If the value of n is 1, each element in the array will shift one position to the left. This means that the second element will become the first, the third element will move to the second position, and so forth. Simultaneously, the original first element will be appended to the end of the array.
ALGORITHM:
- STEP 1: START
- STEP 2: INITIALIZE arr ={1, 2, 3, 4, 5 }.
- STEP 3: length = sizeof(arr1)/sizeof(arr1[0])
- STEP 4: length = sizeof(arr)/sizeof(arr[0])
- STEP 5: SET n =3
- STEP 6: SET i=0. REPEAT STEP 7 to STEP 8 UNTIL i<length
- STEP 7: PRINT arr[i]
- STEP 8: i = i+1
- STEP 9: SET i= 0. REPEAT STEP 10 to STEP 16 UNTIL i<n
- STEP 10: DEFINE j, first.
- STEP 11: first = arr[0]
- STEP 12: SET j=0. REPEAT STEP 13 to STEP 14 UNTIL j<length-1.
- STEP 13: arr[j]= arr[j+1]
- STEP 14: j=j+1.
- STEP 15: arr[j]= first
- STEP 16: i=i+1.
- STEP 17: PRINT new line.
- STEP 18: PRINT "Array after left rotation"
- STEP 19: ENDSET i=0. REPEAT STEP 20 to STEP 21 UNTIL i<length
- STEP 20: PRINT arr[i]
- STEP 21: i=i+1.
- STEP 22: RETURN 0.
- STEP 23: END
PROGRAM:
Example
#include <stdio.h>
int main()
{
//Initialize array
int arr[] = {1, 2, 3, 4, 5};
//Calculate length of array arr
int length = sizeof(arr)/sizeof(arr[0]);
//n determine the number of times an array should be rotated
int n = 3;
//Displays original array
printf("Original array: \n");
for (int i = 0; i < length; i++) {
printf("%d ", arr[i]);
}
//Rotate the given array by n times toward left
for(int i = 0; i < n; i++){
int j, first;
//Stores the first element of the array
first = arr[0];
for(j = 0; j < length-1; j++){
//Shift element of array by one
arr[j] = arr[j+1];
}
//First element of array will be added to the end
arr[j] = first;
}
printf("\n");
//Displays resulting array after rotation
printf("Array after left rotation: \n");
for(int i = 0; i < length; i++){
printf("%d ", arr[i]);
}
return 0;
}
Output:
Output
Original Array:
1 2 3 4 5
Array after left rotation:
4 5 1 2 3
