Find the missing and repeating number
Given an unsorted array of size n. Array elements are in the range of 1 to n. One number from set {1, 2, …n} is missing and one number occurs twice in the array. Find these two numbers.
Examples:
Input: arr[] = {3, 1, 3}
Output: Missing = 2, Repeating = 3
Explanation: In the array, 2 is missing and 3 occurs twiceInput: arr[] = {4, 3, 6, 2, 1, 1}
Output: Missing = 5, Repeating = 1
Below are various methods to solve the problems:
Method 1 (Use Sorting)
Approach:
- Sort the input array.
- Traverse the array and check for missing and repeating.
Time Complexity: O(nLogn)
Thanks to LoneShadow for suggesting this method.
Method 2 (Use count array)
Approach:
- Create a temp array temp[] of size n with all initial values as 0.
- Traverse the input array arr[], and do following for each arr[i]
- if(temp[arr[i]] == 0) temp[arr[i]] = 1;
- if(temp[arr[i]] == 1) output “arr[i]” //repeating
- Traverse temp[] and output the array element having value as 0 (This is the missing element)
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 3 (Use elements as Index and mark the visited places)
Approach:
Traverse the array. While traversing, use the absolute value of every element as an index and make the value at this index as negative to mark it visited. If something is already marked negative then this is the repeating element. To find missing, traverse the array again and look for a positive value.
C++
// C++ program to Find the repeating// and missing elements#include <bits/stdc++.h>using namespace std;void printTwoElements(int arr[], int size){ int i; cout << "The repeating element is "; for (i = 0; i < size; i++) { if (arr[abs(arr[i]) - 1] > 0) arr[abs(arr[i]) - 1] = -arr[abs(arr[i]) - 1]; else cout << abs(arr[i]) << "\n"; } cout << "and the missing element is "; for (i = 0; i < size; i++) { if (arr[i] > 0) cout << (i + 1); }}/* Driver code */int main(){ int arr[] = { 7, 3, 4, 5, 5, 6, 2 }; int n = sizeof(arr) / sizeof(arr[0]); printTwoElements(arr, n);}// This code is contributed by Shivi_Aggarwal |
C
// C program to Find the repeating// and missing elements#include <stdio.h>#include <stdlib.h>void printTwoElements(int arr[], int size){ int i; printf("\nThe repeating element is "); for (i = 0; i < size; i++) { if (arr[abs(arr[i]) - 1] > 0) arr[abs(arr[i]) - 1] = -arr[abs(arr[i]) - 1]; else printf(" %d ", abs(arr[i])); } printf("\nand the missing element is "); for (i = 0; i < size; i++) { if (arr[i] > 0) printf("%d", i + 1); }}// Driver codeint main(){ int arr[] = { 7, 3, 4, 5, 5, 6, 2 }; int n = sizeof(arr) / sizeof(arr[0]); printTwoElements(arr, n); return 0;} |
Java
// Java program to Find the repeating// and missing elementsimport java.io.*;class GFG { static void printTwoElements(int arr[], int size) { int i; System.out.print("The repeating element is "); for (i = 0; i < size; i++) { int abs_val = Math.abs(arr[i]); if (arr[abs_val - 1] > 0) arr[abs_val - 1] = -arr[abs_val - 1]; else System.out.println(abs_val); } System.out.print("and the missing element is "); for (i = 0; i < size; i++) { if (arr[i] > 0) System.out.println(i + 1); } } // Driver code public static void main(String[] args) { int arr[] = { 7, 3, 4, 5, 5, 6, 2 }; int n = arr.length; printTwoElements(arr, n); }}// This code is contributed by Gitanjali |
Python3
