Codility PermMissingElem Java solution

PermMissingElem is a time complexity test in Codility where an array contains a list of integers from 1 to N+1, where N is the size of the array. There is one value missing and the object of the test is to find the missing value in the quickest time possible. This video show my Java solution to PermMissingElem.

[furqanshaikh]

Hi Dave, We could have used Sum of Arithmetic means as well.

class Solution {
public int solution(int[] A) {
double N = A.length + 2;
double sumOfArithematicSeries = N * (N - 1) / 2;
for (int i : A) {
sumOfArithematicSeries -= i;
}
return (int) (sumOfArithematicSeries);
}
}

[Ngonie_]

Hi Dave, how are you? Hope I find you doing good. Just a quick question
(i)On the first for loop, why are we iterating from 1
(ii)On the second for loop, why are we removing the object and not the value of the array index
Thanks

[jabert]

Hi Dave, It was very helpful thanks . if we use some other array may be {23, 21, 24, 25} how to get the missing elements . for that we need do ?

[terrencemoore]

def solution(A):
size_A = len(A)
if size_A == 0:
return 1
set_A = set(A)
range_A = size_A + 1
for i in range(1, range_A):
if i not in set_A:
return i

Why does this keep giving me a runtime error for inputs like [1]? I keep getting a 50%

[jetlavi]

Dave, when we know that missing element is consecutive, why just do not read it in order following sorting with Arrays?
it could be I am wrong but..

import java.util.Arrays;
public class PermMissingElem {
private int solution(int [] arr) {
int result=0;
Arrays.sort(arr);
int lastval = 0;
for(int i=0; i < arr.length; i++) {
if(i==0) { lastval=arr[i]; continue; } // just init first reading
if( arr[i]-lastval==1) continue; // continuing reading
else {
result = arr[i]+1;
break;
}
}
return result;
}

public static void main(String [] args) {
int[] intarr = new int [] {2, 3, 1, 5};
PermMissingElem obj = new PermMissingElem();

}

}

[getinetamare]

Thanks :) return (int)Math.ceil((Y-X) / (double)D);

[testmail]

A 70% solution I got for this on Codility (JavaScript):

function solution ( A ) {
A.sort();
var res = i = 0;

while(i<A.length){
res = (A[i]>1 && res==0) ? 0: A[i];
if(res==0) break;
if(!A.includes(res+1) && (res+1)<A[A.length-1]) break;
i++;
}

return (res+1);
}