Python – Surrounding elements to K

Given Matrix, from each row, get surrounding elements of K, if present.

Input : test_list = [[7, 6, 3, 2], [5, 6], [2, 1], [6, 1, 2]], K = 6 
Output : [[7, 3], [5], [], [1]] 
Explanation : All elements surrounded by 6 are extracted.

Input : test_list = [[7, 6, 3, 2], [5, 6], [2, 1], [6, 1, 2]], K = 1 
Output : [[], [], [2], [6, 2]] 
Explanation : All elements surrounded by 1 are removed. 

Method #1: Using index() + loop + try/except

In this, we check for the index of elements using index(), and then get the required surrounding elements by accessing next and previous elements. If the element is not present, an empty list is returned.

Python3

# Python3 code to demonstrate working of
# Surrounding elements to K
# Using index() + loop + try/except

# initializing list
test_list = [[7, 6, 3, 2], [5, 6], [2, 1], [6, 1, 2]]

# printing original list
print("The original list is : " + str(test_list))

# initializing K
K = 6

res = []
for sub in test_list:

    # getting index
    try :
        idx = sub.index(K)
    except Exception as e :
        res.append([])
        continue

    # appending Surrounding elements
    if idx != 0 and idx != len(sub) - 1:
        res.append([sub[idx - 1], sub[idx + 1]])
    elif idx == 0:
        res.append([sub[idx + 1]])
    else : res.append([sub[idx - 1]])

# printing result
print("The Surrounding elements : " + str(res))

Output:

The original list is : [[7, 6, 3, 2], [5, 6], [2, 1], [6, 1, 2]] The Surrounding elements : [[7, 3], [5], [], [1]]

Time Complexity: O(n)
Auxiliary Space: O(n)

Method #2 : Using index() + in operator + loop

In this, we check if the element is present in a row before application of index(), to avoid using try/except block. Rest all functionalities are the same as the above method.

Python3

# Python3 code to demonstrate working of
# Surrounding elements to K
# Using index() + in operator + loop

# initializing list
test_list = [[7, 6, 3, 2], [5, 6], [2, 1], [6, 1, 2]]

# printing original list
print("The original list is : " + str(test_list))

# initializing K
K = 6

res = []
for sub in test_list:

    # getting index
    # checking for element presence in row
    if K in sub:
        idx = sub.index(K)
    else:
        res.append([])
        continue

    # appending Surrounding elements
    if idx != 0 and idx != len(sub) - 1:
        res.append([sub[idx - 1], sub[idx + 1]])
    elif idx == 0:
        res.append([sub[idx + 1]])
    else:
        res.append([sub[idx - 1]])

# printing result
print("The Surrounding elements : " + str(res))

Output:

The original list is : [[7, 6, 3, 2], [5, 6], [2, 1], [6, 1, 2]] The Surrounding elements : [[7, 3], [5], [], [1]]

Time Complexity: O(n)
Auxiliary Space: O(n)