# Python3 code to Find the repeating # and the missing elementsdef printTwoElements( arr, size): for i in range(size): if arr[abs(arr[i])-1] > 0: arr[abs(arr[i])-1] = -arr[abs(arr[i])-1] else: print("The repeating element is ", abs(arr[i])) for i in range(size): if arr[i]>0: print("and the missing element is ", i + 1)# Driver program to test above function */arr = [7, 3, 4, 5, 5, 6, 2]n = len(arr)printTwoElements(arr, n)# This code is contributed by "Abhishek Sharma 44" |
C#
// C# program to Find the repeating// and missing elementsusing System;class GFG { static void printTwoElements(int[] arr, int size) { int i; Console.Write("The repeating element is "); for (i = 0; i < size; i++) { int abs_val = Math.Abs(arr[i]); if (arr[abs_val - 1] > 0) arr[abs_val - 1] = -arr[abs_val - 1]; else Console.WriteLine(abs_val); } Console.Write("and the missing element is "); for (i = 0; i < size; i++) { if (arr[i] > 0) Console.WriteLine(i + 1); } } // Driver program public static void Main() { int[] arr = { 7, 3, 4, 5, 5, 6, 2 }; int n = arr.Length; printTwoElements(arr, n); }}// This code is contributed by Sam007 |
PHP
<?php// PHP program to Find the repeating// and missing elementsfunction printTwoElements($arr, $size){ $i; echo "The repeating element is", " "; for($i = 0; $i < $size; $i++) { if($arr[abs($arr[$i]) - 1] > 0) $arr[abs($arr[$i]) - 1] = - $arr[abs($arr[$i]) - 1]; else echo ( abs($arr[$i])); } echo "\nand the missing element is "; for($i = 0; $i < $size; $i++) { if($arr[$i] > 0) echo($i + 1); }} // Driver Code $arr = array(7, 3, 4, 5, 5, 6, 2); $n = count($arr); printTwoElements($arr, $n);// This code is contributed by anuj_67.?> |
Javascript
<script>// Program to Find the repeating// and missing elements function printTwoElements(arr,size) { var i; document.write("The repeating element is "); for (i = 0; i < size; i++) { var abs_value = Math.abs(arr[i]); if (arr[abs_value - 1] > 0) arr[abs_value - 1] = -arr[abs_value - 1]; else document.write( abs_value); } document.write("<br> and the missing element is "); for (i = 0; i < size; i++) { if (arr[i] > 0) document.write (i + 1); }} /* Driver code */ arr = new Array ( 7, 3, 4, 5, 5, 6, 2 ); n = arr.length; printTwoElements(arr, n); // This code is contributed by simranarora5sos</script> |
Output
The repeating element is 5 and the missing element is 1
Time Complexity: O(n)
Thanks to Manish Mishra for suggesting this method.
Method 4 (Make two equations)
Approach:
- Let x be the missing and y be the repeating element.
- Get the sum of all numbers using formula S = n(n+1)/2 – x + y
- Get product of all numbers using formula P = 1*2*3*…*n * y / x
- The above two steps give us two equations, we can solve the equations and get the values of x and y.
Time Complexity: O(n)
Thanks to disappearedng for suggesting this solution.
Note: This method can cause arithmetic overflow as we calculate product and sum of all array elements.
Method 5 (Use XOR)
Approach:
- Let x and y be the desired output elements.
- Calculate XOR of all the array elements.
xor1 = arr[0]^arr[1]^arr[2]…..arr[n-1]
- XOR the result with all numbers from 1 to n
xor1 = xor1^1^2^…..^n
- In the result xor1, all elements would nullify each other except x and y. All the bits that are set in xor1 will be set in either x or y. So if we take any set bit (We have chosen the rightmost set bit in code) of xor1 and divide the elements of the array in two sets – one set of elements with the same bit set and other 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 first set, we will get x, and by doing same in other set we will get y.
