Python - Nearest occurrence between two elements in a List

Given a list and two elements, x and y find the nearest occurrence index of element x from element y.

Input  : test_list = [2, 4, 5, 7, 8, 6, 3, 8, 7, 2, 0, 9, 4, 9, 4], x = 4, y = 6 
Output : 1 
Explanation : 4 is found at 1, 12 and 14th index, 6 is at 5th index, nearest is 1st index.

Input  : test_list = [2, 4, 5, 7, 8, 6, 3, 8, 7, 2, 0, 9, 4, 9, 4], x = 7, y = 6 
Output : 3 
Explanation : 7 is found at 3rd and 8th index, 6 is at 5th index, nearest is 3rd index. 

Method : Using list comprehension + loop + index()

In this, we find all indices of y using list comprehension, and then get index of x using index(), post that loop is used to get index difference, and the nearest index is returned as result.

# Python3 code to demonstrate working of
# Nearest occurrence of x from y in List
# Using list comprehension + loop + index()


# Function to find index of nearest
# occurrence between two elements
def nearestOccurrenceIndex(test_list, x, y):

    # checking if both elements are present in list
    if x not in test_list or y not in test_list:
        return -1
    # getting indices of x
    x_idx = [idx for idx in range(len(test_list)) if test_list[idx] == x]

    # getting y index
    y_idx = test_list.index(y)

    # getting min_dist index
    min_dist = 1000000
    res = None
    for ele in x_idx:

        # checking for min ele, and updating index
        if abs(ele - y_idx) < min_dist:
            res = ele
            min_dist = abs(ele - y_idx)
    return res


# initializing list
input_list = [2, 4, 5, 7, 8, 6, 3, 8, 4, 2, 0, 9, 4, 9, 4]

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

# initializing x
x = 4

# initializing y
y = 6

# printing result
print("Minimum distance index: ", nearestOccurrenceIndex(input_list, x, y))

The original list is : [2, 4, 5, 7, 8, 6, 3, 8, 4, 2, 0, 9, 4, 9, 4]
Minimum distance index:  8

Time Complexity: O(n), where n is the length of the list.
Auxiliary Space: O(1)

Method #2: Using linear search

Approach: 

We use linear search to find the nearest occurrence of two elements x and y in the input list. The algorithm iterates through the list using a for loop and checks if the current element is equal to x or y. If the current element is equal to x or y, it checks if min_dist is None or if the distance between the current index and min_idx is less than min_dist. If the distance between the current index and min_idx is less than min_dist or min_dist is None, it updates min_dist and min_idx. At the end of the loop, if both x and y are found in the list, the algorithm returns min_dist. Otherwise, it returns -1 to indicate that one of the elements was not found in the list.

Algorithm:

1. Initialize variables min_dist and min_idx to be None
2. Iterate through the list using a for loop and check if the current element is equal to x or y.
3. If the current element is equal to x or y, check if min_dist is None or if the distance between the current index and min_idx is less than min_dist.
4. If the distance between the current index and min_idx is less than min_dist or min_dist is None, update min_dist and min_idx.
5. If both x and y are found in the list, return min_dist. Otherwise, return -1.
6. End.

def nearest_occurrence_linear(test_list, x, y):
  
    min_dist = None
    min_idx = None
    
    for i in range(len(test_list)):
      
        if test_list[i] == x or test_list[i] == y:
            if min_idx is not None and test_list[i] != test_list[min_idx]:
                dist = i - min_idx
                if min_dist is None or dist < min_dist:
                    min_dist = dist
            min_idx = i

    if min_dist is not None:
        return min_dist
    else:
        return -1


# Input lists 
test_list = [2, 4, 5, 7, 8, 6, 3, 8, 4, 2, 0, 9, 4, 9, 4]

x = 4
y = 6

# Print the answer
print(nearest_occurrence_linear(test_list, x, y))

3

Time Complexity: O(n), where n is the length of the list
Space Complexity: O(1)

Method 3: Using 2 pointers

Steps:

1. Initialize two pointers, one for x and one for y, to -1.
2. Initialize a variable to store the minimum distance to a large value, such as float('inf').
3. Loop through the list, and for each element, update the index of the corresponding pointer if it matches x or y.
4. If both pointers have been updated, calculate the distance between them and update the minimum distance variable if the distance is smaller than the current minimum.
5. Return the index of the pointer that was updated last.

Example:

def nearestOccurrenceIndex(test_list, x, y):

    x_idx = -1
    y_idx = -1
    
    min_dist = float('inf')
    
    for i, val in enumerate(test_list):
      
        if val == x:
            x_idx = i
        elif val == y:
            y_idx = i
            
        if x_idx != -1 and y_idx != -1:
            dist = abs(x_idx - y_idx)
            
            if dist < min_dist:
                min_dist = dist
                if x_idx > y_idx:
                    res = x_idx
                else:
                    res = y_idx
                    
    if min_dist == float('inf'):
        return -1
      
    return res

# Initializing list
input_list = [2, 4, 5, 7, 8, 6, 3, 8, 4, 2, 0, 9, 4, 9, 4]

# Printing original list
print("The original list is : " + str(input_list))

# Initializing x
x = 4

# Initializing y
y = 6

# Printing the result
print("Minimum distance index: ", nearestOccurrenceIndex(input_list, x, y))

The original list is : [2, 4, 5, 7, 8, 6, 3, 8, 4, 2, 0, 9, 4, 9, 4]
Minimum distance index:  8

Time complexity: O(n), since we need to loop through the entire list. 
Auxiliary space: O(1), since we only need to store a few variables that don't depend on the size of the input list.