Find minimum difference between any two elements | Set 2
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
Recommended Posts:
- Find minimum difference between any two elements
- Find an index such that difference between product of elements before and after it is minimum
- Minimize the difference between minimum and maximum elements
- Choose k array elements such that difference of maximum and minimum is minimized
- Minimum absolute difference of adjacent elements in a circular array
- Find set of m-elements with difference of any two elements is divisible by k
- Find k ordered pairs in array with minimum difference d
- Find maximum difference between nearest left and right smaller elements
- Find maximum number of elements such that their absolute difference is less than or equal to 1
- Find the winner by adding Pairwise difference of elements in the array until Possible
- Find the difference of count of equal elements on the right and the left for each element
- Find the first, second and third minimum elements in an array
- Find minimum sum such that one of every three consecutive elements is taken
- Find minimum elements after considering all possible transformations
- Find minimum changes required in an array for it to contain k distinct elements
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.
- Find a pair (n,r) in an integer array such that value of nPr is maximum
- Minimum increment or decrement operations required to make the array sorted
- Find missing elements from an Array
- Count smaller elements on right side and greater elements on left side using Binary Index Tree
- Find a pair (n,r) in an integer array such that value of nCr is maximum
- What is Competitive Programming and How to Prepare for It?
- Find a pair (n,r) in an integer array such that value of nPr is maximum
- Minimum increment or decrement operations required to make the array sorted
- Longest Increasing Subsequence using Longest Common Subsequence Algorithm
- Find a pair (n,r) in an integer array such that value of nCr is maximum