Below is the implementation of the above approach:
C++
// C++ program to Find the repeating// and missing elements#include <bits/stdc++.h>using namespace std;/* The output of this function is stored at*x and *y */void getTwoElements(int arr[], int n, int* x, int* y){ /* Will hold xor of all elements and numbers from 1 to n */ int xor1; /* Will have only single set bit of xor1 */ int set_bit_no; int i; *x = 0; *y = 0; xor1 = arr[0]; /* Get the xor of all array elements */ for (i = 1; i < n; i++) xor1 = xor1 ^ arr[i]; /* XOR the previous result with numbers from 1 to n*/ for (i = 1; i <= n; i++) xor1 = xor1 ^ i; /* Get the rightmost set bit in set_bit_no */ set_bit_no = xor1 & ~(xor1 - 1); /* Now divide elements into two sets by comparing a rightmost set bit of xor1 with the bit at the same position in each element. Also, get XORs of two sets. The two XORs are the output elements. The following two for loops serve the purpose */ for (i = 0; i < n; i++) { if (arr[i] & set_bit_no) /* arr[i] belongs to first set */ *x = *x ^ arr[i]; else /* arr[i] belongs to second set*/ *y = *y ^ arr[i]; } for (i = 1; i <= n; i++) { if (i & set_bit_no) /* i belongs to first set */ *x = *x ^ i; else /* i belongs to second set*/ *y = *y ^ i; } /* *x and *y hold the desired output elements */}/* Driver code */int main(){ int arr[] = { 1, 3, 4, 5, 5, 6, 2 }; int* x = (int*)malloc(sizeof(int)); int* y = (int*)malloc(sizeof(int)); int n = sizeof(arr) / sizeof(arr[0]); getTwoElements(arr, n, x, y); cout << " The missing element is " << *x << " and the repeating" << " number is " << *y; getchar();}// This code is contributed by Code_Mech |
C
// C program to Find the repeating// and missing elements#include <stdio.h>#include <stdlib.h>/* The output of this function is stored at *x and *y */void getTwoElements(int arr[], int n, int* x, int* y){ /* Will hold xor of all elements and numbers from 1 to n */ int xor1; /* Will have only single set bit of xor1 */ int set_bit_no; int i; *x = 0; *y = 0; xor1 = arr[0]; /* Get the xor of all array elements */ for (i = 1; i < n; i++) xor1 = xor1 ^ arr[i]; /* XOR the previous result with numbers from 1 to n*/ for (i = 1; i <= n; i++) xor1 = xor1 ^ i; /* Get the rightmost set bit in set_bit_no */ set_bit_no = xor1 & ~(xor1 - 1); /* Now divide elements in two sets by comparing rightmost set bit of xor1 with bit at same position in each element. Also, get XORs of two sets. The two XORs are the output elements. The following two for loops serve the purpose */ for (i = 0; i < n; i++) { if (arr[i] & set_bit_no) /* arr[i] belongs to first set */ *x = *x ^ arr[i]; else /* arr[i] belongs to second set*/ *y = *y ^ arr[i]; } for (i = 1; i <= n; i++) { if (i & set_bit_no) /* i belongs to first set */ *x = *x ^ i; else /* i belongs to second set*/ *y = *y ^ i; } /* *x and *y hold the desired output elements */}/* Driver program to test above function */int main(){ int arr[] = { 1, 3, 4, 5, 5, 6, 2 }; int* x = (int*)malloc(sizeof(int)); int* y = (int*)malloc(sizeof(int)); int n = sizeof(arr) / sizeof(arr[0]); getTwoElements(arr, n, x, y); printf(" The missing element is %d" " and the repeating number" " is %d", *x, *y); getchar();} |
Java
// Java program to Find the repeating// and missing elementsimport java.io.*;class GFG { static int x, y; static void getTwoElements(int arr[], int n) { /* Will hold xor of all elements and numbers from 1 to n */ int xor1; /* Will have only single set bit of xor1 */ int set_bit_no; int i; x = 0; y = 0; xor1 = arr[0]; /* Get the xor of all array elements */ for (i = 1; i < n; i++) xor1 = xor1 ^ arr[i]; /* XOR the previous result with numbers from 1 to n*/ for (i = 1; i <= n; i++) xor1 = xor1 ^ i; /* Get the rightmost set bit in set_bit_no */ set_bit_no = xor1 & ~(xor1 - 1); /* Now divide elements into two sets by comparing rightmost set bit of xor1 with the bit at the same position in each element. Also, get XORs of two sets. The two XORs are the output elements. The following two for loops serve the purpose */ for (i = 0; i < n; i++) { if ((arr[i] & set_bit_no) != 0) /* arr[i] belongs to first set */ x = x ^ arr[i]; else /* arr[i] belongs to second set*/ y = y ^ arr[i]; } for (i = 1; i <= n; i++) { if ((i & set_bit_no) != 0) /* i belongs to first set */ x = x ^ i; else /* i belongs to second set*/ y = y ^ i; } /* *x and *y hold the desired output elements */ } /* Driver program to test above function */ public static void main(String[] args) { int arr[] = { 1, 3, 4, 5, 1, 6, 2 }; int n = arr.length; getTwoElements(arr, n); System.out.println(" The missing element is " + x + "and the " + "repeating number is " + y); }}// This code is contributed by Gitanjali. |
Python3
# Python3 program to find the repeating # and missing elements # The output of this function is stored # at x and y def getTwoElements(arr, n): global x, y x = 0 y = 0 # Will hold xor of all elements # and numbers from 1 to n xor1 = arr[0] # Get the xor of all array elements for i in range(1, n): xor1 = xor1 ^ arr[i] # XOR the previous result with numbers # from 1 to n for i in range(1, n + 1): xor1 = xor1 ^ i # Will have only single set bit of xor1 set_bit_no = xor1 & ~(xor1 - 1) # Now divide elements into two # sets by comparing a rightmost set # bit of xor1 with the bit at the same # position in each element. Also, # get XORs of two sets. The two # XORs are the output elements. # The following two for loops # serve the purpose for i in range(n): if (arr[i] & set_bit_no) != 0: # arr[i] belongs to first set x = x ^ arr[i] else: # arr[i] belongs to second set y = y ^ arr[i] for i in range(1, n + 1): if (i & set_bit_no) != 0: # i belongs to first set x = x ^ i else: # i belongs to second set y = y ^ i # x and y hold the desired # output elements # Driver codearr = [ 1, 3, 4, 5, 5, 6, 2 ]n = len(arr) getTwoElements(arr, n)print("The missing element is", x, "and the repeating number is", y) # This code is contributed by stutipathak31jan |
C#
// C# program to Find the repeating// and missing elementsusing System;class GFG { static int x, y; static void getTwoElements(int[] arr, int n) { /* Will hold xor of all elements and numbers from 1 to n */ int xor1; /* Will have only single set bit of xor1 */ int set_bit_no; int i; x = 0; y = 0; xor1 = arr[0]; /* Get the xor of all array elements */ for (i = 1; i < n; i++) xor1 = xor1 ^ arr[i]; /* XOR the previous result with numbers from 1 to n*/ for (i = 1; i <= n; i++) xor1 = xor1 ^ i; /* Get the rightmost set bit in set_bit_no */ set_bit_no = xor1 & ~(xor1 - 1); /* Now divide elements in two sets by comparing rightmost set bit of xor1 with bit at same position in each element. Also, get XORs of two sets. The two XORs are the output elements.The following two for loops serve the purpose */ for (i = 0; i < n; i++) { if ((arr[i] & set_bit_no) != 0) /* arr[i] belongs to first set */ x = x ^ arr[i]; else /* arr[i] belongs to second set*/ y = y ^ arr[i]; } for (i = 1; i <= n; i++) { if ((i & set_bit_no) != 0) /* i belongs to first set */ x = x ^ i; else /* i belongs to second set*/ y = y ^ i; } /* *x and *y hold the desired output elements */ } // Driver program public static void Main() { int[] arr = { 1, 3, 4, 5, 1, 6, 2 }; int n = arr.Length; getTwoElements(arr, n); Console.Write(" The missing element is " + x + "and the " + "repeating number is " + y); }}// This code is contributed by Sam007 |
PHP
<?php// PHP program to Find the repeating// and missing elementsfunction getTwoElements(&$arr, $n){ /* Will hold xor of all elements and numbers from 1 to n */ $xor1; /* Will have only single set bit of xor1 */ $set_bit_no; $i; $x = 0; $y = 0; $xor1 = $arr[0]; /* Get the xor of all array elements */ for ($i = 1; $i < $n; $i++) $xor1 = $xor1 ^ $arr[$i]; /* XOR the previous result with numbers from 1 to n*/ for ($i = 1; $i <= $n; $i++) $xor1 = $xor1 ^ $i; /* Get the rightmost set bit in set_bit_no */ $set_bit_no = $xor1 & ~($xor1 - 1); /* Now divide elements in two sets by comparing rightmost set bit of xor1 with bit at same position in each