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

**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:

- If
**a > 0**and**b > 0**then the resultant product will be positive. - If
**a < 0**and**b > 0**then the result will be zero as**a * (a + 1) * … * 0 * … (b – 1) * b = 0**. - 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 <iostream>` `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` |

## Javascript

`<script>` `// Javascript implementation of the approach` `// Function to check whether the product` `// of integers of the range [a, b]` `// is positive, negative or zero` `function` `solve( a, b)` `{` ` ` `// If both a and b are positive then` ` ` `// the product will be positive` ` ` `if` `(a > 0 && b > 0)` ` ` `{` ` ` `document.write( ` `"Positive"` `);` ` ` `}` ` ` `// If a is negative and b is positive then` ` ` `// the product will be zero` ` ` `else` `if` `(a <= 0 && b >= 0)` ` ` `{` ` ` `document.write( ` `"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` ` ` `let n = Math.abs(a - b) + 1;` ` ` `// If n is even then the resultant` ` ` `// product is positive` ` ` `if` `(n % 2 == 0)` ` ` `{` ` ` `document.write( ` `"Positive"` `);` ` ` `}` ` ` ` ` `// If n is odd then the resultant` ` ` `// product is negative` ` ` `else` ` ` `{` ` ` `document.write( ` `"Negative"` `);` ` ` `}` ` ` `}` `}` ` ` `// Driver code` ` ` ` ` `let a = -10;` ` ` `let b = -2;` ` ` ` ` `solve(a, b);` `// This code is contributed by Bobby` `</script>` |

**Output:**

Negative

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