1)Write a C program to find positive and negative elements in array
2)Write a C program to rotate elements in an array left and right
3)Write a C program to insert,delete,display a element in array
4)Write a C program to get elements for 2D array
5)Write a C program to merge two arrays of same size
6)Write a C program to extract the diagonal elements from the matrix and sum it.
Answers
Answer:
Input arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2, n =7
1) Store the first d elements in a temp array
temp[] = [1, 2]
2) Shift rest of the arr[]
arr[] = [3, 4, 5, 6, 7, 6, 7]
3) Store back the d elements
arr[] = [3, 4, 5, 6, 7, 1, 2]
Time complexity : O(n)
Auxiliary Space : O(d)
METHOD 2 (Rotate one by one)
leftRotate(arr[], d, n)
start
For i = 0 to i < d
Left rotate all elements of arr[] by one
end
To rotate by one, store arr[0] in a temporary variable temp, move arr[1] to arr[0], arr[2] to arr[1] …and finally temp to arr[n-1]
Let us take the same example arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2
Rotate arr[] by one 2 times
We get [2, 3, 4, 5, 6, 7, 1] after first rotation and [ 3, 4, 5, 6, 7, 1, 2] after second rotation.
Below is the implementation of the above approach :
// C++ program to rotate an array by
// d elements
#include <bits/stdc++.h>
using namespace std;
/*Function to left Rotate arr[] of
size n by 1*/
void leftRotatebyOne(int arr[], int n)
{
int temp = arr[0], i;
for (i = 0; i < n - 1; i++)
arr[i] = arr[i + 1];
arr[n-1] = temp;
}
/*Function to left rotate arr[] of size n by d*/
void leftRotate(int arr[], int d, int n)
{
for (int i = 0; i < d; i++)
leftRotatebyOne(arr, n);
}
/* utility function to print an array */
void printArray(int arr[], int n)
{
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}
/* Driver program to test above functions */
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function calling
leftRotate(arr, 2, n);
printArray(arr, n);
return 0;
}
Explanation:
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