element. Also, get XORs of two sets. The two XORs are the output elements.The following two for loops serve the purpose */ for ($i = 0; $i < $n; $i++) { if (($arr[$i] & $set_bit_no) != 0) /* arr[i] belongs to first set */ $x = $x ^ $arr[$i]; else /* arr[i] belongs to second set*/ $y = $y ^ $arr[$i]; } for ($i = 1; $i <= $n; $i++) { if (($i & $set_bit_no) != 0) /* i belongs to first set */ $x = $x ^ $i; else /* i belongs to second set*/ $y = $y ^ $i; } /* *x and *y hold the desired output elements */ echo("The missing element is " . $x . " and the repeating number is " . $y);}// Driver Code$arr = array( 1, 3, 4, 5, 1, 6, 2 );$n = sizeof($arr);getTwoElements($arr, $n);// This code is contributed by Code_Mech |
Output
The missing element is 7 and the repeating number is 5
Time Complexity: O(n)
This method doesn’t cause overflow, but it doesn’t tell which one occurs twice and which one is missing. We can add one more step that checks which one is missing and which one is repeating. This can be easily done in O(n) time.
Method 6 (Use a Map)
Approach:
This method involves creating a Hashtable with the help of Map. In this, the elements are mapped to their natural index. In this process, if an element is mapped twice, then it is the repeating element. And if an element’s mapping is not there, then it is the missing element.
Below is the implementation of the above approach:
C++
// C++ program to find the repeating// and missing elements using Maps #include <iostream>#include <unordered_map>using namespace std;int main(){ int arr[] = { 4, 3, 6, 2, 1, 1 }; int n = 6; unordered_map<int, bool> numberMap; for(int i : arr) { if (numberMap.find(i) == numberMap.end()) { numberMap[i] = true; } else { cout << "Repeating = " << i; } } cout << endl; for(int i = 1; i <= n; i++) { if (numberMap.find(i) == numberMap.end()) { cout << "Missing = " << i; } } return 0;}// This code is contributed by RohitOberoi |
Java
// Java program to find the// repeating and missing elements// using Mapsimport java.util.*;public class Test1 { public static void main(String[] args) { int[] arr = { 4, 3, 6, 2, 1, 1 }; Map<Integer, Boolean> numberMap = new HashMap<>(); int max = arr.length; for (Integer i : arr) { if (numberMap.get(i) == null) { numberMap.put(i, true); } else { System.out.println("Repeating = " + i); } } for (int i = 1; i <= max; i++) { if (numberMap.get(i) == null) { System.out.println("Missing = " + i); } } }} |
Python3
# Python3 program to find the # repeating and missing elements # using Mapsdef main(): arr = [ 4, 3, 6, 2, 1, 1 ] numberMap = {} max = len(arr) for i in arr: if not i in numberMap: numberMap[i] = True else: print("Repeating =", i) for i in range(1, max + 1): if not i in numberMap: print("Missing =", i)main()# This code is contributed by stutipathak31jan |
C#
// C# program to find the// repeating and missing elements// using Mapsusing System;using System.Collections.Generic;class GFG{ public static void Main(String[] args) { int[] arr = { 4, 3, 6, 2, 1, 1 }; Dictionary<int, Boolean> numberMap = new Dictionary<int, Boolean>(); int max = arr.Length; foreach (int i in arr) { if (!numberMap.ContainsKey(i)) { numberMap.Add(i, true); } else { Console.WriteLine("Repeating = " + i); } } for (int i = 1; i <= max; i++) { if (!numberMap.ContainsKey(i)) { Console.WriteLine("Missing = " + i); } } }}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// JavaScript program to find the repeating// and missing elements using Maps // driver programlet arr = [ 4, 3, 6, 2, 1, 1 ];let n = 6; let numberMap = new Map(); for(let i of arr) { if (numberMap.has(i) == false) { numberMap.set(i,true); } else { document.write("Repeating = ",i); }} document.write("</br>"); for(let i = 1; i <= n; i++){ if (numberMap.has(i) == false) { document.write("Missing = " ,i); }}// This code is contributed by Shinjanpatra</script> |
Output
Repeating = 1 Missing = 5
Method 7 (Make two equations using sum and sum of squares)
Approach:
- Let x be the missing and y be the repeating element.
