Asked by SG
Given an array in which all numbers except two are repeated once. (i.e. we have 2n+2 numbers and n numbers are occurring twice and remaining two have occurred once). Find those two numbers in the most efficient way.
Method 1(Use Sorting)
First, sort all the elements. In the sorted array, by comparing adjacent elements we can easily get the non-repeating elements. Time complexity of this method is O(nLogn)
Method 2(Use XOR)
Let x and y be the non-repeating elements we are looking for and arr[] be the input array. First, calculate the XOR of all the array elements.
xor = arr[0]^arr[1]^arr[2].....arr[n-1]
All the bits that are set in xor will be set in one non-repeating element (x or y) and not in others. So if we take any set bit of xor and divide the elements of the array in two sets – one set of elements with same bit set and another set with same bit not set. By doing so, we will get x in one set and y in another set. Now if we do XOR of all the elements in the first set, we will get the first non-repeating element, and by doing same in other sets we will get the second non-repeating element.
Let us see an example.
arr[] = {2, 4, 7, 9, 2, 4}
1) Get the XOR of all the elements.
xor = 2^4^7^9^2^4 = 14 (1110)
2) Get a number which has only one set bit of the xor.
Since we can easily get the rightmost set bit, let us use it.
set_bit_no = xor & ~(xor-1) = (1110) & ~(1101) = 0010
Now set_bit_no will have only set as rightmost set bit of xor.
3) Now divide the elements in two sets and do xor of
elements in each set and we get the non-repeating
elements 7 and 9. Please see the implementation for this step.
Approach :
Step 1: Xor all the elements of the array into a variable sum thus all the elements present twice in an array will get removed as for example, 4 = “100” and if 4 xor 4 => “100” xor “100” thus answer will be “000”.
Step 2: Thus in the sum the final answer will be 3 xor 5 as both 2 and 4 are xor with itself giving 0, therefore sum = “011” xor “101” i.e sum = “110” = 6.
Step 3: Now we will take 2’s Complement of sum i.e (-sum) = “010”.
Step 4: Now bitwise And the 2’s of sum with the sum i.e “110” & “010” gives the answer “010” (Aim for bitwise & is that we want to get a number that contains only the rightmost set bit of the sum).
Step 5: bitwise & all the elements of the array with this obtained sum, 2 = “010” & “010” = 2, 3 = “011” & “010” = “010” , 4 = “100” & “010” = “000”, 5 = “101” & “010” = “000”.
Step 6: As we can see that the bitwise & of 2,3 > 0 thus they will be xor with sum1 and bitwise & of 4,5 is resulting into 0 thus they will be xor with sum2.
Step 7: As 2 is present two times so getting xor with sum1 two times only the result 3 is being stored in it and As 4 is also present two times thus getting xor with sum2 will cancel it’s value and thus only 5 will remain there.
Implementation:
C++
// C++ proogram for above approach#include <bits/stdc++.h>using namespace std;/* This function sets the values of *x and *y to non-repeating elementsin an array arr[] of size n*/void get2NonRepeatingNos(int arr[], int n, int *x, int *y) { /* Will hold Xor of all elements */ int Xor = arr[0]; /* Will have only single set bit of Xor */ int set_bit_no; int i; *x = 0; *y = 0; /* Get the Xor of all elements */ for(i = 1; i < n; i++) Xor ^= arr[i]; /* Get the rightmost set bit in set_bit_no */ 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. */ for(i = 0; i < n; i++) { /*Xor of first set */ if(arr[i] & set_bit_no) *x = *x ^ arr[i]; /*Xor of second set*/ else { *y = *y ^ arr[i]; } } } /* Driver code */int main() { int arr[] = {2, 3, 7, 9, 11, 2, 3, 11}; int n = sizeof(arr)/sizeof(*arr); int *x = new int[(sizeof(int))]; int *y = new int[(sizeof(int))]; get2NonRepeatingNos(arr, n, x, y); cout<<"The non-repeating elements are "<<*x<<" and "<<*y; } // This code is contributed by rathbhupendra |
C
// C program for above approach#include <stdio.h>#include <stdlib.h>/* This function sets the values of *x and *y to non-repeating elementsin an array arr[] of size n*/void get2NonRepeatingNos(int arr[], int n, int *x, int *y) { /* Will hold Xor of all elements */ int Xor = arr[0]; /* Will have only single set bit of Xor */ int set_bit_no; int i; *x = 0; *y = 0; /* Get the Xor of all elements */ for(i = 1; i < n; i++) Xor ^= arr[i]; /* Get the rightmost set bit in set_bit_no */ 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. */ for(i = 0; i < n; i++) { /*Xor of first set */ if(arr[i] & set_bit_no) *x = *x ^ arr[i]; /*Xor of second set*/ else { *y = *y ^ arr[i]; } } } /* Driver program to test above function */int main(){ int arr[] = {2, 3, 7, 9, 11, 2, 3, 11}; int *x = (int *)malloc(sizeof(int)); int *y = (int *)malloc(sizeof(int)); get2NonRepeatingNos(arr, 8, x, y); printf("The non-repeating elements are %d and %d", *x, *y); getchar();} |
