Find minimum difference between any two elements | Set 2
Last Updated :
31 Mar, 2023
Given an unsorted array arr[] of size n, the task is to find the minimum difference between any pair in the given array.
Input: arr[] = {1, 2, 3, 4}
Output: 1
The possible absolute differences are:
{1, 2, 3, 1, 2, 1}
Input: arr[] = {10, 2, 5, 4}
Output: 1
Approach:
- Traverse through the array and create a hash array to store the frequency of the array elements.
- Now, traverse through the hash array and calculate the distance between the two nearest elements.
- The frequency was calculated to check whether some element has frequency > 1 which means the absolute distance will be 0 i.e. |arr[i] - arr[i]|.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define MAX 100001
// Function to return the minimum
// absolute difference between any
// two elements of the array
int getMinDiff(int arr[], int n)
{
// To store the frequency of each element
int freq[MAX] = { 0 };
for (int i = 0; i < n; i++) {
// Update the frequency of current element
freq[arr[i]]++;
// If current element appears more than once
// then the minimum absolute difference
// will be 0 i.e. |arr[i] - arr[i]|
if (freq[arr[i]] > 1)
return 0;
}
int mn = INT_MAX;
// Checking the distance between the nearest
// two elements in the frequency array
for (int i = 0; i < MAX; i++) {
if (freq[i] > 0) {
i++;
int cnt = 1;
while ((freq[i] == 0) && (i != MAX - 1)) {
cnt++;
i++;
}
mn = min(cnt, mn);
i--;
}
}
// Return the minimum absolute difference
return mn;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3, 4 };
int n = sizeof(arr) / sizeof(int);
cout << getMinDiff(arr, n);
return 0;
}
// Java implementation of the approach
import java.util.*;
class GFG
{
private static final int MAX = 100001;
// Function to return the minimum
// absolute difference between any
// two elements of the array
static int getMinDiff(int arr[], int n)
{
// To store the frequency of each element
int[] freq = new int[MAX];
for(int i = 0; i < n; i++)
{
freq[i] = 0;
}
for (int i = 0; i < n; i++)
{
// Update the frequency of current element
freq[arr[i]]++;
// If current element appears more than once
// then the minimum absolute difference
// will be 0 i.e. |arr[i] - arr[i]|
if (freq[arr[i]] > 1)
return 0;
}
int mn = Integer.MAX_VALUE;
// Checking the distance between the nearest
// two elements in the frequency array
for (int i = 0; i < MAX; i++)
{
if (freq[i] > 0)
{
i++;
int cnt = 1;
while ((freq[i] == 0) && (i != MAX - 1))
{
cnt++;
i++;
}
mn = Math.min(cnt, mn);
i--;
}
}
// Return the minimum absolute difference
return mn;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 2, 3, 4 };
int n = arr.length;
System.out.println(getMinDiff(arr, n));
}
}
// This code is contributed by nidhi16bcs2007
# Python3 implementation of the approach
MAX = 100001
# Function to return the minimum
# absolute difference between any
# two elements of the array
def getMinDiff(arr, n):
# To store the frequency of each element
freq = [0 for i in range(MAX)]
for i in range(n):
# Update the frequency of current element
freq[arr[i]] += 1
# If current element appears more than once
# then the minimum absolute difference
# will be 0 i.e. |arr[i] - arr[i]|
if (freq[arr[i]] > 1):
return 0
mn = 10**9
# Checking the distance between the nearest
# two elements in the frequency array
for i in range(MAX):
if (freq[i] > 0):
i += 1
cnt = 1
while ((freq[i] == 0) and (i != MAX - 1)):
cnt += 1
i += 1
mn = min(cnt, mn)
i -= 1
# Return the minimum absolute difference
return mn
# Driver code
arr = [ 1, 2, 3, 4]
n = len(arr)
print(getMinDiff(arr, n))
# This code is contributed by Mohit Kumar
// C# implementation of the approach
using System;
class GFG
{
private static int MAX = 100001;
// Function to return the minimum
// absolute difference between any
// two elements of the array
static int getMinDiff(int []arr, int n)
{
// To store the frequency of each element
int[] freq = new int[MAX];
for(int i = 0; i < n; i++)
{
freq[i] = 0;
}
for (int i = 0; i < n; i++)
{
// Update the frequency of current element
freq[arr[i]]++;
// If current element appears more than once
// then the minimum absolute difference
// will be 0 i.e. |arr[i] - arr[i]|
if (freq[arr[i]] > 1)
return 0;
}
int mn = int.MaxValue;
// Checking the distance between the nearest
// two elements in the frequency array
