Find the only positive or only negative number in the given Array

Given an array arr[] of either entirely positive integers or entirely negative integers except for one number. The task is to find that number.

Examples:

Input: arr[] = {3, 5, 2, 8, -7, 6, 9}
Output: -7
Explanation: Except -7 all the numbers in arr[] are positive integers. 

Input: arr[] = {-3,  5,  -9}
Output: 5

Approach: The given problem can be solved by just comparing the first three numbers of arr[]. After that apply Linear Search and find the number. Follow the steps below to solve the problem. 

• If the size of arr[] is smaller than 3, then return 0.
• Initialize the variables Cp and Cn.
• Where Cp = Count of positive integers and Cn = Count of negative integers.
• Iterate over the first 3 element
  • If (arr[i]>0), Increment Cp by 1.
  • Else Increment Cn by 1.
• Cp can be 2 or 3 and similarly, Cn can also either 2 or 3.
• If Cp<Cn, then the single element is positive, apply linear search and find it.
• If Cn<Cp, then the single element is negative, apply linear search and find it.

Below is the implementation of the above approach: 

C++

// C++ program for the above approach
#include <iostream>
using namespace std;

// Function to return the single element
int find(int* a, int N)
{
    // Size can not be smaller than 3
    if (N < 3)
        return 0;

    // Initialize the variable
    int i, Cp = 0, Cn = 0;

    // Check the single element is
    // positive or negative
    for (i = 0; i < 3; i++) {

        if (a[i] > 0)
            Cp++;
        else
            Cn++;
    }

    // Check for positive single element
    if (Cp < Cn) {
        for (i = 0; i < N; i++)
            if (a[i] > 0)
                return a[i];
    }

    // Check for negative single element
    else {
        for (i = 0; i < N; i++)
            if (a[i] < 0)
                return a[i];
    }
}

// Driver Code
int main()
{
    int arr[] = { 3, 5, 2, 8, -7, 6, 9 };
    int N = sizeof(arr) / sizeof(arr[0]);

    cout << find(arr, N);
}

Java

// Java program for the above approach
import java.util.*;
public class GFG {

// Function to return the single element
static int find(int []a, int N)
{
    // Size can not be smaller than 3
    if (N < 3)
        return 0;

    // Initialize the variable
    int i, Cp = 0, Cn = 0;

    // Check the single element is
    // positive or negative
    for (i = 0; i < 3; i++) {

        if (a[i] > 0)
            Cp++;
        else
            Cn++;
    }

    // Check for positive single element
    if (Cp < Cn) {
        for (i = 0; i < N; i++)
            if (a[i] > 0)
                return a[i];
    }

    // Check for negative single element
    else {
        for (i = 0; i < N; i++)
            if (a[i] < 0)
                break;
    }
    return a[i];
}

// Driver Code
public static void main(String args[])
{
    int []arr = { 3, 5, 2, 8, -7, 6, 9 };
    int N = arr.length;
    System.out.println(find(arr, N));
}
}

// This code is contributed by Samim Hossain Mondal.

Python3

# python program for the above approach

# Function to return the single element
def find(a, N):

    # Size can not be smaller than 3
    if (N < 3):
        return 0

    # Initialize the variable
    Cp = 0
    Cn = 0

    # Check the single element is
    # positive or negative
    for i in range(0, 3):

        if (a[i] > 0):
            Cp += 1
        else:
            Cn += 1

        # Check for positive single element
    if (Cp < Cn):
        for i in range(0, N):
            if (a[i] > 0):
                return a[i]

        # Check for negative single element
    else:
        for i in range(0, N):
            if (a[i] < 0):
                return a[i]

# Driver Code
if __name__ == "__main__":

    arr = [3, 5, 2, 8, -7, 6, 9]
    N = len(arr)

    print(find(arr, N))

    # This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
class GFG {

// Function to return the single element
static int find(int []a, int N)
{
    // Size can not be smaller than 3
    if (N < 3)
        return 0;

    // Initialize the variable
    int i, Cp = 0, Cn = 0;

    // Check the single element is
    // positive or negative
    for (i = 0; i < 3; i++) {

        if (a[i] > 0)
            Cp++;
        else
            Cn++;
    }

    // Check for positive single element
    if (Cp < Cn) {
        for (i = 0; i < N; i++)
            if (a[i] > 0)
                return a[i];
    }

    // Check for negative single element
    else {
        for (i = 0; i < N; i++)
            if (a[i] < 0)
                break;
    }
    return a[i];
}

// Driver Code
public static void Main()
{
    int []arr = { 3, 5, 2, 8, -7, 6, 9 };
    int N = arr.Length;
    Console.Write(find(arr, N));
}
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
// Javascript program for the above approach

// Function to return the single element
function find(a, N)
{
    // Size can not be smaller than 3
    if (N < 3)
        return 0;

    // Initialize the variable
    let i, Cp = 0, Cn = 0;

    // Check the single element is
    // positive or negative
    for (i = 0; i < 3; i++) {

        if (a[i] > 0)
            Cp++;
        else
            Cn++;
    }

    // Check for positive single element
    if (Cp < Cn) {
        for (i = 0; i < N; i++)
            if (a[i] > 0)
                return a[i];
    }

    // Check for negative single element
    else {
        for (i = 0; i < N; i++)
            if (a[i] < 0)
                return a[i];
    }
}

// Driver Code
let arr = [ 3, 5, 2, 8, -7, 6, 9 ];
let N = arr.length;

document.write(find(arr, N));

// This code is contributed Samim Hossain Mondal.
</script>

Output

-7

Time Complexity: O(N) 
Auxiliary Space: O(1)