Java
//Java Program for above approachpublic class UniqueNumbers { // This function sets the values of // *x and *y to non-repeating elements // in an array arr[] of size n public static void UniqueNumbers2(int[] arr,int n) { int sum =0; for(int i = 0;i<n;i++) { // Xor all the elements of the array // all the elements occuring twice will // cancel out each other remaining // two unnique numbers will be xored sum = (sum^arr[i]); } // Bitwise & the sum with it's 2's Complement // Bitwise & will give us the sum containing // only the rightmost set bit sum = (sum&-sum); // sum1 and sum2 will contains 2 unique // elements elements initialized with 0 box // number xored with 0 is number itself int sum1 = 0; int sum2 = 0; // traversing the array again for(int i = 0;i<arr.length;i++) { // Bitwise & the arr[i] with the sum // Two possibilities either result == 0 // or result > 0 if((arr[i]&sum) > 0) { // if result > 0 then arr[i] xored // with the sum1 sum1 = (sum1^arr[i]); } else { // if result == 0 then arr[i] // xored with sum2 sum2 = (sum2^arr[i]); } } // print the the two unique numbers System.out.println("The non-repeating elements are "+ sum1+" and "+sum2); } public static void main(String[] args) { int[] arr = new int[]{2, 3, 7, 9, 11, 2, 3, 11}; int n = arr.length; UniqueNumbers2(arr,n); }}// This code is contributed by Parshav Nahta |
Python3
# Python3 program for above approach# This function sets the values of# *x and *y to non-repeating elements# in an array arr[] of size ndef UniqueNumbers2(arr, n): sums = 0 for i in range(0, n): # Xor all the elements of the array # all the elements occuring twice will # cancel out each other remaining # two unnique numbers will be xored sums = (sums ^ arr[i]) # Bitwise & the sum with it's 2's Complement # Bitwise & will give us the sum containing # only the rightmost set bit sums = (sums & -sums) # sum1 and sum2 will contains 2 unique # elements elements initialized with 0 box # number xored with 0 is number itself sum1 = 0 sum2 = 0 # Traversing the array again for i in range(0, len(arr)): # Bitwise & the arr[i] with the sum # Two possibilities either result == 0 # or result > 0 if (arr[i] & sums) > 0: # If result > 0 then arr[i] xored # with the sum1 sum1 = (sum1 ^ arr[i]) else: # If result == 0 then arr[i] # xored with sum2 sum2 = (sum2 ^ arr[i]) # Print the the two unique numbers print("The non-repeating elements are ", sum1 ," and ", sum2)# Driver Codeif __name__ == "__main__": arr = [ 2, 3, 7, 9, 11, 2, 3, 11 ] n = len(arr) UniqueNumbers2(arr, n)# This code is contributed by akhilsaini |
C#
// C# program for above approachusing System;class GFG{// This function sets the values of// *x and *y to non-repeating elements// in an array arr[] of size nstatic void UniqueNumbers2(int[] arr, int n){ int sum = 0; for(int i = 0; i < n; i++) { // Xor all the elements of the array // all the elements occuring twice will // cancel out each other remaining // two unnique numbers will be xored sum = (sum ^ arr[i]); } // Bitwise & the sum with it's 2's Complement // Bitwise & will give us the sum containing // only the rightmost set bit sum = (sum & -sum); // sum1 and sum2 will contains 2 unique // elements elements initialized with 0 box // number xored with 0 is number itself int sum1 = 0; int sum2 = 0; // Traversing the array again for(int i = 0; i < arr.Length; i++) { // Bitwise & the arr[i] with the sum // Two possibilities either result == 0 // or result > 0 if ((arr[i] & sum) > 0) { // If result > 0 then arr[i] xored // with the sum1 sum1 = (sum1 ^ arr[i]); } else { // If result == 0 then arr[i] // xored with sum2 sum2 = (sum2 ^ arr[i]); } } // Print the the two unique numbers Console.WriteLine("The non-repeating " + "elements are " + sum1 + " and " + sum2);}// Driver Codestatic public void Main(){ int[] arr = { 2, 3, 7, 9, 11, 2, 3, 11 }; int n = arr.Length; UniqueNumbers2(arr, n);}}// This code is contributed by akhilsaini |
Output:
The non-repeating elements are 7 and 9
Time Complexity: O(n)
Auxiliary Space: O(1)
Please refer below post for detailed explanation :
Find the two numbers with odd occurrences in an unsorted array
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