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Close Scanner to avoid resource leak (#5077)

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SOZEL
2024-03-13 01:49:58 +07:00
committed by GitHub
parent 47a9b1b647
commit ab371843ac
6 changed files with 255 additions and 237 deletions
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113
src/main/java/com/thealgorithms/maths/NonRepeatingElement.java
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@ -10,61 +10,70 @@ import java.util.Scanner;
public class NonRepeatingElement {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int i, res = 0;
System.out.println("Enter the number of elements in the array");
int n = sc.nextInt();
if ((n & 1) == 1) {
// Not allowing odd number of elements as we are expecting 2 non repeating numbers
System.out.println("Array should contain even number of elements");
return;
}
int[] arr = new int[n];
System.out.println("Enter " + n + " elements in the array. NOTE: Only 2 elements should not repeat");
for (i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
try {
sc.close();
} catch (Exception e) {
System.out.println("Unable to close scanner" + e);
}
// Find XOR of the 2 non repeating elements
for (i = 0; i < n; i++) {
res ^= arr[i];
}
// Finding the rightmost set bit
res = res & (-res);
int num1 = 0, num2 = 0;
for (i = 0; i < n; i++) {
if ((res & arr[i]) > 0) { // Case 1 explained below
num1 ^= arr[i];
} else {
num2 ^= arr[i]; // Case 2 explained below
try (Scanner sc = new Scanner(System.in)) {
int i, res = 0;
System.out.println("Enter the number of elements in the array");
int n = sc.nextInt();
if ((n & 1) == 1) {
// Not allowing odd number of elements as we are expecting 2 non repeating
// numbers
System.out.println("Array should contain even number of elements");
return;
}
}
int[] arr = new int[n];
System.out.println("The two non repeating elements are " + num1 + " and " + num2);
sc.close();
System.out.println("Enter " + n + " elements in the array. NOTE: Only 2 elements should not repeat");
for (i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
try {
sc.close();
} catch (Exception e) {
System.out.println("Unable to close scanner" + e);
}
// Find XOR of the 2 non repeating elements
for (i = 0; i < n; i++) {
res ^= arr[i];
}
// Finding the rightmost set bit
res = res & (-res);
int num1 = 0, num2 = 0;
for (i = 0; i < n; i++) {
if ((res & arr[i]) > 0) { // Case 1 explained below
num1 ^= arr[i];
} else {
num2 ^= arr[i]; // Case 2 explained below
}
}
System.out.println("The two non repeating elements are " + num1 + " and " + num2);
sc.close();
}
}
/*
Explanation of the code:
let us assume we have an array [1,2,1,2,3,4]
Property of XOR: num ^ num = 0.
If we XOR all the elemnets of the array we will be left with 3 ^ 4 as 1 ^ 1 and 2 ^ 2 would give
0. Our task is to find num1 and num2 from the result of 3 ^ 4 = 7. We need to find two's
complement of 7 and find the rightmost set bit. i.e. (num & (-num)) Two's complement of 7 is 001
and hence res = 1. There can be 2 options when we Bitise AND this res with all the elements in our
array
1. Result will come non zero number
2. Result will be 0.
In the first case we will XOR our element with the first number (which is initially 0)
In the second case we will XOR our element with the second number(which is initially 0)
This is how we will get non repeating elements with the help of bitwise operators.
* Explanation of the code:
* let us assume we have an array [1,2,1,2,3,4]
* Property of XOR: num ^ num = 0.
* If we XOR all the elemnets of the array we will be left with 3 ^ 4 as 1 ^ 1
* and 2 ^ 2 would give
* 0. Our task is to find num1 and num2 from the result of 3 ^ 4 = 7. We need to
* find two's
* complement of 7 and find the rightmost set bit. i.e. (num & (-num)) Two's
* complement of 7 is 001
* and hence res = 1. There can be 2 options when we Bitise AND this res with
* all the elements in our
* array
* 1. Result will come non zero number
* 2. Result will be 0.
* In the first case we will XOR our element with the first number (which is
* initially 0)
* In the second case we will XOR our element with the second number(which is
* initially 0)
* This is how we will get non repeating elements with the help of bitwise
* operators.
*/
}