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