Picking Numbers


Problem Statement :


Given an array of integers, find the longest subarray where the absolute difference between any two elements is less than or equal to 1.

Example
a = [1, 1, 2, 2, 4, 4, 5, 5, 5]

There are two subarrays meeting the criterion: [1, 1, 2, 2] and [4, 4, 5, 5, 5]. The maximum length subarray has 5 elements.


Function Description

Complete the pickingNumbers function in the editor below.

pickingNumbers has the following parameter(s):

int a[n]: an array of integers

Returns

int: the length of the longest subarray that meets the criterion


Input Format

The first line contains a single integer n, the size of the array a.
The second line contains n space-separated integers, each an a[i].


Constraints
2 <= n <= 100
0 < a[i] < 100
The answer will be  >= 2


Solution :



title-img 


                            Solution in C :

python 3  :

#!/bin/python3

import sys


n = int(input().strip())
a = [int(a_temp) for a_temp in input().strip().split(' ')]

from collections import Counter
d = Counter(a)
best = 0
for i in range(99):
    best = max(d[i] + d[i+1], best)
print(best)









Java  :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] a = new int[n];
        for(int a_i=0; a_i < n; a_i++){
            a[a_i] = in.nextInt();
        }
        
        int[] freq = new int[100];
        for(int i = 0; i < a.length; ++i)
        {
            freq[a[i]]++;
        }
        
        int curBest = 0;
        for(int i = 0; i < 99; ++i)
        {
            curBest = Math.max(curBest, freq[i]+freq[i+1]);
        }
        System.out.println(curBest);
    }
}












C++  :

#include <bits/stdc++.h>

using namespace std;

int N;
int A[1000];

int main()
{
    scanf("%d", &N);
    for(int i=0; i<N; i++)
    {
        int a;
        scanf("%d", &a);
        A[a]++;
    }
    int ans=0;
    for(int i=1; i<1000; i++)
        ans=max(ans, A[i-1]+A[i]);
    printf("%d\n", ans);
    return 0;
}








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