Minimum distance between any special pair in the given array
Last Updated :
15 Jul, 2025
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:
- Store the initial indices of array elements in a Map.
- Sort the given array arr[].
- Now, find the distance between the indices of adjacent elements of the sorted array using the Map.
- Maintain the minimum distance for each pair of adjacent elements in the step above.
- 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;
}
// Java program for the above approach
import java.util.*;
class Main {
// Function that finds the minimum
// difference between two vectors
public static int Mindist(List<Integer> left, List<Integer> right) {
int res = Integer.MAX_VALUE;
for (int i = 0; i < left.size(); i++) {
int num = left.get(i);
// Find lower bound of the index
int index = right.indexOf(right.stream().filter(x -> x <= num).findFirst().orElse(right.get(right.size() - 1)));
// Find two adjacent indices
// to take difference
if (index == 0) {
res = Math.min(res, Math.abs(num - right.get(index)));
} else if (index == right.size()) {
res = Math.min(
res,
Math.min(
Math.abs(num - right.get(index - 1)),
Math.abs(num - right.get(index))
)
);
}
}
// Return the result
return res;
}
// Function to find the minimum distance
// between index of special pairs
public static int SpecialPairs(int[] nums) {
// Stores the index of each element
// in the array arr[]
Map<Integer, Integer> m = new HashMap<Integer, Integer>();
List<Integer> vals = new ArrayList<Integer>();
for (int i = 0; i < nums.length; i++) {
m.put(nums[i], i);
}
for (int p : m.keySet()) {
vals.add(p);
}
int res = Integer.MAX_VALUE;
for (int i = 1; i < vals.size(); i++) {
List<Integer> vec = new ArrayList<Integer>();
vec.add(m.get(vals.get(i)));
// Take adjacent difference
// of same values
for (int j = 1; j < vec.size(); j++) {
res = Math.min(res, Math.abs(vec.get(j) - vec.get(j - 1)));
}
if (i > 0) {
int a = vals.get(i);
// Left index array
List<Integer> left = new ArrayList<Integer>();
left.add(m.get(a));
int b = vals.get(i - 1);
// Right index array
List<Integer> right = new ArrayList<Integer>();
right.add(m.get(b));
// Find the minimum gap between
// the two adjacent different
// values
res = Math.min(res, Mindist(left, right)) + 1;
}
}
return res;
}
// Driver Code
public static void main(String[] args) {
int[] arr = {0, -10, 5, -5, 1};
System.out.println(SpecialPairs(arr));
}
}
// This code is contributed by princekumaras
# 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
// Function that finds the minimum
// difference between two vectors
function mindist(left, right) {
let res = Number.MAX_SAFE_INTEGER;
for (let i = 0; i < left.length; i++) {
const num = left[i];
// Find lower bound of the index
const index = right.findIndex((i) => num >= i);
// Find two adjacent indices
// to take difference
if (index === 0) {
res = Math.min(res, Math.abs(num - right[index]));
} else if (index === right.length) {
res = Math.min(
res,
Math.min(
Math.abs(num - right[index - 1]),
Math.abs(num - right[index])
)
);
}
}
// Return the result
return res;
}
// Function to find the minimum distance
// between index of special pairs
function specialPairs(nums) {
// Stores the index of each element
// in the array arr[]
const m = new Map();
const vals = [];
for (let i = 0; i < nums.length; i++) {
m.set(nums[i], i);
}
for (const p of m.keys()) {
vals.push(p);
}
let res = Number.MAX_SAFE_INTEGER;
for (let i = 1; i < vals.length; i++) {
const vec = [m.get(vals[i])];
// Take adjacent difference
// of same values
for (let j = 1; j < vec.length; j++) {
res = Math.min(res, Math.abs(vec[j] - vec[j - 1]));
}
if (i) {
const a = vals[i];
// Left index array
const left = [m.get(a)];
const b = vals[i - 1];
// Right index array
const right = [m.get(b)];
// Find the minimum gap between
// the two adjacent different
// values
res = Math.min(res, mindist(left, right)) + 1;
}
}
return res;
}
// Driver Code
const arr = [0, -10, 5, -5, 1];
console.log(specialPairs(arr));
// This code is contributed by Aditya Sharma
using System;
using System.Collections.Generic;
using System.Linq;
class Program {
// Function that finds the minimum
// difference between two vectors
static int mindist(List<int> left, List<int> right)
{
int res = int.MaxValue;
for (int i = 0; i < left.Count; i++) {
int num = left[i];
// Find lower bound of the index
int index = right.IndexOf(
right.Where(i = > num >= i).Min());
// Find two adjacent indices
// to take difference
if (index == 0)
res = Math.Min(
res, Math.Abs(num - right[index]));
else if (index == right.Count)
res = Math.Min(
res,
Math.Min(
Math.Abs(num - right[index - 1]),
Math.Abs(num - right[index])));
}
// Return the result
return res;
}
// Function to find the minimum distance
// between index of special pairs
static int specialPairs(List<int> nums)
{
// Stores the index of each element
// in the array arr[]
Dictionary<int, int> m = new Dictionary<int, int>();
List<int> vals = new List<int>();
for (int i = 0; i < nums.Count; i++)
m[nums[i]] = i;
vals = m.Keys.ToList();
int res = int.MaxValue;
for (int i = 1; i < vals.Count; i++) {
List<int> vec = new List<int>();
vec.Add(m[vals[i]]);
// Take adjacent difference
// of same values
for (int j = 1; j < vec.Count; j++)
res = Math.Min(
res, Math.Abs(vec[j] - vec[j - 1]));
if (i > 0) {
int a = vals[i];
// Left index array
List<int> left = new List<int>();
left.Add(m[a]);
int b = vals[i - 1];
// Right index array
List<int> right = new List<int>();
right.Add(m[b]);
// Find the minimum gap between
// the two adjacent different
// values
res = Math.Min(res, mindist(left, right))
+ 1;
}
}
return res;
}
static void Main(string[] args)
{
// Given array
List<int> arr
= new List<int>() { 0, -10, 5, -5, 1 };
// Function call
Console.Write(specialPairs(arr));
}
}
// This code is contributed by Prince Kumar
Time Complexity: O(N*log N)
Auxiliary Space: O(N) , since N extra space has been taken.
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