B) its a small peice of data that tracks your preferances on diffrent websites
Have a great day !
Rest your fingers gently on the home row or home keys.
Rest your palms on the keyboard.
Relax your fingers.
All the above are proper keyboard techniques apart from slouch in your chair. It is always recommended to sit up straight with your feet positioned on the floor for balance. Do not cross. Center your body to the H key and have your elbows at sides and bent about 90 degrees. Use correct fingering and deploy touch typing. These and many others will help develop optimal speed and accuracy and help prevent the development of stress injury.
Money that can be promptly and easily appraised falls under the M1 Money classification.
<h3>What are broad and narrow money, respectively?</h3>
Broad money typically refers to M2, M3, and/or M4. The most liquid kinds of money, such as currency (banknotes and coins), as well as bank account balances that may be instantly changed into currency or used for cashless transactions, are generally referred to as "narrow money" (overnight deposits, checking accounts).
<h3>Describe Narrow Money.</h3>
All of the actual money that the central bank has falls under the category of "narrow money," which is a subset of the money supply. Demand deposits, money, and other liquid assets are included. In the US, "narrow money" is referred to as M1 (M0 plus demand accounts).
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Answer:
You can simplify the problem down by recognizing that you just need to keep track of the integers you've seen in array that your given. You also need to account for edge cases for when the array is empty or the value you get would be greater than your max allowed value. Finally, you need to ensure O(n) complexity, you can't keep looping for every value you come across. This is where the boolean array comes in handy. See below -
public static int solution(int[] A)
{
int min = 1;
int max = 100000;
boolean[] vals = new boolean[max+1];
if(A.length == 0)
return min;
//mark the vals array with the integers we have seen in the A[]
for(int i = 0; i < A.length; i++)
{
if(A[i] < max + 1)
vals[A[i]] = true;
}
//start at our min val and loop until we come across a value we have not seen in A[]
for (int i = 1; i < max; i++)
{
if(vals[i] && min == i)
min++;
else if(!vals[i])
break;
}
if(min > max)
return max;
return min;
}
Answer:
You are gonna waste money and it might not be the best idea
Explanation:
