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FindDuplicate.java
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76 lines (61 loc) · 2.11 KB
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// Find Duplicate in an array
// You have been given an integer array/list(ARR)of size N which contains numbers from 0 to(N-2).Each number is present at least once.That is,if N=5,the array/list constitutes values ranging from 0 to 3 and among these,there is a single integer value that is present twice.You need to find and return that duplicate number present in the array.
// Note:
// Duplicate number is always present in the given array/list.
// Input format:
// The first line contains an Integer't'which denotes the number of test cases or queries to be run.Then the test cases follow.
// First line of each test case or query contains an integer'N'representing the size of the array/list.
// Second line contains'N'single space separated integers representing the elements in the array/list.
// Output Format:
// For each test case,print the duplicate element in the array/list.
// Output for every test case will be printed in a separate line.
// Constraints:
// 1<=t<=10^2
// 0<=N<=10^3
// Time Limit:1 sec
// Sample Input 1:
// 1
// 9 0 7 2 5 4 7 1 3 6
// Sample Output 1:
// 7
// Sample Input 2:
// 2
// 5
// 0 2 1 3 1
// 7
// 0 3 1 5 4 3 2
// Sample Output 2:
// 1
// 3
package Array;
import java.util.Scanner;
public class FindDuplicate {
public static int duplicateElement(int[] arr) {
int n = arr.length;
for (int i = 0; i < n; i++) {
int j = 0;
for (; j < n; j++) {
if (j != i) {
if (arr[i] == arr[j]) {
return arr[i];
}
}
}
}
return Integer.MIN_VALUE;
}
public static int[] takingInput() {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
return arr;
}
public static void main(String[] args) {
int[] arr = takingInput();
int n = duplicateElement(arr);
System.out.println(n);
}
}