for (int i = 0; i < MAX; i++)
{
if (freq[i] > 0)
{
i++;
int cnt = 1;
while ((freq[i] == 0) && (i != MAX - 1))
{
cnt++;
i++;
}
mn = Math.Min(cnt, mn);
i--;
}
}
// Return the minimum absolute difference
return mn;
}
// Driver code
public static void Main()
{
int []arr = { 1, 2, 3, 4 };
int n = arr.Length;
Console.WriteLine(getMinDiff(arr, n));
}
}
// This code is contributed by AnkitRai01
<script>
// JavaScript implementation of the approach
let MAX = 100001;
// Function to return the minimum
// absolute difference between any
// two elements of the array
function getMinDiff(arr,n)
{
// To store the frequency of each element
let freq=[];
for(let i = 0; i < n; i++)
{
freq[i] = 0;
}
for (let i = 0; i < n; i++)
{
// Update the frequency of current element
freq[arr[i]]++;
// If current element appears more than once
// then the minimum absolute difference
// will be 0 i.e. |arr[i] - arr[i]|
if (freq[arr[i]] > 1)
return 0;
}
let mn = Number.MAX_VALUE;
// Checking the distance between the nearest
// two elements in the frequency array
for (let i = 0; i < MAX; i++)
{
if (freq[i] > 0)
{
i++;
let cnt = 1;
while ((freq[i] == 0) && (i != MAX - 1))
{
cnt++;
i++;
}
mn = Math.min(cnt, mn);
i--;
}
}
// Return the minimum absolute difference
return mn;
}
// Driver code
let arr = [ 1, 2, 3, 4 ];
let n = arr.length;
document.write(getMinDiff(arr, n));
//contributed by 171fa07058
</script>
Time Complexity : O(n)
Explanation: In the above case we have taken MAX is equal to 100001, but we can take max as a maximum element in the array
Auxiliary Space: O(MAX)
Explanation: We have taken a frequency array of size MAX.
Alternate Shorter Implementation :
C++
#include<iostream>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
int main() {
vector<int> arr = {1, 2, 3, 4};
vector<int> diff_list;
// Get the combinations of numbers
for(int i = 0; i < arr.size(); i++) {
for(int j = i+1; j < arr.size(); j++) {
// Find the absolute difference
diff_list.push_back(abs(arr[i] - arr[j]));
}
}
// Find the minimum difference
int min_diff = *min_element(diff_list.begin(), diff_list.end());
cout << min_diff << endl;
return 0;
}
# Python3 implementation of the approach
import itertools
arr = [1,2,3,4]
diff_list = []
# Get the combinations of numbers
for n1, n2 in list(itertools.combinations(arr, 2)):
# Find the absolute difference
diff_list.append(abs(n1-n2))
print(min(diff_list))
# This code is contributed by mailprakashindia
let arr = [1, 2, 3, 4];
let diffList = [];
// Get the combinations of numbers
for (let i = 0; i < arr.length; i++) {
for (let j = i + 1; j < arr.length; j++) {
// Find the absolute difference
diffList.push(Math.abs(arr[i] - arr[j]));
}
}
// Find the minimum difference
let minDiff = Math.min(...diffList);
console.log(minDiff);
import java.util.ArrayList;
import java.util.Collections;
public class Main {
public static void main(String[] args)
{
ArrayList<Integer> arr = new ArrayList<Integer>();
arr.add(1);
arr.add(2);
arr.add(3);
arr.add(4);
ArrayList<Integer> diff_list = new ArrayList<
Integer>(); // Get the combinations of numbers
for (int i = 0; i < arr.size(); i++) {
for (int j = i + 1; j < arr.size(); j++) {
// Find the absolute difference
diff_list.add(
Math.abs(arr.get(i) - arr.get(j)));
}
}
// Find the minimum difference
int min_diff = Collections.min(diff_list);
System.out.println(min_diff);
}
}
using System;
using System.Collections.Generic;
using System.Linq;
class Program {
static void Main(string[] args)
{
List<int> arr = new List<int>{ 1, 2, 3, 4 };
List<int> diff_list = new List<int>();
// Get the combinations of numbers
for (int i = 0; i < arr.Count; i++) {
for (int j = i + 1; j < arr.Count; j++) {
// Find the absolute difference
diff_list.Add(Math.Abs(arr[i] - arr[j]));
}
}
// Find the minimum difference
int min_diff = diff_list.Min();
Console.WriteLine(min_diff);
}
}
Time Complexity: O (r* ( n C r ) )
Explanation: Here r is 2 because we are making combinations of two elements using the iterator tool and n is the length of the given array, so if we calculate it, it will be O (n*(n-1) which will be O (n*n)
Auxiliary Space: O ( ( n C r ) ) ( Here, r=2)
Explanation: We have used the list to store all the combinations of arrays and we have made diff_list array to store absolute difference
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