87 lines
2.0 KiB
Java
87 lines
2.0 KiB
Java
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/* Given an array of positive integers. Find the length of the longest sub-sequence such that elements in the subsequence are consecutive integers,
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the consecutive numbers can be in any order. */
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/* Time Complexity: O(n)
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Space Complexity: O(n) */
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import java.math.*;
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import java.util.*;
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import java.io.*;
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class Driverclass {
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static class FastReader{
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BufferedReader br;
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StringTokenizer st;
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public FastReader(){
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br = new BufferedReader(new InputStreamReader(System.in));
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}
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String next(){
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while (st == null || !st.hasMoreElements()){
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try{ st = new StringTokenizer(br.readLine()); } catch (IOException e){ e.printStackTrace(); }
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}
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return st.nextToken();
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}
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String nextLine(){
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String str = "";
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try{ str = br.readLine(); } catch (IOException e) { e.printStackTrace(); }
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return str;
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}
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Integer nextInt(){
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return Integer.parseInt(next());
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}
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}
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public static void main(String args[]) {
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FastReader sc = new FastReader();
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PrintWriter out = new PrintWriter(System.out);
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int t = sc.nextInt();
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while(t>0) {
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int n = sc.nextInt();
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int a[] = new int[n];
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for(int i=0; i<n; i++)
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a[i] = sc.nextInt();
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out.println(new Solution().findLongestConseqSubseq(a, n));
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t--;
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}
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out.flush();
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}
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}
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class Solution {
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// arr[] : the input array
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// N : size of the array arr[]
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static int findLongestConseqSubseq(int arr[], int n){
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Set<Integer> set = new HashSet<>();
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int ans=0;
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for(int num: arr)
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set.add(num);
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for(int i=0; i<n; i++){
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if(!set.contains(arr[i]-1)){
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int j = arr[i];
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while(set.contains(j))
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j++;
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ans = Math.max(ans, j-arr[i]);
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}
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}
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return ans;
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}
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}
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/*
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Input:
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N = 7
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a[] = {2,6,1,9,4,5,3}
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Output:
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6
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Explanation:
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The consecutive numbers here
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are 1, 2, 3, 4, 5, 6. These 6
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numbers form the longest consecutive
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subsquence.
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*/
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