Detect if two integers have opposite signs

Given two signed integers, write a function that returns true if the signs of given integers are different, otherwise false. For example, the function should return true -1 and +100, and should return false for -100 and -200. The function should not use any of the arithmetic operators. 
Let the given integers be x and y. The sign bit is 1 in negative numbers, and 0 in positive numbers. The XOR of x and y will have the sign bit as 1 if they have opposite sign. In other words, XOR of x and y will be negative number if x and y have opposite signs. The following code use this logic. 

C++

// C++ Program to Detect
// if two integers have opposite signs.
#include<iostream>
using namespace std;

bool oppositeSigns(int x, int y)
{
    return ((x ^ y) < 0);
}

int main()
{
    int x = 100, y = -100;
    if (oppositeSigns(x, y) == true)
    cout << "Signs are opposite";
    else
    cout << "Signs are not opposite";
    return 0;
}

// this code is contributed by shivanisinghss2110

C

// C++ Program to Detect
// if two integers have opposite signs.
#include<stdbool.h>
#include<stdio.h>

bool oppositeSigns(int x, int y)
{
    return ((x ^ y) < 0);
}

int main()
{
    int x = 100, y = -100;
    if (oppositeSigns(x, y) == true)
       printf ("Signs are opposite");
    else
      printf ("Signs are not opposite");
    return 0;
}

Java

// Java Program to Detect
// if two integers have opposite signs.

class GFG {

    static boolean oppositeSigns(int x, int y)
    {
        return ((x ^ y) < 0);
    }

    public static void main(String[] args)
    {
        int x = 100, y = -100;
        if (oppositeSigns(x, y) == true)
            System.out.println("Signs are opposite");
        else
            System.out.println("Signs are not opposite");
    }
}

// This code is contributed by prerna saini.

Python3

# Python3 Program to Detect
# if two integers have
# opposite signs.
def oppositeSigns(x, y):
    return ((x ^ y) < 0);

x = 100
y = 1

if (oppositeSigns(x, y) == True):
    print ("Signs are opposite")
else:
    print ("Signs are not opposite")

# This article is contributed by Prerna Saini.

C#

// C# Program to Detect
// if two integers have
// opposite signs.
using System;

class GFG {

    // Function to detect signs
    static bool oppositeSigns(int x, int y)
    {
        return ((x ^ y) < 0);
    }

    // Driver Code
    public static void Main()
    {
        int x = 100, y = -100;
        if (oppositeSigns(x, y) == true)
            Console.Write("Signs are opposite");
        else
            Console.Write("Signs are not opposite");
    }
}

// This code is contributed by Nitin Mittal.

PHP

<?php
// PHP Program to Detect if two
// integers have opposite signs.

function oppositeSigns($x, $y)
{
    return (($x ^ $y) < 0);
}

    // Driver Code
    $x = 100;
    $y = -100;
    if (oppositeSigns($x, $y) == true)
    echo ("Signs are opposite");
    else
    echo ("Signs are not opposite");

// This code is contributed by vt_m.
?>

Javascript

<script>

// Javascript Program to Detect
// if two integers have opposite signs.

function oppositeSigns(x, y)
    {
        return ((x ^ y) < 0);
    }

// Driver Code

    let x = 100, y = -100;
    if (oppositeSigns(x, y) == true)
         document.write("Signs are opposite");
    else
         document.write("Signs are not opposite");

</script>

Output

Signs are opposite

Time Complexity: O(1)

Auxiliary Space: O(1)

Source: Detect if two integers have opposite signs
We can also solve this by using two comparison operators. See the following code. 

CPP

bool oppositeSigns(int x, int y)
{
    return (x < 0)? (y >= 0): (y < 0);
}

C

//C program to detect whether two integers
//have different sign or not

#include <stdio.h>
#include<stdbool.h>

bool oppositeSigns(int x, int y)
{
    return (x < 0)? (y >= 0): (y < 0);
}

//this code is contributed by shruti456rawal

Java

class GFG{
static boolean oppositeSigns(int x, int y)
{
    return (x < 0)? (y >= 0): (y < 0);
}
}

// This code contributed by umadevi9616

Python3

def oppositeSigns(x, y):

    return (y >= 0) if (x < 0) else (y < 0);

# This code is contributed by shivanisingjss2110

C#

using System;
class GFG{
static boo oppositeSigns(int x, int y)
{
    return (x < 0)? (y >= 0): (y < 0);
}
}

// This code contributed by shivanisinghss2110

Javascript

<script>
function oppositeSigns(x, y)
{
    return (x < 0)? (y >= 0): (y < 0);
}


// This code contributed by shivanisinghss2110
</script>

Time Complexity: O(1)

The first method is more efficient. The first method uses a bitwise XOR and a comparison operator. The second method uses two comparison operators and a bitwise XOR operation is more efficient compared to a comparison operation. 
We can also use following method. It doesn’t use any comparison operator. The method is suggested by Hongliang and improved by gaurav. 

