[i]) {
ans[i] = -1;
}
}
return ans;
}
}Code Editor (Testing phase)
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