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February 10, 2024

NPTEL Programming, Data Structures And Algorithms Using Python (Week 3 Programming Assignment) 2024

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Programming Assignment Solutions

Programming Assignment Week 3

Problem 1: remdup(l)

Define a Python function remdup(l) that takes a nonempty list of integers l and removes all duplicates in l, keeping only the first occurrence of each number. For instance:


>>> remdup([3,1,3,5])
[3, 1, 5]

>>> remdup([7,3,-1,-5])
[7, 3, -1, -5]

>>> remdup([3,5,7,5,3,7,10])
[3, 5, 7, 10]
                

Problem 2: sumsquare(l)

Write a Python function sumsquare(l) that takes a nonempty list of integers and returns a list [odd,even], where odd is the sum of squares all the odd numbers in l and even is the sum of squares of all the even numbers in l.


>>> sumsquare([1,3,5])
[35, 0]

>>> sumsquare([2,4,6])
[0, 56]

>>> sumsquare([-1,-2,3,7])
[59, 4]
                

Problem 3: transpose(m)

A two-dimensional matrix can be represented in Python row-wise, as a list of lists: each inner list represents one row of the matrix. For instance, the matrix:
1 2 3 4
5 6 7 8
would be represented as [[1, 2, 3, 4], [5, 6, 7, 8]].
The transpose of a matrix converts each row into a column. The transpose of the matrix above is:
1 5
2 6
3 7
4 8
which would be represented as [[1, 5], [2, 6], [3, 7], [4, 8]].
Write a Python function transpose(m) that takes as input a two-dimensional matrix m and returns the transpose of m. The argument m should remain undisturbed by the function.

Examples:


>>> transpose([[1,2,3],[4,5,6]])
[[1, 4], [2, 5], [3, 6]]

>>> transpose([[1],[2],[3]])
[[1, 2, 3]]

>>> transpose([[3]])
[[3]]
                

Programming Assignment Solutions

Python


def remdup(l):
  L=[]
  for ips in l:
    if ips not in L:
      L.append(ips)
  return(L)

def sumsquare(l):
  odd_sum=0
  even_sum=0
  for ats in l:
    if ats%2!=0:
      odd_sum+=ats*ats
    else:
      even_sum+=ats*ats
  return([odd_sum,even_sum])

def transpose(m):
  ans=list()
  for ind in range(len(m[0])):
    a=[]
    for j in range(len(m)):
      a.append(m[j][ind])
    ans.append(a)
  return(ans)  
                

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