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AnnyKZ [126]
3 years ago
9

Write a method that returns a version of the given array where all the 10's have been removed. The remaining elements should shi

ft left towards the start of the array as needed, and the empty spaces at the end of the array should be set to 0.
Ex: {1, 10, 10, 2} yields {1, 2, 0, 0}. You may modify and return the given array or make a new array.

Code:

public class RemoveTen {
public static void main(String[] args) {
int[] nums = {1, 10, 10, 2};
int[] result = removeTen(nums);
for(int i = 0; i < result.length; i++)
System.out.print(result[i] + " ");
}

public static int[] removeTen(int[] nums) {
/*FIXME Complete the implementation of the removeTen method*/
}
}
Computers and Technology
1 answer:
Harlamova29_29 [7]3 years ago
8 0

Answer:

The removeTens method is as follows

public static int[] removeTen(int[] nums) {

int [] arr = nums;

   int index = 0;

   while(index < arr.length && arr[index] != 10){

       index++;

   for(int j = index + 1; j < arr.length; j++) {

       if(arr[j] != 10) {

           arr[index] = arr[j];

           arr[j] = 10;

           index++;        }    }

   for( ; index < arr.length; index++){

       arr[index] = 0;}}

   return arr;

}

Explanation:

This defines the removeTens method; it receives array nums as its parameter

public static int[] removeTen(int[] nums) {

This creates and initializes a new array

int [] arr = nums;

This initializes index to 0

   int index = 0;

The following is repeated when array element is not 10 and if array index is still valid

   while(index < arr.length && arr[index] != 10){

This increments the index by 1

       index++;

This iterates through the array

   for(int j = index + 1; j < arr.length; j++) {

This checks if array element is not 10

       if(arr[j] != 10) {

If yes:

           arr[index] = arr[j];

The current array element is set to 10

           arr[j] = 10;

The index is incremented by 1        

  index++;        }    }

This iterates through the array again and move 0s to the back

   for( ; index < arr.length; index++){

       arr[index] = 0;}}

This returns the new array

  return arr;

}

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Likurg_2 [28]

Answer:

The program to this question as follows:

Program:

targetValue = 0.3333 #defining variable targetValue and assign value

sensorReading = 0.0 #defining variable sensorReading and assign value

sensorReading = 1.0/3.0 #calculate value in sensorReading variable

Val=sensorReading - targetValue

#calculate the difference and store in Val variable

if (Val < 0.0001): #use of if block to check condition

  print ("Equal")  #print value

else:  #else block

  print ("Not equal") #print value

Output:

Equal

Explanation:

In the above Python program code, there are two variables "targetValue and sensorReading" is defined, in which targetValue store a value, that is "0.3333", and sensorReading holds a value, that is "0.0".

  • In the next step, the "Val" variable is defined, that calculate the difference between both variable, that conditional statement is used.
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Answer:

Here the given problem is modeled as a Graph problem.

Explanation:

Input:-  n, k and the function d(pi,pj) which outputs an integer between 1 and n2

Algorithm:-We model each student as a node. So, there would be n nodes. We make a foothold between two nodes u and v (where u and v denote the scholars pu and pv respectively) iff d(pu,pv) > k. Now, Let's call the graph G(V, E) where V is that the vertex set of the graph ( total vertices = n which is that the number of students), and E is that the edge set of the graph ( where two nodes have edges between them if and only the drama between them is bigger than k).

We now need to partition the nodes of the graph into two sets S1 and S2 such each node belongs to precisely one set and there's no edge between the nodes within the same set (if there's a foothold between any two nodes within the same set then meaning that the drama between them exceeds k which isn't allowed). S1 and S2 correspond to the partition of scholars into two buses.

The above formulation is akin to finding out if the graph G(V,E) is a bipartite graph. If the Graph G(V, E) is bipartite then we have a partition of the students into sets such that the total drama <= k else such a partition doesn't exist.

Now, finding whether a graph is bipartite or not is often found using BFS (Breadth First algorithm) in O(V+E) time. Since V = n and E = O(n2) , the worst-case time complexity of the BFS algorithm is O(n2). The pseudo-code is given as

PseudoCode:

// Input = n,k and a function d(pi,pj)

// Edges of a graph are represented as an adjacency list

1. Make V as a vertex set of n nodes.

2. for each vertex  u ∈ V

\rightarrow  for each vertex v ∈ V

\rightarrow\rightarrowif( d(pu, pj) > k )

\rightarrow\rightarrow\rightarrow add vertex u to Adj[v]   // Adj[v] represents adjacency list of v

\rightarrow\rightarrow\rightarrow add vertex v to Adj[u] // Adj[u] represents adjacency list of u

3.  bool visited[n] // visited[i] = true if the vertex i has been visited during BFS else false

4. for each vertex u ∈ V

\rightarrowvisited[u] = false

5. color[n] // color[i] is binary number used for 2-coloring the graph  

6. for each vertex u ∈ V  

\rightarrow if ( visited[u] == false)

\rightarrow\rightarrow color[u] = 0;

\rightarrow\rightarrow isbipartite = BFS(G,u,color,visited)  // if the vertices reachable from u form a bipartite graph, it returns true

\rightarrow\rightarrow if (isbipartite == false)

\rightarrow\rightarrow\rightarrow print " No solution exists "

\rightarrow\rightarrow\rightarrow exit(0)

7.  for each vertex u ∈V

\rightarrow if (color[u] == 0 )

\rightarrow\rightarrow print " Student u is assigned Bus 1"

\rightarrowelse

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BFS(G,s,color,visited)  

1. color[s] = 0

2. visited[s] = true

3. Q = Ф // Q is a priority Queue

4. Q.push(s)

5. while Q != Ф {

\rightarrow u = Q.pop()

\rightarrow for each vertex v ∈ Adj[u]

\rightarrow\rightarrow if (visited[v] == false)

\rightarrow\rightarrow\rightarrow color[v] = (color[u] + 1) % 2

\rightarrow\rightarrow\rightarrow visited[v] = true

\rightarrow\rightarrow\rightarrow Q.push(v)

\rightarrow\rightarrow else

\rightarrow\rightarrow\rightarrow if (color[u] == color[v])

\rightarrow\rightarrow\rightarrow\rightarrow return false // vertex u and v had been assigned the same color so the graph is not bipartite

}

6. return true

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Answer/Explanation:

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