Jim wants to buy a car, but he’ll probably only need it for a couple of years. He has a short commute to work, so he won’t be pu
1 answer:
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Explanation:
Let, DG is the datagram so, DG= 2400.
Let, FV is the Value of Fragment and F is the Flag and FO is the Fragmentation Offset.
Let, M is the MTU so, M=700.
Let, IP is the IP header so, IP= 20.
Let, id is the identification number so, id=422
Required numbers of the fragment =
Insert values in the formula =
Then, = =
The generated numbers of the fragment is 4
- If FV = 1 then, bytes in data field of DG=
and id=422 and FO=0 and F=1.
- If FV = 2 then, bytes in data field of DG=
and F=1.
- If FV = 3 then, bytes in data field of DG=
and F=1.
- If FV = 4 then, bytes in data field of DG=
and id=422 and FO=255
and F=0.
Answer:
Check the explanation
Explanation:
the js code is as follows
function count_max_group(ar) {
if (ar.length < 2) return ar.length; // if no of elements are 2 then max groups are 2
let st = [0]; // stack to store the groups
for (let i = 1; i < ar.length; i++) {
if (ar[i] >= ar[i - 1]) st.push(i); // pushing elements if if ar[i] is greater than the previous element
for (let j = i; j > 0 && ar[j] < ar[j - 1]; j--) { // finding the max number of grps
swap(ar, j, j - 1); // swapping
if (j <= st[st.length - 1]) st.pop();
}
}
return st.length;
}
function swap(a, i, j) { // function to swap two variables
let t = a[i];
a[i] = a[j];
a[j] = t;
}
//sample solutions
console.log("the sample solutions are\n");
console.log("[2,1,6,4,3,7]\n");
console.log(count_max_group([2,1,6,4,3,7]));
console.log("[4,3,2,6,1]\n") ;
console.log(count_max_group([4,3,2,6,1]));
console.log("[2,4,1,6,5,9,7]\n");
console.log(count_max_group([2,4,1,6,5,9,7]));