- Let N is the size of array.
- Get the sum of all numbers using formula S = N(N+1)/2
- Get the sum of square of all numbers using formula Sum_Sq = N(N+1)(2N+1)/6
- Iterate through a loop from i=1….N
- S -= A[i]
- Sum_Sq -= (A[i]*A[i])
- It will give two equations
x-y = S – (1)
x^2 – y^2 = Sum_sq
x+ y = (Sum_sq/S) – (2)
Time Complexity: O(n)
C++
#include <bits/stdc++.h> using namespace std;vector<int>repeatedNumber(const vector<int> &A) { long long int len = A.size(); long long int Sum_N = (len * (len+1) ) /2, Sum_NSq = (len * (len +1) *(2*len +1) )/6; long long int missingNumber=0, repeating=0; for(int i=0;i<A.size(); i++){ Sum_N -= (long long int)A[i]; Sum_NSq -= (long long int)A[i]*(long long int)A[i]; } missingNumber = (Sum_N + Sum_NSq/Sum_N)/2; repeating= missingNumber - Sum_N; vector <int> ans; ans.push_back(repeating); ans.push_back(missingNumber); return ans; }int main(void){ std::vector<int> v = {4, 3, 6, 2, 1, 6,7}; vector<int> res = repeatedNumber(v); for(int x: res){ cout<< x<<" "; } cout<<endl;} |
Java
import java.util.*;import java.math.BigInteger;class GFG { static Vector<Integer> repeatedNumber(int[] a) { BigInteger n=BigInteger.valueOf(a.length); BigInteger s=BigInteger.valueOf(0); BigInteger ss=BigInteger.valueOf(0); for(int x : a) { s= s.add(BigInteger.valueOf(x)); ss= ss.add(BigInteger.valueOf(x).multiply(BigInteger.valueOf(x))); } BigInteger as= n.multiply(n.add(BigInteger.valueOf(1))).divide(BigInteger.valueOf(2)); BigInteger ass= as.multiply(BigInteger.valueOf(2).multiply(n).add(BigInteger.valueOf(1))).divide(BigInteger.valueOf(3)); BigInteger sub=as.subtract(s); BigInteger add=(ass.subtract(ss)).divide(sub); //(ass-ss)/sub; int b = sub.add(add).divide(BigInteger.valueOf(2)).intValue(); //(sub+add)/2; int A = BigInteger.valueOf(b).subtract(sub).intValue(); Vector<Integer> ans = new Vector<>(); ans.add(A); ans.add(b); return ans; } // Driver Code public static void main(String[] args) { int[] v = { 4, 3, 6, 2, 1, 6, 7 }; Vector<Integer> res = repeatedNumber(v); for (int x : res) { System.out.print(x + " "); } }}// This code is contributed by Rajput-Ji |
Python3
def repeatedNumber(A): length = len(A) Sum_N = (length * (length + 1)) // 2 Sum_NSq = ((length * (length + 1) * (2 * length + 1)) // 6) missingNumber, repeating = 0, 0 for i in range(len(A)): Sum_N -= A[i] Sum_NSq -= A[i] * A[i] missingNumber = (Sum_N + Sum_NSq // Sum_N) // 2 repeating = missingNumber - Sum_N ans = [] ans.append(repeating) ans.append(missingNumber) return ans# Driver codev = [ 4, 3, 6, 2, 1, 6, 7 ]res = repeatedNumber(v)for i in res: print(i, end = " ")# This code is contributed by stutipathak31jan |
C#
using System;using System.Collections.Generic;class GFG { static List<int> repeatedNumber(int[] A) { int len = A.Length; int Sum_N = (len * (len + 1)) / 2; int Sum_NSq = (len * (len + 1) * (2 * len + 1)) / 6; int missingNumber = 0, repeating = 0; for (int i = 0; i < A.Length; i++) { Sum_N -= A[i]; Sum_NSq -= A[i] * A[i]; } missingNumber = (Sum_N + Sum_NSq / Sum_N) / 2; repeating = missingNumber - Sum_N; List<int> ans = new List<int>(); ans.Add(repeating); ans.Add(missingNumber); return ans; } // Driver Code public static void Main(String[] args) { int[] v = { 4, 3, 6, 2, 1, 6, 7 }; List<int> res = repeatedNumber(v); foreach (int x in res) { Console.Write(x + " "); } }}// This code is contributed by PrinciRaj1992 |
