Print all Increasing Subsequence of a List

Given a list or array of integer, the task is to print all such subsequences of this list such in which the elements are arranged in increasing order.

A Subsequence of the list is an ordered subset of that list’s element having same sequential ordering as the original list.

Examples:

Input: arr = {1, 2]}
Output:
2
1
1 2

Input: arr = {1, 3, 2}
Output:
2
3
1
1 2
1 3

Approach:

  • Here, we will use recursion to find the desired output.
  • The function will take two lists as an argument and the base condition will be until the list is empty.
  • At each step of recursion, we will make the decision whether to include or exclude a particular element of the original list.
  • For achieving this, we will maintain two list namely inp and out, the input and output list of that step.
  • While including an element in output list, we will check whether that element is greater than the last element in output list, if yes then we will include that element.
  • When the length of the input list becomes zero then the output list will contain the desired output. This is also a base condition too.

Below is the implementation of the above approach:

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# Python3 implementation
# To store all increasing subsequence of the given list
st = []
  
# Method to find increasing subsequence
def find(inp, out) :
    if len(inp)== 0 :
        if len(out) != 0 :
            # storing result
            st.append(out)
        return
  
    # excluding 1st element
    find(inp[1:], out[:])
  
    # including first element
    # checking the condition
    # for increasing subsequence
    if len(out)== 0:
        find(inp[1:], inp[:1])
    elif inp[0] > out[-1] :
        out.append(inp[0])
        find(inp[1:], out[:])
  
# The input list
ls1 = [1, 3, 2]
ls2 = []
  
# Calling the function
find(ls1, ls2)
  
# Printing the lists
for i in st:
    print(*i)

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Output:

2
3
1
1 2
1 3

Time Complexity: O(2^{n}).



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Improved By : nidhi_biet