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Given an array arr and a number n, you have to find whether there exist a subset in arr whose sum is n. You have to print length of the subset.
1. There exists only one subset like that
2. All number in arr are positive

Suppose you are a stock trader and you can do as many trades but if you stop you can't do another trade. You can start with any trade. Given an array of profits/loss of trades and find the maximum profit you can make.

Input:
Number of trades
Profit/loss in each trade

Output:
Max Profit

Ex:

Input:

7 
1 2 3 4 -2 -3 1

Output:

10

Explaination: Trade of [1,2,3,4]

Input:

5
-2 -3 -4 1 2

Output:

3

P.S: Any solution than Brute-Force.

Given two integer arrays A and B.
B contains exactly same numbers as A except two additional numbers. Find the two elements with minimum time and space complexity.
for ex: A ={1, 4, 2, 6, 3}
B = {4, 0,7, 6, 3, 2, 1}
ans: 0 7
I came with this solution:

Arrays.sort(A);
Arrays.sort(B);
int i=0, j=0;
while(j<=i+2 || i<A.length){
if(A[i]==B[j]){
++i; 
++j}
else{
System.out.println(b[j]);
j++;
}
}
if(j==A.length+1){
System.out.println(B[j++]+" "B[j]);
}
if(j==B.length)
System.out.println)(B[j]);

How will you dictionary sort integers without converting them to strings.
For ex: 1 2 10 20 100 110
Ans: 1 10 100 110 2 20.

Given an input n, find the smallest number x such that x % n = 0 and sum of the digits in x is equal to n.
Note: x is so large even long long int can't keep it.

It is obvious that find the sum of digits using (x/10) gives timeout, so how can its complexity be reduced.

Given an input n, find the smallest number x such that x % n = 0 and sum of the digits in x is equal to n.
Note: x is so large even long long int can't keep it.

It is obvious that find the sum of digits using (x/10) gives timeout, so how can its complexity be reduced.