# Find Unique pair in an array with pairs of numbers

Given an array where every element appears twice except a pair (two elements). Find the elements of this unique pair.

Examples:

```Input  : 6, 1, 3, 5, 1, 3, 7, 6
Output : 5 7
All elements appear twice except 5 and 7

Input  : 1 3 4 1
Output : 3 4
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is based on below post.

Find Two Missing Numbers | Set 2 (XOR based solution)

1. XOR each element of the array and you will left with the XOR of two different elements which are going to be our result. Let this XOR be “XOR
2. Now find a set bit in XOR.
3. Now divide array elements in two groups. One group that has the bit found in step 2 as set and other group that has the bit as 0.
4. XOR of elements present in first group would be our first element. And XOR of elements present in second group would be our second element.

## C++

 `// C program to find a unique pair in an array ` `// of pairs. ` `#include ` ` `  `void` `findUniquePair(``int` `arr[], ``int` `n) ` `{ ` `    ``// XOR each element and get XOR of two unique  ` `    ``// elements(ans) ` `    ``int` `XOR = arr; ` `    ``for` `(``int` `i = 1; i < n; i++)  ` `        ``XOR = XOR ^ arr[i]; ` `   `  `    ``// Now XOR has XOR of two missing elements. Any set ` `    ``// bit in it must be set in one missing and unset in ` `    ``// other missing number ` `  `  `    ``// Get a set bit of XOR (We get the rightmost set bit) ` `    ``int` `set_bit_no = XOR & ~(XOR-1); ` `  `  `    ``// Now divide elements in two sets by comparing rightmost ` `    ``// set bit of XOR with bit at same position in each element. ` `    ``int` `x = 0, y = 0; ``// Initialize missing numbers ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``if` `(arr[i] & set_bit_no) ` `            ``x = x ^ arr[i]; ``/*XOR of first set in arr[] */` `        ``else` `            ``y = y ^ arr[i]; ``/*XOR of second set in arr[] */` `    ``} ` ` `  `    ``printf``(``"The unique pair is (%d, %d)"``, x, y); ` `     `  `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a[] = { 6, 1, 3, 5, 1, 3, 7, 6 }; ` `    ``int` `n = ``sizeof``(a)/``sizeof``(a); ` `    ``findUniquePair(a, n);  ` `    ``return` `0; ` `} `

## Java

 `// Java program to find a unique pair  ` `// in an array of pairs. ` `class` `GFG ` `{ ` `    ``static` `void` `findUniquePair(``int``[] arr, ``int` `n) ` `    ``{ ` `        ``// XOR each element and get XOR of two ` `        ``// unique elements(ans) ` `        ``int` `XOR = arr[``0``]; ` `         `  `        ``for` `(``int` `i = ``1``; i < n; i++)  ` `            ``XOR = XOR ^ arr[i]; ` ` `  `        ``// Now XOR has XOR of two missing elements. ` `        ``// Any set bit in it must be set in one  ` `        ``// missing and unset in other missing number ` ` `  `        ``// Get a set bit of XOR (We get the  ` `        ``// rightmost set bit) ` `        ``int` `set_bit_no = XOR & ~(XOR-``1``); ` ` `  `        ``// Now divide elements in two sets by  ` `        ``// comparing rightmost set bit of XOR with  ` `        ``// bit at same position in each element. ` `        ``// Initialize missing numbers ` `        ``int` `x = ``0``, y = ``0``;  ` `         `  `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` `            ``if` `((arr[i] & set_bit_no)>``0``) ` `             `  `                ``/*XOR of first set in arr[] */` `                ``x = x ^ arr[i];  ` `            ``else` `                ``/*XOR of second set in arr[] */` `                ``y = y ^ arr[i];  ` `        ``} ` ` `  `        ``System.out.println(``"The unique pair is ("` `+  ` `                               ``x + ``","` `+ y + ``")"``); ` ` `  `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args) { ` `    ``int``[] a = { ``6``, ``1``, ``3``, ``5``, ``1``, ``3``, ``7``, ``6` `}; ` `    ``int` `n = a.length; ` `    ``findUniquePair(a, n);  ` `    ``} ` ` `  `} ` ` `  `/* This code is contributed by Mr. Somesh Awasthi */`

