Can someone please help me out with this problem:

Sort the odd elements in a list (ODDSORT) Input: A[1...n] a list of integers.

Output: List A[1...n] such that A[i] = A[i] if A[i] is even, and A[i] A[j] for all 1

i j n such that A[i] and A[j] are both odd.

For this item consider the problem of sorting the odd elements in a list of integers while keeping the even ones unchanged in their original places. Sort the odd elements in a list (ODDSORT) Input: A[l...n] a list of integers. Output: List A'[l...n] such that A'[i] = A[i] if A'[i] is even, and A'[i] = A'[j] for all 1 sis jsn such that A'[i] and A'[j] are both odd. Let OS(A[a ... b]) represent the output of the ODDSORT problem on input A[a...b]. Let a || | denote the concatenation of lists a and B, and let minx(A[a...b]) denote the index (in range [a... b]) of the minimum odd element from list Aa...b]. (1 point) 1. Below is a self-reduction for the ODDSORT problem. State a recursive algorithm using pseudocode for solving the ODDSORT problem based on this self-reduction. OS(A[a... b]) = A[a] || OS(A[a+1... b]) A[k] || OS(A[a+l...k - 1] || A[a] || A[k +1... b]) if a 2 b if a Y, otherwise returns X ||Y. Below is another self-reduction for the ODDSORT problem. State a recursive algorithm using pseudocode for solving the ODDSORT problem based on this self-reduction. if a b if b= a +1 if b> a+1 SOS(A[a... b]) = { with A' A" A" - - - pairsort(A[a], A[b]) A" SOS(A[a... t2l) || A[t2 +1...b] A' [a...t1] || SOS(A'[t +1...b]) SOS(A [a ... tz]) || A" [t2 + 1...b] For this item consider the problem of sorting the odd elements in a list of integers while keeping the even ones unchanged in their original places. Sort the odd elements in a list (ODDSORT) Input: A[l...n] a list of integers. Output: List A'[l...n] such that A'[i] = A[i] if A'[i] is even, and A'[i] = A'[j] for all 1 sis jsn such that A'[i] and A'[j] are both odd. Let OS(A[a ... b]) represent the output of the ODDSORT problem on input A[a...b]. Let a || | denote the concatenation of lists a and B, and let minx(A[a...b]) denote the index (in range [a... b]) of the minimum odd element from list Aa...b]. (1 point) 1. Below is a self-reduction for the ODDSORT problem. State a recursive algorithm using pseudocode for solving the ODDSORT problem based on this self-reduction. OS(A[a... b]) = A[a] || OS(A[a+1... b]) A[k] || OS(A[a+l...k - 1] || A[a] || A[k +1... b]) if a 2 b if a Y, otherwise returns X ||Y. Below is another self-reduction for the ODDSORT problem. State a recursive algorithm using pseudocode for solving the ODDSORT problem based on this self-reduction. if a b if b= a +1 if b> a+1 SOS(A[a... b]) = { with A' A" A" - - - pairsort(A[a], A[b]) A" SOS(A[a... t2l) || A[t2 +1...b] A' [a...t1] || SOS(A'[t +1...b]) SOS(A [a ... tz]) || A" [t2 + 1...b]