Given an array of positive integers. All numbers occur even number of times except one number which occurs an odd number of times. Find the number in O(n) time & constant space.

**Examples :**

Input : arr = {1, 2, 3, 2, 3, 1, 3} Output : 3 Input : arr = {5, 7, 2, 7, 5, 2, 5} Output : 5

## Recommended: Please solve it on “__PRACTICE__ ” first, before moving on to the solution.

__PRACTICE__## Java

`// Java program to find the element occurring` `// odd number of times` `class` `OddOccurrence {` ` ` ` ` `// function to find the element occurring odd` ` ` `// number of times` ` ` `static` `int` `getOddOccurrence(` `int` `arr[], ` `int` `arr_size)` ` ` `{` ` ` `int` `i;` ` ` `for` `(i = ` `0` `; i < arr_size; i++) {` ` ` `int` `count = ` `0` `;` ` ` `for` `(` `int` `j = ` `0` `; j < arr_size; j++) {` ` ` `if` `(arr[i] == arr[j])` ` ` `count++;` ` ` `}` ` ` `if` `(count % ` `2` `!= ` `0` `)` ` ` `return` `arr[i];` ` ` `}` ` ` `return` `-` `1` `;` ` ` `}` ` ` ` ` `// driver code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `arr[] = ` `new` `int` `[] { ` `2` `, ` `3` `, ` `5` `, ` `4` `, ` `5` `, ` `2` `, ` `4` `, ` `3` `, ` `5` `, ` `2` `, ` `4` `, ` `4` `, ` `2` `};` ` ` `int` `n = arr.length;` ` ` `System.out.println(getOddOccurrence(arr, n));` ` ` `}` `}` `// This code has been contributed by Kamal Rawal` |

**Output:**

5

A **Better Solution** is to use Hashing. Use array elements as key and their counts as value. Create an empty hash table. One by one traverse the given array elements and store counts. Time complexity of this solution is O(n). But it requires extra space for hashing.

**Program :**

## Java

`// Java program to find the element occurring odd` `// number of times` `import` `java.io.*;` `import` `java.util.HashMap;` ` ` `class` `OddOccurrence {` ` ` `// function to find the element occurring odd` ` ` `// number of times` ` ` `static` `int` `getOddOccurrence(` `int` `arr[], ` `int` `n)` ` ` `{` ` ` `HashMap<Integer, Integer> hmap = ` `new` `HashMap<>();` ` ` ` ` `// Putting all elements into the HashMap` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) {` ` ` `if` `(hmap.containsKey(arr[i])) {` ` ` `int` `val = hmap.get(arr[i]);` ` ` ` ` `// If array element is already present then` ` ` `// increase the count of that element.` ` ` `hmap.put(arr[i], val + ` `1` `);` ` ` `}` ` ` `else` ` ` ` ` `// if array element is not present then put` ` ` `// element into the HashMap and initialize` ` ` `// the count to one.` ` ` `hmap.put(arr[i], ` `1` `);` ` ` `}` ` ` ` ` `// Checking for odd occurrence of each element present` ` ` `// in the HashMap` ` ` `for` `(Integer a : hmap.keySet()) {` ` ` `if` `(hmap.get(a) % ` `2` `!= ` `0` `)` ` ` `return` `a;` ` ` `}` ` ` `return` `-` `1` `;` ` ` `}` ` ` ` ` `// driver code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `arr[] = ` `new` `int` `[] { ` `2` `, ` `3` `, ` `5` `, ` `4` `, ` `5` `, ` `2` `, ` `4` `, ` `3` `, ` `5` `, ` `2` `, ` `4` `, ` `4` `, ` `2` `};` ` ` `int` `n = arr.length;` ` ` `System.out.println(getOddOccurrence(arr, n));` ` ` `}` `}` `// This code is contributed by Kamal Rawal` |

**Output:**

5

## Java

`// Java program to find the element occurring odd number of times` ` ` `class` `OddOccurance {` ` ` `int` `getOddOccurrence(` `int` `ar[], ` `int` `ar_size)` ` ` `{` ` ` `int` `i;` ` ` `int` `res = ` `0` `;` ` ` `for` `(i = ` `0` `; i < ar_size; i++) {` ` ` `res = res ^ ar[i];` ` ` `}` ` ` `return` `res;` ` ` `}` ` ` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `OddOccurance occur = ` `new` `OddOccurance();` ` ` `int` `ar[] = ` `new` `int` `[] { ` `2` `, ` `3` `, ` `5` `, ` `4` `, ` `5` `, ` `2` `, ` `4` `, ` `3` `, ` `5` `, ` `2` `, ` `4` `, ` `4` `, ` `2` `};` ` ` `int` `n = ar.length;` ` ` `System.out.println(occur.getOddOccurrence(ar, n));` ` ` `}` `}` `// This code has been contributed by Mayank Jaiswal` |

**Output:**

5

Please refer complete article on Find the Number Occurring Odd Number of Times for more details!

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