Javascript
<script>function repeatedNumber(A){ let length = A.length let Sum_N = Math.floor((length * (length + 1)) / 2) let Sum_NSq = Math.floor((length * (length + 1) * (2 * length + 1))/6) let missingNumber = 0 let repeating = 0 for(let i=0;i<A.length;i++){ Sum_N -= A[i] Sum_NSq -= A[i] * A[i] } missingNumber = Math.floor(Math.floor(Sum_N + Sum_NSq / Sum_N) / 2) repeating = missingNumber - Sum_N let ans = [] ans.push(repeating) ans.push(missingNumber)return ans}// Driver codelet v = [ 4, 3, 6, 2, 1, 6, 7 ]let res = repeatedNumber(v)for(let i of res) document.write(i," ")// This code is contributed by shinjanpatra</script> |
Output
6 5
Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.
Method 8 (Using OR Operator):
Thanks to Anish Shaha for suggesting this method.
Approach:
Given an input array
- Performing OR operation on input array.
- At the same time checking if that number has occurred before, by determining if the position is already set or not. We will get the repeating number in this step.
- To find missing value we have to check the bit containing 0 using OR again.
C++
#include <bits/stdc++.h>using namespace std;int main(){ // Input: vector<int> arr = {4, 3, 6, 2, 1, 1}; int n = arr.size(); // Declaring output variables // Note : arr[i]-1 is used instead of arr[i] as we want to use all 64 bits int bitOr = (1 << (arr[0]-1)); int repeating, missing; // Performing XOR as well as Checking repeating number for(int i=1; i<n; i++){ // If OR operation with 1 gives same output that means, we already have 1 at that position if((bitOr | (1 << (arr[i]-1))) == bitOr) { repeating = arr[i]; continue; } bitOr = (bitOr | (1 << (arr[i]-1))); } // Checking missing number for(int i=0; i<n; i++){ // property: OR with 0 yield 1 hence value of bitOr changes if((bitOr | (1 << i)) != bitOr) { missing = i+1; break; } } cout << "Repeating : " << repeating << "\nMissing : " << missing; return 0;} |
Java
import java.util.*;class GFG{ public static void main(String[] args) { // Input: int[] arr = {4, 3, 6, 2, 1, 1}; int n = arr.length; // Declaring output variables // Note : arr[i]-1 is used instead of arr[i] as we want to use all 64 bits int bitOr = (1 << (arr[0]-1)); int repeating = 0, missing = 0; // Performing XOR as well as Checking repeating number for(int i=1; i<n; i++){ // If OR operation with 1 gives same output that means, we already have 1 at that position if((bitOr | (1 << (arr[i]-1))) == bitOr) { repeating = arr[i]; continue; } bitOr = (bitOr | (1 << (arr[i]-1))); } // Checking missing number for(int i = 0; i < n; i++) { // property: OR with 0 yield 1 hence value of bitOr changes if((bitOr | (1 << i)) != bitOr) { missing = i+1; break; } } System.out.print("Repeating : " + repeating+ "\nMissing : " + missing); }}// This code is contributed by Rajput-Ji |
C#
using System;public class GFG{ public static void Main(String[] args) { // Input: int[] arr = { 4, 3, 6, 2, 1, 1 }; int n = arr.Length; // Declaring output variables // Note : arr[i]-1 is used instead of arr[i] as we want to use all 64 bits int bitOr = (1 << (arr[0] - 1)); int repeating = 0, missing = 0; // Performing XOR as well as Checking repeating number for (int i = 1; i < n; i++) { // If OR operation with 1 gives same output that means, we already have 1 at // that position if ((bitOr | (1 << (arr[i] - 1))) == bitOr) { repeating = arr[i]; continue; } bitOr = (bitOr | (1 << (arr[i] - 1))); } // Checking missing number for (int i = 0; i < n; i++) { // property: OR with 0 yield 1 hence value of bitOr changes if ((bitOr | (1 << i)) != bitOr) { missing = i + 1; break; } } Console.Write("Repeating : " + repeating + "\nMissing : " + missing); }}// This code is contributed by Rajput-Ji |
