Write a function that returns a chessboard pattern ("B" for black squares, "W" for white squares). The function takes a number N
as input and generates the corresponding board.
Input The first line of the input consists of an integer- num, representing the size of the chessboard.
Output Print N lines consist of N space-separated characters representing the chessboard pattern.
Input The first line of the input consists of an integer- num, representing the size of the chessboard.
Output Print N lines consist of N space-separated characters representing the chessboard pattern.
1 answer:
Answer:
Here is the C++ program.
#include <iostream> //to use input output functions
using namespace std; // to access objects like cin cout
void chessboard(int N){ // function that takes a number N as parameter generates corresponding board
int i, j;
string black = "B"; // B for black squares
string white = "W"; // W for white squares
for(i=1;i<=N;i++){ //iterates through each column of the board
for(j=1;j<=N;j++){ //iterates through each row of the board
if((i+j)%2==0) // if sum of i and j is completely divisible by 2
cout<<black<<" "; //displays B when above if condition is true
else //if (i+j)%2 is not equal to 0
cout<<white<<" "; } // displays W when above if condition is false
cout<<endl; } } //prints the new line
int main(){ //start of the main function
int num; //declares an integer num
cout << "Enter an integer representing the size of the chessboard: "; //prompts user to enter size of chess board
cin >> num; //reads value of num from user
chessboard(num); } //calls chessboard function to display N lines consist of N space-separated characters representing the chessboard pattern
Explanation:
The function works as follows:
Lets say that N = 2
two string variables black and white are declared. The value of black is set to "B" and value of white is set to "W" to return a chessboard pattern in B and W squares.
The outer loop for(i=1;i<=N;i++) iterates through each column of the chess board. The inner loop for(j=1;j<=N;j++) iterates through each row of chess board.
At first iteration of outer loop:
N = 2
outer loop:
i=1
i<=N is true because i=1 and 1<=2
So the program enters the body of the outer for loop. It has an inner for loop:
for(j=1;j<=N;j++)
At first iteration of inner loop:
j = 1
j<=N is true because j=1 and 1<=2
So the program enters the body of the inner for loop. It has an if statement:
if((i+j)%2==0) this statement works as:
if(1+1) % 2 == 0
(1+1 )% 2 takes the mod of 1+1 with 2 which gives the remainder of the division.
2%2 As 2 is completely divisible by 2 so 2%2 == 0
Hence the if condition evaluates to true. So the statement in if part executes:
cout<<black<<" ";
This prints B on the output screen with a space.
B
The value of j is incremented to 1.
j = 2
At second iteration of inner loop:
j = 2
j<=N is true because j=2 and 2=2
So the program enters the body of the inner for loop. It has an if statement:
if((i+j)%2==0) this statement works as:
if(1+2) % 2 == 0
(1+2 )% 2 takes the mod of 1+2 with 2 which gives the remainder of the division.
3%2 As 3 is not completely divisible by 2
Hence the if condition evaluates to false. So the statement in else part executes:
cout<<white<<" ";
This prints W on the output screen with a space.
B W
The value of j is incremented to 1.
j = 3
Now
j<=N is false because j=3 and 3>2
So the loop breaks and program moves to the outer loop to iterate through the next row.
At second iteration of outer loop:
N = 2
outer loop:
i=2
i<=N is true because i=2 and 2 = 2
So the program enters the body of the outer for loop. It has an inner for loop:
for(j=1;j<=N;j++)
At first iteration of inner loop:
j = 1
j<=N is true because j=1 and 1<=2
So the program enters the body of the inner for loop. It has an if statement:
if((i+j)%2==0) this statement works as:
if(2+1) % 2 == 0
(2+1 )% 2 takes the mod of 2+1 with 2 which gives the remainder of the division.
3%2 As 3 is not completely divisible by 2
Hence the if condition evaluates to false. So the statement in else part executes:
cout<<white<<" ";
This prints W on the output screen with a space.
B W
W
The value of j is incremented to 1.
j = 2
At second iteration of inner loop:
j = 2
j<=N is true because j=2 and 2=2
So the program enters the body of the inner for loop. It has an if statement:
if((i+j)%2==0) this statement works as:
if(2+2) % 2 == 0
(2+2 )% 2 takes the mod of 2+2 with 2 which gives the remainder of the division.
4%2 As 4 is completely divisible by 2 so 4%2 == 0
Hence the if condition evaluates to false. So the statement in if part executes:
cout<<black<<" ";
This prints B on the output screen with a space.
B W
W B
The value of j is incremented to 1.
j = 3
Now
j<=N is false because j=3 and 3>2
So the loop breaks and program moves to the outer loop. The value of outer loop variable i is incremented to 1 so i = 3
N = 2
outer loop:
i=3
i<=N is false because i=3 and 3>2
So this outer loop ends.
Now the output of this program is:
B W
W B
Screenshot of this program along with its output is attached.
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Answer:
Explanation:
Problem statement:
to simulate a binary search algorithm on an array of random values.
Binary Search: Search a sorted array by repeatedly dividing the search interval in half. Begin with an interval covering the whole array. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. Otherwise narrow it to the upper half. Repeatedly check until the value is found or the interval is empty.
Input/output description
Input:
Size of array: 4
Enter array:10 20 30 40
Enter element to be searched:40
The Output will look like this:
Element is present at index 3
Algorithm and Flowchart:
We basically ignore half of the elements just after one comparison.
Compare x with the middle element.
If x matches with middle element, we return the mid index.
Else If x is greater than the mid element, then x can only lie in right half subarray after the mid element. So we recur for right half.
Else (x is smaller) recur for the left half.
The Flowchart can be seen in the first attached image below:
Program listing:
// C++ program to implement recursive Binary Search
#include <bits/stdc++.h>
using namespace std;
// A recursive binary search function. It returns
// location of x in given array arr[l..r] is present,
// otherwise -1
int binarySearch(int arr[], int l, int r, int x)
{
if (r >= l) {
int mid = l + (r - l) / 2;
// If the element is present at the middle
// itself
if (arr[mid] == x)
return mid;
// If element is smaller than mid, then
// it can only be present in left subarray
if (arr[mid] > x)
return binarySearch(arr, l, mid - 1, x);
// Else the element can only be present
// in right subarray
return binarySearch(arr, mid + 1, r, x);
}
// We reach here when element is not
// present in array
return -1;
}
int main(void)
{ int n,x;
cout<<"Size of array:\n";
cin >> n;
int arr[n];
cout<<"Enter array:\n";
for (int i = 0; i < n; ++i)
{ cin >> arr[i]; }
cout<<"Enter element to be searched:\n";
cin>>x;
int result = binarySearch(arr, 0, n - 1, x);
(result == -1) ? cout << "Element is not present in array"
: cout << "Element is present at index " << result;
return 0;
}
The Sample test run of the program can be seen in the second attached image below.
Time(sec) :
0
Memory(MB) :
3.3752604177856
The Output:
Size of array:4
Enter array:10 20 30 40
Enter element to be searched:40
Element is present at index 3
Conclusions:
Time Complexity:
The time complexity of Binary Search can be written as
T(n) = T(n/2) + c
The above recurrence can be solved either using Recurrence T ree method or Master method. It falls in case II of Master Method and solution of the recurrence is Theta(Logn).
Auxiliary Space: O(1) in case of iterative implementation. In case of recursive implementation, O(Logn) recursion call stack space.
