Explain: The execute part of fetch decode execute cycle
2 answers:
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Answer:
Here is the JAVA code:
public class Main{
public static int solution(int[] A) { //method that takes non-empty array A consisting of 0s and 1s
int N = A.length; // number of 0s and 1s in array A
int r = 0; //result of adjacency
for (int i = 0; i < N - 1; i++; ) { // iterates through A
if (A[i] == A[i + 1]) //if i-th element of A is equal to (i+1)th element
r = r + 1; } //add 1 to the count of r
if (r == N-1) //for test cases like {1,1}
{return r-1; }
int max = 0; //to store maximum possible adjacency
for (int i = 0; i <N; i++) { //iterates through array
int c = 0;
if (i > 0) { //starts from 1 and covering the last
if (A[i-1] != A[i]) //checks if i-1 element of A is not equal to ith element of A
c = c + 1; //adds 1 to counter variable
else
c = c - 1; } //decrements c by 1
if (i < N - 1) {//starting with 0
if (A[i] != A[i + 1]) //checks if ith element of A is not equal to i+1th element of A
c = c + 1; //adds 1 to counter variable
else
c = c - 1; } //decrements c by 1
max = Math.max(max,c); } //finds the maximum of max and c
return r + max; } //returns result + maximum result
public static void main(String[] args) {
int[] A = {1, 1, 0, 1, 0, 0}; //sample array to test the method
System.out.println(solution(A));} } //calls the method passing array to it
Explanation:
The program works as follows:
A[] = {1, 1, 0, 1, 0, 0}
N = A.length
N = 6
The A has the following elements:
A[0] = 1
A[1] = 1
A[2] = 0
A[3] = 1
A[4] = 0
A[5] = 0
Program iterates through array A using for loop. Loop variable i is initialized to 0
if condition if (A[i] == A[i + 1]) checks
if (A[0] == A[0 + 1])
A[0] = 1
A[0 + 1] = A[1] = 1
They both are 1 so they are equal
r = r + 1;
Since the above if condition is true so 1 is added to the value of r Hence
r = 1
At each iteration of the loop the if condition checks whether the adjacent elements of A are equal. If true then r is incremented to 1 otherwise not.
So after all the iterations value of r = 2
if (r == N-1) evaluates to false because r=2 and N-1 = 5
So program moves to the statement:
for (int i = 0; i <N; i++)
This loop iterates through the array A
if (i > 0) condition checks if value of i is greater than 0. This evaluates to false and program control moves to statement:
if (i < N - 1) which makes if(0<5) This evaluates to true and program control moves to statement
if (A[i] != A[i + 1]) which means:
if (A[0] != A[0 + 1]) -> if (A[0] != A[1])
We know that
A[0] = 1
A[1] = 1
So this evaluates to false and else part is executed:
value of c is decremented to 1. So c=-1
max = Math.max(max,c) statement returns the max of max and c
max = 0
c = -1
So max = 0
value of i is incremented to 1 so i = 1
At next step:
if (i < N - 1) which makes if(1<5) This evaluates to true and program control moves to statement
if (A[i] != A[i + 1]) which means:
if (A[1] != A[1 + 1]) -> if (A[1] != A[2])
A[1] = 1
A[2] = 0
So the statement evaluates to true and following statement is executed
c = c + 1; The value of c is incremented to 1. So
c = -1 + 1
Hence
Hence c= 0, max = 0 and i = 2
next step:
if (i < N - 1) which makes if(2<5) This evaluates to true and program control moves to statement
if (A[i] != A[i + 1]) which means:
if (A[2] != A[2 + 1]) -> if (A[2] != A[3])
A[2] = 0
A[3] = 1
So the statement evaluates to true and following statement is executed
c = c + 1; The value of c is incremented to 1. So
c = 0 + 1
c = 1
Hence
The statement max = Math.max(max,c) returns the max of max and c
max = 0
c = 1
So max = 1
Hence c= 1, max = 1 and i = 3
next step:
if (i < N - 1) which makes if(3<5) This evaluates to true and program control moves to statement
if (A[i] != A[i + 1]) which means:
if (A[3] != A[3 + 1]) -> if (A[3] != A[4])
A[3] = 1
A[4] = 0
So the statement evaluates to true and following statement is executed
c = c + 1; The value of c is incremented to 1. So
c = 1 + 1
c = 2
Hence
The statement max = Math.max(max,c) returns the max of max and c
max = 1
c = 2
So max = 2
Hence c= 2, max = 2 i = 4
next step:
if (i < N - 1) which makes if(4<5) This evaluates to true and program control moves to statement
if (A[i] != A[i + 1]) which means:
if (A[4] != A[4+ 1]) -> if (A[4] != A[5])
A[4] = 0
A[5] = 0
So this evaluates to false and else part is executed:
value of c is decremented to 1. So c=1
max = Math.max(max,c) statement returns the max of max and c
max = 2
c = 1
So max = 2
value of i is incremented to 1 so i = 5
next step:
if (i < N - 1) which makes if(5<5) This evaluates to false
if (i > 0) evaluates to true so following statement executes:
if (A[i-1] != A[i])
if (A[5-1] != A[5])
if (A[4] != A[5])
A[4] = 0
A[5] = 0
This statement evaluates to false so else part executes and value of c is decremented to 1
Hence
max = 2
c = 0
So max = 2
value of i is incremented to 1 so i = 6
The loop breaks because i <N evaluates to false.
Program control moves to the statement:
return r + max;
r = 2
max = 2
r + max = 2+2 = 4
So the output of the above program is:
4