CPP

bool oppositeSigns(int x, int y)
{
    return ((x ^ y) >> 31);
}

C

//C program to detect whether two integers
//have different sign or not

#include <stdio.h>
#include<stdbool.h>

bool oppositeSigns(int x, int y)
{
    return ((x ^ y) >> 31);
}

//this code is contributed by shruti456rawal

Java

import java.io.*;

class GFG {
static boolean oppositeSigns(int x, int y)
{
    return ((x ^ y) >> 31);
}

    }

// This code is contributed by shivanisinghss2110

Python3

def oppositeSigns(x, y):

    return ((x ^ y) >> 31)

# this code is contributed by shivanisinghss2110

C#

using System;

class GFG {
static bool oppositeSigns(int x, int y)
{
    return ((x ^ y) >> 31);
}

    }

// This code is contributed by shivanisinghss2110

Javascript

<script>
function oppositeSigns(x, y)
{
    return ((x ^ y) >> 31);
}


// This code is contributed by shivanisinghss2110
</script>

Time Complexity: O(1)

The function is written only for compilers where size of an integer is 32 bit. The expression basically checks sign of (x^y) using bitwise operator ‘>>’. As mentioned above, the sign bit for negative numbers is always 1. The sign bit is the leftmost bit in binary representation. So we need to checks whether the 32th bit (or leftmost bit) of x^y is 1 or not. We do it by right shifting the value of x^y by 31, so that the sign bit becomes the least significant bit. If sign bit is 1, then the value of (x^y)>>31 will be 1, otherwise 0. 

C++

// C++ Program to detect if two integers have opposite
// signs.
#include <iostream>
using namespace std;

bool oppositeSigns(int x, int y)
{
    long long product = 1ll * x * y;
    return (product < 0);
}

int main()
{
    int x = 100, y = -100;
    if (oppositeSigns(x, y) == true)
        cout << "Signs are opposite";
    else
        cout << "Signs are not opposite";
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

C

// C Program to detect
// if two integers have opposite signs.
#include <stdbool.h>
#include <stdio.h>

bool oppositeSigns(int x, int y)
{
    long long product = 1ll * x * y;
    return (product < 0);
}

int main()
{
    int x = 100, y = -100;
    if (oppositeSigns(x, y) == true)
        printf("Signs are opposite");
    else
        printf("Signs are not opposite");
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// Java program for the above approach
import java.util.*;

class GFG {

  static boolean oppositeSigns(int x, int y)
  {
    long product = 1*x*y;
    return (product<0);
  }

  // Driver Code
  public static void main(String[] args)
  {
    int x = 100, y = -100;
    if (oppositeSigns(x, y) == true)
      System.out.print( "Signs are opposite");
    else
      System.out.print("Signs are not opposite");
  }
}

// This code is contributed by sanjoy_62.

Python3

# Python Program to detect
# if two integers have opposite signs.

def oppositeSigns(x,y):
    product = x*y
    return (product<0)

# driver code
x = 100
y = -100
if(oppositeSigns(x, y) == True):
  print("Signs are opposite")
else :
  print("Signs are not opposite")

# this code is contributed by shinjanpatra

C#

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

  static bool oppositeSigns(int x, int y)
  {
    long product = 1*x*y;
    return (product<0);
  }

// Driver Code
public static void Main(String[] args)
{
    int x = 100, y = -100;
    if (oppositeSigns(x, y) == true)
      Console.WriteLine( "Signs are opposite");
    else
      Console.WriteLine("Signs are not opposite");
}
}

// This code is contributed by avijitmondal1998.

Javascript

// JavaScript Program to detect
// if two integers have opposite signs.

function oppositeSigns(x,y)
{
    const product = Number(x)*Number(y);
    return (product<0);
}

// driver code
let x = 100, y = -100;
if(oppositeSigns(x, y) == true)
{
    console.log("Signs are opposite");
}
else console.log("Signs are not opposite");

// this code is contributed by shinjanpatra

Output

Signs are opposite

Approach : The basic approach is to calculate the product the two integers, and as we know, two integers having opposite signs will always produce a negative integer, we need to just find out whether the product is negative or not.

Time Complexity: O(1)

Space Complexity: O(1)

Please write comments if you find any of the above codes/algorithms incorrect, or find other ways to solve the same problem.