Minimum distance between any special pair in the given array

Given an array arr[] of N integers, the task is to find the minimum possible absolute difference between indices of a special pair.

A special pair is defined as a pair of indices (i, j) such that if arr[i] ≤ arr[j], then there is no element X (where arr[i] < X < arr[j]) present in between indices i and j. 
For example:
arr[] = {1, -5, 5} 
Here, {1, 5} forms a special pair as there are no elements in the range (1 to 5) in between arr[0] and arr[2].

Print the minimum absolute difference abs(j – i) such that pair (i, j) forms a special pair.

Examples:

Input: arr[] = {0, -10, 5, -5, 1} 
Output: 2 
Explanation: 
The elements 1 and 5 forms a special pair since there is no elements X in the range 1 < X < 5 in between them, and they are 2 indices away from each other.
Input: arr[] = {3, 3} 
Output: 1

Naive Approach: The simplest approach is to consider every pair of elements of the array and check if they form a special pair or not. If found to be true, then print the minimum distance among all the pairs formed.

Time Complexity: O(N³) 
Auxiliary Space: O(1) 

Efficient Approach: To optimize the above approach, the idea is to observe that we have to only consider the distance between the position of adjacent elements in the sorted array since those pair of elements will not have any value X between it. Below are the steps:

1. Store the initial indices of array elements in a Map.
2. Sort the given array arr[].
3. Now, find the distance between the indices of adjacent elements of the sorted array using the Map.
4. Maintain the minimum distance for each pair of adjacent elements in the step above.
5. After the completing the above steps, print the minimum distance formed.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function that finds the minimum
// difference between two vectors
int mindist(vector<int>& left,
            vector<int>& right)
{
    int res = INT_MAX;
    for (int i = 0; i < left.size(); ++i) {

        int num = left[i];

        // Find lower bound of the index
        int index
            = lower_bound(right.begin(),
                        right.end(), num)
            - right.begin();

        // Find two adjacent indices
        // to take difference
        if (index == 0)
            res = min(res,
                    abs(num
                        - right[index]));

        else if (index == right.size())
            res = min(res,
                    abs(num
                        - right[index - 1]));
        else
            res = min(res,
                    min(abs(num
                            - right[index - 1]),
                        abs(num
                            - right[index])));
    }

    // Return the result
    return res;
}

// Function to find the minimum distance
// between index of special pairs
int specialPairs(vector<int>& nums)
{
    // Stores the index of each element
    // in the array arr[]
    map<int, set<int> > m;
    vector<int> vals;

    // Store the indexes
    for (int i = 0;
        i < nums.size(); ++i) {
        m[nums[i]].insert(i);
    }

    // Get the unique values in list
    for (auto p : m) {
        vals.push_back(p.first);
    }

    int res = INT_MAX;

    for (int i = 0;
        i < vals.size(); ++i) {

        vector<int> vec(m[vals[i]].begin(),
                        m[vals[i]].end());

        // Take adjacent difference
        // of same values
        for (int i = 1;
            i < vec.size(); ++i)
            res = min(res,
                    abs(vec[i]
                        - vec[i - 1]));

        if (i) {
            int a = vals[i];

            // Left index array
            vector<int> left(m[a].begin(),
                            m[a].end());

            int b = vals[i - 1];

            // Right index array
            vector<int> right(m[b].begin(),
                            m[b].end());

            // Find the minimum gap between
            // the two adjacent different
            // values
            res = min(res,
                    mindist(left, right));
        }
    }
    return res;
}

// Driver Code
int main()
{
    // Given array
    vector<int> arr{ 0, -10, 5, -5, 1 };

    // Function Call
    cout << specialPairs(arr);

    return 0;
}

Python3

# Python3 program for the above approach
import sys

# Function that finds the minimum
# difference between two vectors
def mindist(left, right):

    res = sys.maxsize

    for i in range(len(left)):
        num = left[i]

        # Find lower bound of the index
        index = right.index(min(
                [i for i in right if num >= i]))

        # Find two adjacent indices
        # to take difference
        if (index == 0):
            res = min(res,
                  abs(num - right[index]))
        elif (index == len(right)):
            res = min(res,
                  min(abs(num - right[index -1]),
                      abs(num - right[index])))

    # Return the result
    return res

# Function to find the minimum distance
# between index of special pairs
def specialPairs(nums):

    # Stores the index of each element
    # in the array arr[]
    m = {}
    vals = []

    for i in range(len(nums)):
        m[nums[i]] = i

    for p in m:
        vals.append(p)

    res = sys.maxsize

    for i in range(1, len(vals)):
        vec = [m[vals[i]]]

        # Take adjacent difference
        # of same values
        for i in range(1, len(vec)):
            res = min(res,
                      abs(vec[i] - vec[i - 1]))

        if (i):
            a = vals[i]

            # Left index array
            left = [m[a]]

            b = vals[i - 1]

            # Right index array
            right = [m[b]]

            # Find the minimum gap between
            # the two adjacent different
            # values
            res = min(res,
                      mindist(left, right)) + 1

    return res

# Driver Code
if __name__ == "__main__":

    # Given array
    arr = [ 0, -10, 5, -5, 1 ]

    # Function call
    print(specialPairs(arr))

# This code is contributed by dadi madhav

Output:

2

Time Complexity: O(N*log N) 
Auxiliary Space: O(N) , since N extra space has been taken.