# Program to find minimum possible difference of indices of adjacent elements in Python

PythonServer Side ProgrammingProgramming

Suppose we have a list of numbers nums, we can say that two numbers nums[i] ≤ nums[j] are adjacent when there is no number in between (nums[i], nums[j]) in nums. We have to find the minimum possible |j - i| such that nums[j] and nums[i] are adjacent.

So, if the input is like nums = [1, -9, 6, -6, 2], then the output will be 2, as we can see that 2 and 6 are adjacent and they are 2 indices away from each other.

To solve this, we will follow these steps −

• indexes := a new map

• for each index i and value x in A, do

• insert i at the end of indexes[x]

• ans := size of A

• for each row in the list of all values of indexes, do

• for i in range 0 to size of row - 2, do

• ans := minimum of ans and (row[i + 1] - row[i])

• vals := sort the list indexes

• for k in range 0 to size of vals - 2, do

• r1 := indexes[vals[k]]

• r2 := indexes[vals[k + 1]]

• i := j := 0

• while i < size of r1 and j < size of r2, do

• ans := minimum of ans and |r1[i] - r2[j]|

• if r1[i] < r2[j], then

• i := i + 1

• otherwise,

• j := j + 1

• return ans

## Example (Python)

Let us see the following implementation to get better understanding −

Live Demo

from collections import defaultdict
class Solution:
def solve(self, A):
indexes = defaultdict(list)
for i, x in enumerate(A):
indexes[x].append(i)
ans = len(A)
for row in indexes.values():
for i in range(len(row) - 1):
ans = min(ans, row[i + 1] - row[i])
vals = sorted(indexes)
for k in range(len(vals) - 1):
r1 = indexes[vals[k]]
r2 = indexes[vals[k + 1]]
i = j = 0
while i < len(r1) and j < len(r2):
ans = min(ans, abs(r1[i] - r2[j]))
if r1[i] < r2[j]:
i += 1
else:
j += 1
return ans
ob = Solution()
nums = [1, -9, 6, -6, 2]
print(ob.solve(nums))

## Input

[1, -9, 6, -6, 2]

## Output

2
Published on 21-Dec-2020 18:32:47