Answer:
javac Welcome.java
Explanation:
In order to the file to compile with the default compiler in the JDK, the instruction must be of the form:
javac filename
(Note: the filename with the .java extension)
Answer: provided in the explanation section
Explanation:
Given that:
Assume D(k) =║ true it is [1 : : : k] is valid sequence words or false otherwise
now the sub problem s[1 : : : k] is a valid sequence of words IFF s[1 : : : 1] is a valid sequence of words and s[ 1 + 1 : : : k] is valid word.
So, from here we have that D(k) is given by the following recorance relation:
D(k) = ║ false maximum (d[l]∧DICT(s[1 + 1 : : : k]) otherwise
Algorithm:
Valid sentence (s,k)
D [1 : : : k] ∦ array of boolean variable.
for a ← 1 to m
do ;
d(0) ← false
for b ← 0 to a - j
for b ← 0 to a - j
do;
if D[b] ∧ DICT s([b + 1 : : : a])
d (a) ← True
(b). Algorithm Output
if D[k] = = True
stack = temp stack ∦stack is used to print the strings in order
c = k
while C > 0
stack push (s [w(c)] : : : C] // w(p) is the position in s[1 : : : k] of the valid world at // position c
P = W (p) - 1
output stack
= 0 =
cheers i hope this helps !!!
Answer:
Explanation:
Since all of the items in the array would be integers sorting them would not be a problem regardless of the difference in integers. O(n) time would be impossible unless the array is already sorted, otherwise, the best runtime we can hope for would be such a method like the one below with a runtime of O(n^2)
static void sortingMethod(int arr[], int n)
{
int x, y, temp;
boolean swapped;
for (x = 0; x < n - 1; x++)
{
swapped = false;
for (y = 0; y < n - x - 1; y++)
{
if (arr[y] > arr[y + 1])
{
temp = arr[y];
arr[y] = arr[y + 1];
arr[y + 1] = temp;
swapped = true;
}
}
if (swapped == false)
break;
}
}
Probably c or a with the question your asking
