Answer:
<em>public static void printLarger(double sales1, double sales2){</em>
<em> if (sales1>sales2){</em>
<em> System.out.println(sales1+" is larger");</em>
<em> }</em>
<em> else {</em>
<em> System.out.println(sales2+" is larger");</em>
<em> }</em>
<em> }</em>
<em>A complete code calling the printLarger method is given below</em>
Explanation:
<em>public class Larger{</em>
<em> public static void main(String[] args) {</em>
<em> int sales1 = 100;</em>
<em> int sales2 = 120;</em>
<em> printLarger(sales1,sales2);</em>
<em> }</em>
<em>public static void printLarger(int sales1, int sales2){</em>
<em> if (sales1>sales2){</em>
<em> System.out.println(sales1+" is larger");</em>
<em> }</em>
<em> else {</em>
<em> System.out.println(sales2+" is larger");</em>
<em> }</em>
<em> }</em>
}
Answer:
Question is incomplete.
Assuming the below info to complete the question
You have a collection of n lockboxes and m gold keys. Each key unlocks at most one box. Without a matching key, the only way to open a box is to smash it with a hammer. Your baby brother has locked all your keys inside the boxes! Luckily, you know which keys (if any) are inside each box.
Detailed answer is written in explanation field.
Explanation:
We have to find the reachability using the directed graph G = (V, E)
In this V are boxes are considered to be non empty and it may contain key.
Edges E will have keys .
G will have directed edge b1b2 if in-case box b1 will have key to box b2 and box b1 contains one key in it.
Suppose if a key opens empty box or doesn’t contain useful key means can’t open anything , then it doesn’t belongs to any edge.
Now, If baby brother has chosen box B, then we have to estimate for other boxes reachability from B in Graph G.
If and only if all other boxes have directed path from box B then just by smashing box B we can get the key to box b1 till last box and we can unlock those.
After first search from B we can start marking all other vertex of graph G.
So algorithm will be O ( V +E ) = O (n+m) time.
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