A kernel manages the whole computer including hardware. In Unix, all processes are launched from systemd/launchd.
Answer:
Replace the power supply
Explanation:
Usually, Capacitors burn up because of different factors.
- Exceeding operation voltage
- High Inverse voltage if is an electrolytic capacitor.
A tentative answer could be the battery, but usually the batteries damages are because wear out of them, that is, cycles of charging and discharging, a problem that could arise in the battery is related with the charge protection, this circuit cares that the battery only get the charge that it needs, in Li-po batteries is 3.7V, in some laptops is 24 V, if so the battery could explode or leaking acid.
The mother-board is the "brain" and the Random Access Memory (RAM), they consume a lot of energy and usually heat up, but doesn’t produce increasing of voltage and its feed it by voltage regulators.
The only valid option is the power supply because the energy comes from a rectifier (made with diodes) and a voltage regulator that step-down the DC voltage output if by any chance the voltage increase or a diode burns up in the power supply the coupling capacitors or input capacitors in the computer will fail.
Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.
public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}
So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.
A quick explanation:
We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.
The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.
Hahahahaha I wanna was the day I wanna was the last time I got to
Graphic images can be stored in a variety of formats including JPEG and GIF. Hope this helped!
