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:
"resume.docx" is located in "Documents"
Explanation:
[<drive-letter:: represents drive name>]:/Main_Directory/Sub_Directory/ETC..
Answer:
int main() {
int _2dArray[32][32];
for (int i = 0; i < 32; i++) {
for (int j = 0; j < 32; j++) {
_2dArray[i][j] = j + i * 32;
}
}
return 0;
}
Explanation:
Here is a generic C/C++ 2d array traversal and main function example. The rest you'll have to figure out based on what kind of app you're making.
Good luck!
