# Check whether product of integers from a to b is positive , negative or zero

Given two integers a and b, the task is to check whether the product of integers from the rage v[a, b] i.e. a * (a + 1) * (a + 2) * … * b is positive, negative or zero.

Examples:

Input: a = -10, b = -2
Output: Negative

Input: a = -10, b = 2
Output: Zero

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive approach: We can run a loop from a to b and multiply all the numbers starting from a to b and check whether the product is positive negative or zero. This solution will fail for large values of a and b and will result in overflow.

Efficient approach: There are three possible case:

1. If a > 0 and b > 0 then the resultant product will be positive.
2. If a < 0 and b > 0 then the result will be zero as a * (a + 1) * … * 0 * … (b – 1) * b = 0.
3. If a < 0 and b < 0 then the result will depend on the count of numbers (as all the numbers are negative)
• If the count of negative numbers is even then the result will be positive.
• Else the result will be negative.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to check whether the product ` `// of integers of the range [a, b] ` `// is positive, negative or zero ` `void` `solve(``long` `long` `int` `a, ``long` `long` `int` `b) ` `{ ` ` `  `    ``// If both a and b are positive then ` `    ``// the product will be positive ` `    ``if` `(a > 0 && b > 0) { ` `        ``cout << ``"Positive"``; ` `    ``} ` ` `  `    ``// If a is negative and b is positive then ` `    ``// the product will be zero ` `    ``else` `if` `(a <= 0 && b >= 0) { ` `        ``cout << ``"Zero"` `<< endl; ` `    ``} ` ` `  `    ``// If both a and b are negative then ` `    ``// we have to find the count of integers ` `    ``// in the range ` `    ``else` `{ ` ` `  `        ``// Total integers in the range ` `        ``long` `long` `int` `n = ``abs``(a - b) + 1; ` ` `  `        ``// If n is even then the resultant ` `        ``// product is positive ` `        ``if` `(n % 2 == 0) { ` `            ``cout << ``"Positive"` `<< endl; ` `        ``} ` `        ``// If n is odd then the resultant ` `        ``// product is negative ` `        ``else` `{ ` `            ``cout << ``"Negative"` `<< endl; ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a = -10, b = -2; ` ` `  `    ``solve(a, b); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to check whether the product ` `// of integers of the range [a, b] ` `// is positive, negative or zero ` `static` `void` `solve(``long` `a, ``long` `b) ` `{ ` ` `  `    ``// If both a and b are positive then ` `    ``// the product will be positive ` `    ``if` `(a > ``0` `&& b > ``0``)  ` `    ``{ ` `        ``System.out.println( ``"Positive"``); ` `    ``} ` ` `  `    ``// If a is negative and b is positive then ` `    ``// the product will be zero ` `    ``else` `if` `(a <= ``0` `&& b >= ``0``) ` `    ``{ ` `        ``System.out.println( ``"Zero"` `); ` `    ``} ` ` `  `    ``// If both a and b are negative then ` `    ``// we have to find the count of integers ` `    ``// in the range ` `    ``else`  `    ``{ ` ` `  `        ``// Total integers in the range ` `        ``long` `n = Math.abs(a - b) + ``1``; ` ` `  `        ``// If n is even then the resultant ` `        ``// product is positive ` `        ``if` `(n % ``2` `== ``0``)  ` `        ``{ ` `            ``System.out.println( ``"Positive"``); ` `        ``} ` `         `  `        ``// If n is odd then the resultant ` `        ``// product is negative ` `        ``else` `        ``{ ` `            ``System.out.println( ``"Negative"``); ` `        ``} ` `    ``} ` `} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `a = -``10``, b = -``2``; ` `     `  `        ``solve(a, b); ` `    ``} ` `} ` ` `  `// This code is contributed by anuj_67.. `

## Python3

 `# Python 3 implementation of the approach ` ` `  `# Function to check whether the product ` `# of integers of the range [a, b] ` `# is positive, negative or zero ` `def` `solve(a,b): ` `     `  `    ``# If both a and b are positive then ` `    ``# the product will be positive ` `    ``if` `(a > ``0` `and` `b > ``0``): ` `        ``print``(``"Positive"``) ` ` `  `    ``# If a is negative and b is positive then ` `    ``# the product will be zero ` `    ``elif` `(a <``=` `0` `and` `b >``=` `0``): ` `        ``print``(``"Zero"``) ` ` `  `    ``# If both a and b are negative then ` `    ``# we have to find the count of integers ` `    ``# in the range ` `    ``else``: ` `         `  `        ``# Total integers in the range ` `        ``n ``=` `abs``(a ``-` `b) ``+` `1` ` `  `        ``# If n is even then the resultant ` `        ``# product is positive ` `        ``if` `(n ``%` `2` `=``=` `0``): ` `            ``print``(``"Positive"``) ` `             `  `        ``# If n is odd then the resultant ` `        ``# product is negative ` `        ``else``: ` `            ``print``(``"Negative"``) ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``a ``=` `-``10` `    ``b ``=` `-``2` ` `  `    ``solve(a, b) ` `     `  `# This code is contributed by ` `# Surendra_Gangwar `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` `     `  `    ``// Function to check whether the product  ` `    ``// of integers of the range [a, b]  ` `    ``// is positive, negative or zero  ` `    ``static` `void` `solve(``long` `a, ``long` `b)  ` `    ``{  ` `     `  `        ``// If both a and b are positive then  ` `        ``// the product will be positive  ` `        ``if` `(a > 0 && b > 0)  ` `        ``{  ` `            ``Console.WriteLine( ``"Positive"``);  ` `        ``}  ` `     `  `        ``// If a is negative and b is positive then  ` `        ``// the product will be zero  ` `        ``else` `if` `(a <= 0 && b >= 0)  ` `        ``{  ` `            ``Console.WriteLine( ``"Zero"` `);  ` `        ``}  ` `     `  `        ``// If both a and b are negative then  ` `        ``// we have to find the count of integers  ` `        ``// in the range  ` `        ``else` `        ``{  ` `     `  `            ``// Total integers in the range  ` `            ``long` `n = Math.Abs(a - b) + 1;  ` `     `  `            ``// If n is even then the resultant  ` `            ``// product is positive  ` `            ``if` `(n % 2 == 0)  ` `            ``{  ` `                ``Console.WriteLine( ``"Positive"``);  ` `            ``}  ` `             `  `            ``// If n is odd then the resultant  ` `            ``// product is negative  ` `            ``else` `            ``{  ` `                ``Console.WriteLine( ``"Negative"``);  ` `            ``}  ` `        ``}  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main ()  ` `    ``{  ` `        ``int` `a = -10, b = -2;  ` `     `  `        ``solve(a, b);  ` `    ``} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```Negative
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