## Python 3

 `# Python 3 program to find a unique ` `# pair in an array of pairs. ` `def` `findUniquePair(arr, n): ` ` `  `    ``# XOR each element and get XOR ` `    ``# of two unique elements(ans) ` `    ``XOR ``=` `arr[``0``] ` `    ``for` `i ``in` `range``(``1``, n):  ` `        ``XOR ``=` `XOR ^ arr[i] ` ` `  `    ``# Now XOR has XOR of two missing ` `    ``# elements. Any set bit in it  ` `    ``# must be set in one missing and ` `    ``# unset in other missing number ` ` `  `    ``# Get a set bit of XOR (We get  ` `    ``# the rightmost set bit) ` `    ``set_bit_no ``=` `XOR & ~(XOR ``-` `1``) ` ` `  `    ``# Now divide elements in two sets ` `    ``# by comparing rightmost set bit ` `    ``# of XOR with bit at same position ` `    ``# in each element. ` `    ``x ``=` `0` `    ``y ``=` `0` `# Initialize missing numbers ` `    ``for` `i ``in` `range``(``0``, n): ` `         `  `        ``if` `(arr[i] & set_bit_no): ` `             `  `            ``# XOR of first set in ` `            ``# arr[] ` `            ``x ``=` `x ^ arr[i]  ` `        ``else``: ` `             `  `            ``# XOR of second set ` `            ``# in arr[] ` `            ``y ``=` `y ^ arr[i]  ` `     `  ` `  `    ``print``(``"The unique pair is ("``, x, ` `             ``", "``, y, ``")"``, sep ``=` `"") ` `     `  `# Driver code ` `a ``=` `[``6``, ``1``, ``3``, ``5``, ``1``, ``3``, ``7``, ``6` `] ` `n ``=` `len``(a) ` `findUniquePair(a, n)  ` ` `  `# This code is contributed by Smitha. `

## C#

 `// C# program to find a unique pair  ` `// in an array of pairs. ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``static` `void` `findUniquePair(``int``[] arr, ``int` `n) ` `    ``{ ` `         `  `        ``// XOR each element and get XOR of two ` `        ``// unique elements(ans) ` `        ``int` `XOR = arr; ` `         `  `        ``for` `(``int` `i = 1; i < n; i++)  ` `            ``XOR = XOR ^ arr[i]; ` ` `  `        ``// Now XOR has XOR of two missing  ` `        ``// elements. Any set bit in it must ` `        ``// be set in one missing and unset ` `        ``// in other missing number ` ` `  `        ``// Get a set bit of XOR (We get the  ` `        ``// rightmost set bit) ` `        ``int` `set_bit_no = XOR & ~(XOR - 1); ` ` `  `        ``// Now divide elements in two sets by  ` `        ``// comparing rightmost set bit of XOR ` `        ``// with bit at same position in each ` `        ``// element. Initialize missing numbers ` `        ``int` `x = 0, y = 0;  ` `         `  `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``if` `((arr[i] & set_bit_no) > 0) ` `             `  `                ``/*XOR of first set in arr[] */` `                ``x = x ^ arr[i];  ` `            ``else` `             `  `                ``/*XOR of second set in arr[] */` `                ``y = y ^ arr[i];  ` `        ``} ` ` `  `        ``Console.WriteLine(``"The unique pair is ("` `                           ``+ x + ``", "` `+ y + ``")"``); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main () ` `    ``{ ` `        ``int``[] a = { 6, 1, 3, 5, 1, 3, 7, 6 }; ` `        ``int` `n = a.Length; ` `         `  `        ``findUniquePair(a, n);  ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output:

`The unique pair is (7, 5)`

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