Minimum Difference in an Array.cpp

//  Minimum Difference in an Array
// Easy
// 0/40
// Asked in companies
// Problem statement
// Given an array of integers, print the minimum of the absolute difference of all possible pairs of elements.

// Example :
// N = 5
// ARR = [ 3, -6, 7, -7, 0 ]
// Out of all pairs, (-7,-6) have a difference of ‘1’, and no other pair has less difference. So ‘ANS’ is ‘1’.    
// Detailed explanation ( Input/output format, Notes, Images )
// Constraints :
// 1 <= T <= 10
// 2 <= N  <= 10^5
// -10^8 <= ARR[i] <= 10^8
// Sum of N <= 10^5

// Time Limit: 1 sec
// Sample Input 1 :
// 2
// 4
// 1 8 3 10   
// 2
// 5 5
// Sample Output 1 :
// 2
// 0
// Explanation Of Sample Input 1 :
// For test case 1, 
// Out of all pairs (1,3) and (8,10) have the minimum difference ‘2’ so the answer is ‘2’.
// For test case 2,
// There is only one possible pair (5,5) so the answer is ‘0’.
// Sample Input 2 :
// 2
// 3
// 8 1 8
// 2
// -3 3
// Sample Output 2 :
// 0
// 6
// C++ (g++ 11)
// 00
//  : 
// 00
//  : 
// 22
// 12345678910
// #include <bits/stdc++.h> 
// int minDiff(int n, vector < int > arr) {
//    // Write your code here.
//    sort(arr.begin(),arr.end());
//    int minD=INT_MAX;
//    for(int i=1;i<n;i++){
//       minD=min(minD,arr[i]-arr[i-1]);
//    }
//    return minD;
// }


#include <bits/stdc++.h> 
int minDiff(int n, vector < int > arr) {
   // Write your code here.
   sort(arr.begin(),arr.end());
   int minD=INT_MAX;
   for(int i=1;i<n;i++){
      minD=min(minD,arr[i]-arr[i-1]);
   }
   return minD;
}