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.
Answer:
7
Explanation:
Because the q.length is a inbuilt function in the programming which used to get the length of the array. In the array, there are 7 values are store. Therefore, the size 7 store in the variable z.
For example:
int[] array={1,2};
int x = array.length;
the answer of above code is 2, because the elements present in the array is 2.
Answer:
The program to this question can be described as follows:
Program:
#include <iostream> //defining header file
using namespace std;
int main() //defining main method
{
int x1,rem,i=1; //defining integer variable
long Num=0;// defining long variable
cout << "Input a decimal number: "; // print message
cin >> x1; //input value by user
while (x1!=0) //loop for calculate binary number
{
//calculating binary value
rem= x1%2;
x1=x1/2;
Num =Num +rem*i;//holding calculate value
i=i*10;
}
cout <<Num;//print value
return 0;
}
Output:
Input a decimal number: 76
1001100
Explanation:
In the above code, four variable "x1, rem, i, and Num" is declared, in which three "x1, rem, and i" is an integer variable, and one "Num" is a long variable.
- In the x1 variable, we take input by the user and use the while loop to calculate its binary number.
- In the loop, first, we check x1 variable value is not equal to 0, inside we calculate it binary number that store in long "Num" variable, after calculating its binary number the print method "cout" is used to prints its value.
The bachelor's program at Eth Zurich is 3 years.
I hope this helps!
:-)
