Part $3(35$ points$)!!$PLEASE SOLVE IT IN C LANGUAGE!!

Implement the following function in skeleton code lab$1$part$3.$c:

$//$ Circularly shift elements of arr from right to left until sequence of values in

$//$ arr becomes the smallest considering all sequence of values obtained by circularly

$//$ shifting elements in arr.

void circularShiftUntilMinArr$($int arr$[],$ int n$)$;

$$ You need to use circularShiftFromRightToLeft function and compareTwoArrays function

while implementing circularShiftUntilMinArr function.

$$ Print out sum of the elements of two arrays one by one.

Assume that arr can have at most $500$ elements in it$.$ Hint: If you need to declare a local array in function

circularShiftUntilMinArr, you can set its size as $500.$

Your task in this part to fill in the missing function definitions in skeleton code lab$1$part$3.$c$.$

main function will stay as it is$.$ Sample Run:

Enter the number of elements:

$6$

Enter $6$ elements:

$485169$

Array elements: $485169$

After circularShiftUntilMinArr:

Array elements: $169485$

Sum of original and shifted elements:

Element $0$: $5$

Element $1$: $14$

Element $2$: $14$

Element $3$: $5$

Element $4$: $14$

Element $5$: $14$ Here is the skeleton code lab$1$part$3.$c code structure: #include

#define SIZE $500$

$//$ reads numbers from the standard input into arr,

$//$ and stores the number of read elements in the memory cell pointed to by nPtr

void readInput$($int arr$[],$ int $*$nPtr$)$;

$//$ prints the elements in arr$[0..($n$-1)]$

void printNumbers$($const int arr$[],$ int n$)$;

$//$ Circularly shift elements of arr from right to left where last element of arr is

$//$ shifted the first position of arr.

$//$ Size of arr is n$.$

void circularShiftFromRightToLeft$($int arr$[],$ int n$)$;

$//$ Let n be the minimum of n$1$ and n$2$ where n$1$ and n$2$ are the number of elements in

$//$ arr$1$ and arr$2,$ respectively.

$//$ Compare corresponding elements of arr$1$ and arr$2$ at each of the first n positions.

$//$

$//$ If arr$1$ and arr$2$ has the same value in each position:

$//$ Return $0$ if n$1==$ n$2$

$//$ Return $1$ if n$1>$ n$2$

$//$ Return $-1$ if n$1<$ n$2$

$//$

$//$ If arr$1$ has a larger value than arr$2$ considering the first position from the

$//$ beginning at which arr$1$ and arr$2$ have a different value:

$//$ Return $1$

$//$

$//$ If arr$1$ has a smaller value than arr$2$ considering the first position from the

$//$ beginning at which arr$1$ and arr$2$ have a different value:

$//$ Return $-1$

int compareTwoArrays$($const int arr$1[],$ const int arr$2[],$ int n$1,$ int n$2)$;

$//$ Circularly shift elements of arr from right to left until sequence of values in

$//$ arr becomes the smallest considering all sequence of values obtained by circularly

$//$ shifting elements in arr.

void circularShiftUntilMinArr$($int arr$[],$ int n$)$;

int main$()$

$\{$

int arr$[$SIZE$]$;

int n;

readInput$($arr$,$ &n$)$;

printNumbers$($arr$,$ n$)$;

circularShiftUntilMinArr$($arr$,$ n$)$;

printf$("$

After circularShiftUntilMinArr:

$")$;

printNumbers$($arr$,$ n$)$;

$//$ Calculate the sum of the elements after shifting for two arrays

$//----$FILL HERE$----//$

return $0$;

$\}$

void readInput$($int arr$[],$ int $*$nPtr$)$

$\{$

$//$ fill here

$\}$

void printNumbers$($const int arr$[],$ int n$)$

$\{$

$//$ fill here

$\}$

void circularShiftFromRightToLeft$($int arr$[],$ int n$)$

$\{$

$//$ fill here

$\}$

int compareTwoArrays$($const int arr$1[],$ const int arr$2[],$ int n$1,$ int n$2)$

$\{$

$//$ fill here

$\}$

void circularShiftUntilMinArr$($int arr$[],$ int n$)$

$\{$

$//$ fill here

$\}$