- Given an array containing n-1 distinct positive numbers range from 1 to n.
- e.g suppose n = 5, array will contain 4 unique (or no duplicate) numbers.
- some of the possible array values are
- int elements[] = { 1, 3, 2, 5 }
- int elements1[] = { 1, 2, 4, 5 }
- int elements2[] = { 4, 3, 2, 1 }
- We would like to find out the missing number in an array.
- We will discuss couple of methods to find out the missing number
Method 1: find a missing number in an integer array using Sum (java)
- Calculate the sum of n numbers ( 1 to n).
- Sum = n * (n + 1) /2, say sumOfNumbers
- Calculate the sum of all elements of input array, say arraySum
- MissingNumber = sumOfNumbers – arraySum
Program: find a missing number using sum formula in java
private static int missingNumberUsingSum(int[] elements) {
int arraySum = 0;
// Sum of array
for (int element : elements) {
arraySum += element;
}
// sum of number from 1 to n
int n = elements.length + 1; // as one number is missing
int sumOfNumbers = (n * (n + 1)) / 2;
return sumOfNumbers - arraySum;
}
Method 2: find a missing number in an integer array using XOR in java
- We will use the properties of XOR gate.
- Truth table of “Exclusive-OR” or XOR gate is :
X |
Y |
Output |
0 |
0 |
0 |
0 |
1 |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
- Let us apply XOR gate on elements[] = { 1, 3, 2, 5 }
- Calculate XOR of n numbers i.e. 1 to 5 and elements of array
- Missing Number = (1 ^ 2 ^ 3 ^ 4 ^ 5) ^ (1 ^ 3 ^ 2 ^ 5)
- We know XOR (P,P) = 0, let us rearrange same numbers
- i.e. (1 ^ 1) ^ ( 2 ^ 2) ^ (3 ^ 3) ^ (4) ^ (5 ^ 5)
- MissingNumber will be 4.
Program: find a missing number in an integer array using XOR in java
private static int missingNumberUsingXOR(int[] elements) {
int n = elements.length + 1;
int missingNumber = elements[0];
for (int index = 1; index <= n; index++) {
if (index < elements.length) {
missingNumber ^= elements[index];
}
missingNumber ^= index;
}
return missingNumber;
}
Complete program: find a missing number in an integer array (Sum & XOR)
package org.learn.arrays;
import java.util.Arrays;
public class MissingNumber {
public static void main(String[] args) {
int elements[] = { 1, 6, 3, 2, 7, 5 };
String array = Arrays.toString(elements);
int missingNumber = missingNumberUsingSum(elements);
System.out.printf("1. Missing number in %s using sum is: %d", array, missingNumber);
missingNumber = missingNumberUsingXOR(elements);
System.out.printf("\n2. Missing number in %s using XOR is: %d", array, missingNumber);
int elements1[] = { 1, 2, 4, 5, 6, 7 };
array = Arrays.toString(elements1);
missingNumber = missingNumberUsingSum(elements1);
System.out.printf("\n3. Missing number in %s using sum is: %d", array, missingNumber);
missingNumber = missingNumberUsingXOR(elements1);
System.out.printf("\n4. Missing number in %s using XOR is: %d", array, missingNumber);
int elements2[] = { 7, 6, 4, 3, 2, 1 };
array = Arrays.toString(elements2);
missingNumber = missingNumberUsingSum(elements2);
System.out.printf("\n5. Missing number in %s using sum is: %d", array, missingNumber);
missingNumber = missingNumberUsingXOR(elements2);
System.out.printf("\n6. Missing number in %s using XOR is: %d", array, missingNumber);
}
private static int missingNumberUsingXOR(int[] elements) {
int n = elements.length + 1;
int missingNumber = elements[0];
for (int index = 1; index <= n; index++) {
if (index < elements.length) {
missingNumber ^= elements[index];
}
missingNumber ^= index;
}
return missingNumber;
}
private static int missingNumberUsingSum(int[] elements) {
int arraySum = 0;
// Sum of array
for (int element : elements) {
arraySum += element;
}
// sum of number from 1 to n
int n = elements.length + 1; // as one number is missing
int sumOfNumbers = (n * (n + 1)) / 2;
return sumOfNumbers - arraySum;
}
}
Output: find missing number in an array of distinct integers
1. Missing number in [1, 6, 3, 2, 7, 5] using sum is: 4
2. Missing number in [1, 6, 3, 2, 7, 5] using XOR is: 4
3. Missing number in [1, 2, 4, 5, 6, 7] using sum is: 3
4. Missing number in [1, 2, 4, 5, 6, 7] using XOR is: 3
5. Missing number in [7, 6, 4, 3, 2, 1] using sum is: 5
6. Missing number in [7, 6, 4, 3, 2, 1] using XOR is: 5