Javascript
<script>// JavaScript code for the above approach// Driver Code // Input: let arr = [4, 3, 6, 2, 1, 1]; let n = arr.length; // Declaring output variables // Note : arr[i]-1 is used instead of arr[i] as we want to use all 64 bits let bitOr = (1 << (arr[0]-1)); let repeating = 0, missing = 0; // Performing XOR as well as Checking repeating number for(let i=1; i<n; i++){ // If OR operation with 1 gives same output that means, we already have 1 at that position if((bitOr | (1 << (arr[i]-1))) == bitOr) { repeating = arr[i]; continue; } bitOr = (bitOr | (1 << (arr[i]-1))); } // Checking missing number for(let i = 0; i < n; i++) { // property: OR with 0 yield 1 hence value of bitOr changes if((bitOr | (1 << i)) != bitOr) { missing = i+1; break; } } document.write("Repeating : " + repeating+ "<br/>" + "Missing : " + missing);// This code is contributed by code_hunt.</script> |
Output
Repeating : 1 Missing : 5
Time Complexity: O(n)
Auxiliary Space O(1)
Limitations of the approach: it only works on size of array <= 64 if we use long and size of array <= 32
Method 9: (Placing every element in its correct position)
Approach: It is clear from the observation that if we sort an array, then arr[i] == i+1. If all elements in an array satisfy this condition, means this is an ideal case. So the idea is to sort the given array and traverse on it and check if arr[i] == i + 1 if so then increment i (because this element is at its correct position), otherwise place this element (arr[i]) at its correct position (arr[arr[i] – 1) by swapping the arr[i] and arr[arr[i] -1]. This swapping will put the element arr[i] at its correct position (i.e arr[arr[i]-1]). After doing this operation gets over for all elements of given array then again travese over the array and check if arr[i] != i + 1 if so, then this is the duplicate element and i + 1 is the missing element.
Below is the implementation of the above approach.
C++
// C++ program to find the missing// and repeating element#include <bits/stdc++.h>using namespace std;void getTwoElements(int arr[], int n){ int repeating(0), missing(0); int i = 0; // Traverse on the array while (i < n) { // If the element is on its correct position if (arr[i] == arr[arr[i] - 1]) i++; else { // If it is not at its correct position // then palce it to its correct position if(arr[i] != arr[arr[i] - 1]) swap(arr[i], arr[arr[i] - 1]); else i++; } } // Find repeating and missing for (int i = 0; i < n; i++) { // If any element is not in its correct position if (arr[i] != i + 1) { repeating = arr[i]; missing = i + 1; break; } } // Print answer cout << "Repeating: " << repeating << endl << "Missing: " << missing << endl;}// Driver codeint main(){ int arr[] = { 2, 3, 1, 5, 1 }; int n = sizeof(arr) / sizeof(int); getTwoElements(arr, n); return 0;}// This code is contributed by Tapesh (tapeshdua420) |
Output
Repeating: 1 Missing: 4
Time Complexity: O(n)
Auxiliary Space: O(1)
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