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lesya [120]
2 years ago
14

Write a program to find a peak in an array of ints. Suppose the array is {-1, 0, 2, 5, 6, 8, 7}. The output should be "A peak is

at array index 5 and the value is 8." This is because the value 8 is larger than its predecessor 6 and its successor 7 in the given array. Note that 8 occurs at index 5. (The array starts at index 0.) A number at index i in an array X is considered a peak if: X[i]>=X[i-1] and X[i]>=X[i+1]. If i is at the beginning of the array, then peak is if X[i]>=X[i+1]. If i is at end of array, then peak is if X[i]>=X[i-1].

Computers and Technology
1 answer:
Sergeeva-Olga [200]2 years ago
5 0

Answer:

Following are the code to this question:

#include<iostream>//declaring header file  

using namespace std;

int main()//main method

{

int n= 6,j=0;//declaring integer variable

int X[n];//defining an array

for(j=0;j<=n;j++)//defining a loop for input value

cin>>X[j];//input value from the user

if(j==0) //defining if block that checks value at beginning

{

if(X[j]>=X[j+1])//defining if block to check to compare first and second value  

{

cout<<"A peak is at array index "<<j<<" and the value is "<<X[j];//use print method to largest value with index number  

}

}

else//defining else block

{

for(j=0;j<=n;j++)//defining for loop for compare other value

{

if(j==n-1) //use if block that checks next index  

{

if(X[j]>=X[j-1])//use if block to compare value  

cout<<"A peak is at array index "<<j<<" and the value is "<<X[j];//use print method to largest value with index number

}

else

{

if(X[j]>=X[j-1] && X[j]>=X[j+1])//comapre value

cout<<"A peak is at array index "<<j<<" and the value is "<<X[j];//use print method to largest value with index number

}

}

}

return 0;

}

Output:

please find the attached file.

Explanation:

In the given code, inside the main method two integer variable "n and j", is declared, in the next step, an array "x"is defined which input the value from the user end.

  • In the next step, multiple if block is used, in the first if block it comapre the first and second value if it grater then it will print the value with its index number.
  • In the next if block, it comapre is next value and if it grater then it will print the value with its index number.  

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Answer:

Out of the two methods The 2nd method is a better option for implementing several stacks and arrays using the single array system this is because Method 2 makes efficient use of the available space

Explanation:

A) Method 1 ( Divide the given single array in the size of n/k.)

  • How to use method 1 to implement several stacks involves :

i)  To implement several stacks through an array (x) is by dividing the array in n/k parts.

ii) The k represents the slots in which the different stacks will be placed and the n represents the size of the array x.

iii) If you need to implement at least two stacks place the first stack in the slot of a [0] to a [n/k - 1], and another stack in the slot of a[n/k] to a[2n/k-1].

note: The  disadvantage of this method is that the use of the space of the array is not much efficient. therefore method 2 is better

  • How to use method 1 to implement queues involves :

i) queues can also be implemented through an array. applying the same method above

ii) Divide the array in slots and place the queues in that slots.

This method has a problem with  the efficient utilization of the space.therefore method 2 is preferred

B) Method 2 ( uses the space efficiently ) uses two more arrays to implement stacks which are : Top_array and Next_array

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ii)The initial actual array is x[] and this will store the stacks.

iii)Simultaneously with several stacks, the stack which contain the free slots in the array x[] will also be maintained.

iv)  The entries of the Top_array[] will be initialized to -1. This implies that all the stacks are empty.

v)  Firstly the entries of the array Next_array[i] will be initialized to i+1, since all the slots initially are free and are pointed to the next slot.

vi) Initialize The top of the free stack that is maintaining the free slots  as 0.

vii)  The complexity of push () (method to insert an element) and pop () (method to delete an element) operations by using this method is O (1).

  • How to use method 2 to implement several queues

The same method applicable to implementing several stacks is used here but with a difference is the presence of three extra arrays which are :

Front_array[] = indicates the number of queues. This array stores the indexes of the front elements of the stacks.

Rear_array[] = determines the sizeof the array k . This array stores the indexes of the last elements of the stacks.

Next_array[] = The array n indicates the size of the single array say x. This array stores the indexes of the next items that is being pushed.

i) The initial actual array is a[] which will store the queues. The free slots will also be maintained.

ii)The entries of the Front_array[] will be initialized to -1. This means  that all the queues are empty initially

ii) Initially the entries of the array Next_array[i] will be initialized to i+1, since all the slots initially are free and are pointed to the next slot.

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Answer:

See explaination

Explanation:

StackExample.java

public class StackExample<T> {

private final static int DEFAULT_CAPACITY = 100;

private int top;

private T[] stack = (T[])(new Object[DEFAULT_CAPACITY]);

/**

* Returns a reference to the element at the top of this stack.

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* atreturn element on top of stack

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*/

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if (isEmpty())

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*/

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//please replace "at" with the at symbol

Note:

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