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 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; } |
chevron_right
filter_none
Java
// 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 |
chevron_right
filter_none
Python3
# 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 |
chevron_right
filter_none
C#
// 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 |
chevron_right
filter_none
Output:
1
Output:
1
Alternate Shorter Implementation :
Python3
# 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 |
chevron_right
filter_none
Output:
1
Recommended Posts:
- Find minimum difference between any two elements
- Find set of m-elements with difference of any two elements is divisible by k
- Minimum distance between any two equal elements in an Array
- Remove Minimum coins such that absolute difference between any two piles is less than K
- Minimum difference between any two primes from the given range
- Place the prisoners into cells to maximize the minimum difference between any two
- Minimum difference between any two weighted nodes in Sum Tree of the given Tree
- Find last two remaining elements after removing median of any 3 consecutive elements repeatedly
- Program to find the maximum difference between the index of any two different numbers
- Minimum Bitwise AND operations to make any two array elements equal
- Minimum Bitwise OR operations to make any two array elements equal
- Minimum Bitwise XOR operations to make any two array elements equal
- Make all elements zero by decreasing any two elements by one at a time
- Longest subarray in which absolute difference between any two element is not greater than X
- Count of numbers upto N having absolute difference of at most K between any two adjacent digits
- Longest subarray with difference exactly K between any two distinct values
- Minimize count of array elements to be removed to maximize difference between any pair up to K
- Find the shortest distance between any pair of two different good nodes
- Find any one of the multiple repeating elements in read only array | Set 2
- Find an index such that difference between product of elements before and after it is minimum
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
