Find the only one occurring odd and find the missing terms. Important GFG

 package com.company;

public class CWR_oddNumberArrayContainer {


    //     my method Time complexity -n*n
    public static void count(int[] arr) {
        for (int i = 0; i < arr.length; i++) {
            int count = 0;
            for (int j = 0; j < arr.length; j++) {
                if (arr[i] == arr[j]) {
                    count++;
                }
            }
            if (count % 2 != 0) {
                System.out.println(arr[i]);
            }
        }
    }

    //    Method 2
    public static int findOdd(int[] arr, int n) {
        int res = 0;
        for (int i = 0; i < n; i++) {
            res = res ^ arr[i];
        }
        return res;
    }

    //    Find the missing number
    public static int findMissingNo(int[] arr, int n) {
        int res = 0;
        for (int i = 0; i < n; i++) {
            res = res ^ arr[i];
        }
        for(int j = 1; j <= n+1; j++){
            res = res ^ j;
        }
        return res;
    }

    public static void main(String[] args) {
        int[] a = {1, 2, 3, 5, 6, 7, 8, 9, 10};
        count(a);
//        System.out.println(findOdd(a, 5));
//        System.out.println(findMissingNo(a, 9));
    }
}