Occurrences of Consecutive 3 Numbers

Given an integer array arr[], count the number of distinct special integers. An integer x is called a special integer if x - 1, x, and x + 1 are all present in the array. Return the number of distinct special integers.

Examples:

Input: arr[] = [1, 2, 3, 3, 4]
Output: 2
Explanation: The special integers in this array are 2 and 3.

Input: arr[] = [2, 3, 5, 7]
Output: 0
Explanation: There is no special integer in this array.

[Naive Approach] Use Linear Search - O(n^2) Time and O(1) Space

The idea is to check every element x and use linear search to determine whether x - 1 and x + 1 are present in the array. If both are present, x is a special integer.

Working of the Approach:

• Traverse each element x in the array.
• Search for x - 1 in the array.
• Search for x + 1 in the array.
• If both are present, increment the count.
• Return the count.

#include <bits/stdc++.h>
using namespace std;

bool isPresent(vector<int>& arr, int value) {
    for (int x : arr) {
        if (x == value)
            return true;
    }

    return false;
}

bool seenBefore(vector<int>& arr, int idx) {
    for (int i = 0; i < idx; i++) {
        if (arr[i] == arr[idx])
            return true;
    }

    return false;
}

int specialIntegers(vector<int>& arr) {
    int count = 0;

    for (int i = 0; i < arr.size(); i++) {
        if (seenBefore(arr, i))
            continue;

        int x = arr[i];

        if (isPresent(arr, x - 1) && isPresent(arr, x + 1))
            count++;
    }

    return count;
}

int main() {
    vector<int> arr = {1, 2, 3, 4};

    cout << specialIntegers(arr);

    return 0;
}

class GFG {

    static boolean isPresent(int[] arr, int value) {
        for (int x : arr) {
            if (x == value)
                return true;
        }

        return false;
    }

    static boolean seenBefore(int[] arr, int idx) {
        for (int i = 0; i < idx; i++) {
            if (arr[i] == arr[idx])
                return true;
        }

        return false;
    }

    static int specialIntegers(int[] arr) {
        int count = 0;

        for (int i = 0; i < arr.length; i++) {
            if (seenBefore(arr, i))
                continue;

            int x = arr[i];

            if (isPresent(arr, x - 1) && isPresent(arr, x + 1))
                count++;
        }

        return count;
    }

    public static void main(String[] args) {
        int[] arr = {1, 2, 3, 4};

        System.out.println(specialIntegers(arr));
    }
}

def isPresent(arr, value):
    for x in arr:
        if x == value:
            return True

    return False


def seenBefore(arr, idx):
    for i in range(idx):
        if arr[i] == arr[idx]:
            return True

    return False


def specialIntegers(arr):
    count = 0

    for i in range(len(arr)):
        if seenBefore(arr, i):
            continue

        x = arr[i]

        if isPresent(arr, x - 1) and isPresent(arr, x + 1):
            count += 1

    return count


if __name__ == "__main__":
    arr = [1, 2, 3, 4]

    print(specialIntegers(arr))

using System;

class GFG
{
    static bool isPresent(int[] arr, int value)
    {
        foreach (int x in arr)
        {
            if (x == value)
                return true;
        }

        return false;
    }

    static bool seenBefore(int[] arr, int idx)
    {
        for (int i = 0; i < idx; i++)
        {
            if (arr[i] == arr[idx])
                return true;
        }

        return false;
    }

    static int specialIntegers(int[] arr)
    {
        int count = 0;

        for (int i = 0; i < arr.Length; i++)
        {
            if (seenBefore(arr, i))
                continue;

            int x = arr[i];

            if (isPresent(arr, x - 1) && isPresent(arr, x + 1))
                count++;
        }

        return count;
    }

    static void Main()
    {
        int[] arr = { 1, 2, 3, 4 };

        Console.WriteLine(specialIntegers(arr));
    }
}

function isPresent(arr, value) {
    for (let x of arr) {
        if (x === value)
            return true;
    }

    return false;
}

function seenBefore(arr, idx) {
    for (let i = 0; i < idx; i++) {
        if (arr[i] === arr[idx])
            return true;
    }

    return false;
}

function specialIntegers(arr) {
    let count = 0;

    for (let i = 0; i < arr.length; i++) {
        if (seenBefore(arr, i))
            continue;

        let x = arr[i];

        if (isPresent(arr, x - 1) && isPresent(arr, x + 1))
            count++;
    }

    return count;
}

// Driver Code
let arr = [1, 2, 3, 4];

console.log(specialIntegers(arr));

2

[Expected Approach] Use a Hash Set - O(n) Time and O(n) Space

The idea is to store all elements of the array in a hash set. This allows us to check whether a number exists in the array in average O(1) time. For each distinct element x, if both x - 1 and x + 1 are present in the set, then x is a special integer.

Working of the Approach:

• Insert all elements of the array into a hash set.
• Traverse the distinct elements of the array.
• For each element x, check whether x - 1 and x + 1 are present in the set.
• If both are present, increment the count.
• Return the count.

#include <bits/stdc++.h>
using namespace std;

int specialIntegers(vector<int>& arr) {
    unordered_set<int> st(arr.begin(), arr.end());
    int count = 0;

    for (int x : st) {
        if (st.count(x - 1) && st.count(x + 1))
            count++;
    }

    return count;
}

int main() {
    vector<int> arr = {1, 2, 3, 4};
    cout << specialIntegers(arr);

    return 0;
}

import java.util.HashSet;

class GFG {

    static int specialIntegers(int[] arr) {
        HashSet<Integer> set = new HashSet<>();

        for (int x : arr)
            set.add(x);

        int count = 0;

        for (int x : set) {
            if (set.contains(x - 1) && set.contains(x + 1))
                count++;
        }

        return count;
    }

    public static void main(String[] args) {
        int[] arr = {1, 2, 3, 4};
        System.out.println(specialIntegers(arr));
    }
}

def specialIntegers(arr):
    st = set(arr)
    count = 0

    for x in st:
        if x - 1 in st and x + 1 in st:
            count += 1

    return count


if __name__ == "__main__":
    arr = [1, 2, 3, 4]
    print(specialIntegers(arr))

using System;
using System.Collections.Generic;

class GFG
{
    static int specialIntegers(int[] arr)
    {
        HashSet<int> set = new HashSet<int>(arr);
        int count = 0;

        foreach (int x in set)
        {
            if (set.Contains(x - 1) && set.Contains(x + 1))
                count++;
        }

        return count;
    }

    static void Main()
    {
        int[] arr = { 1, 2, 3, 4 };
        Console.WriteLine(specialIntegers(arr));
    }
}

function specialIntegers(arr) {
    let st = new Set(arr);
    let count = 0;

    for (let x of st) {
        if (st.has(x - 1) && st.has(x + 1))
            count++;
    }

    return count;
}

// Driver Code
let arr = [1, 2, 3, 4];
console.log(specialIntegers(arr));

2