Check if array elements are consecutive








Input: {5,4,1,2,3}
Output: yes

 Explanation: Max element is 5 and min element id 1, Array contains all elements from 1 to 5
Input: {2,1,0,-3,-2,-1}
Output: yes

 Explanation: Array contains elements from -3 to 2.
Concept used:

 If the array elements are consecutive and distinct then the array is in AP.
So we compare array sum with AP sum.
ap_sum = n/2 * [2a +(n-1)*d] where d = 1, since elements are consecutive.

 Code:

 import java.util.*;

 public class Solution {

     
    public static void main(String[] args) {
       
        int[] a = {5, 4, 2, 1, 3};
        int n = a.length;
        int sum = 0;
        boolean result = false;
        
        int first = a[0];
        for(int i=1; i<n; i++){
            if(a[i]<first)
                first = a[i];
        }
        
        int apSum = (n * (2*first + (n-1)) / 2);
        
        for(int i=0; i<n; i++){
            sum += a[i];
        }
        
        if(apSum == sum)
            result = true;
        
        System.out.println("apsum "+apSum+" array sum "+sum);
        
        if(result)
            System.out.print("Yes");
        else
            System.out.print("No");
        
    }
}

 Output:

 
apsum 15 array sum 15
Yes

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