# Find ratio of zeroes, positive numbers and negative numbers in the Array

Given an array a of integers of size N integers, the task is to find the ratio of positive numbers, negative numbers and zeros in the array up to four decimal places.

Examples:

Input: a[] = {2, -1, 5, 6, 0, -3}
Output: 0.5000 0.3333 0.1667
There are 3 positive, 2 negative, and 1 zero. Their ratio would be positive: 3/6 = 0.5000, negative: 2/6 = 0.3333, and zero: 1/6 = 0.1667.

Input: a[] = {4, 0, -2, -9, -7, 1}
Output: 0.3333 0.5000 0.1667
There are 2 positive, 3 negative, and 1 zero. Their ratio would be positive: 2/6 = 0.3333, negative: 3/6 = 0.5000, and zero: 1/6 = 0.1667.

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

Approach:

1. Count the total number of positive elements in the array.
2. Count the total number of negative elements in the array.
3. Count the total number of zero elements in the array.
4. Divide the total number of positive elements, negative elements, and zero elements by the size of the array, to get the ratio.
5. Print the ratio of positive, negative, and zero elements in the array up to four decimal places.

Below is the implementation of the above approach.

 `// C++ program to find the ratio of positive, ` `// negative, and zero elements in the array. ` `#include ` `using` `namespace` `std; ` ` `  ` `  `// Function to find the ratio of ` `// positive, negative, and zero elements ` `void` `positiveNegativeZero(``int` `arr[], ``int` `len) ` `{ ` `    ``// Initialize the postiveCount, negativeCount, and ` `    ``// zeroCountby 0 which will count the total number ` `    ``// of positive, negative and zero elements ` `    ``float` `positiveCount = 0; ` `    ``float` `negativeCount = 0; ` `    ``float` `zeroCount = 0; ` ` `  `    ``// Traverse the array and count the total number of ` `    ``// positive, negative, and zero elements. ` `    ``for` `(``int` `i = 0; i < len; i++) { ` `        ``if` `(arr[i] > 0) { ` `            ``positiveCount++; ` `        ``} ` `        ``else` `if` `(arr[i] < 0) { ` `            ``negativeCount++; ` `        ``} ` `        ``else` `if` `(arr[i] == 0) { ` `            ``zeroCount++; ` `        ``} ` `    ``} ` ` `  `    ``// Print the ratio of positive, ` `    ``// negative, and zero elements ` `    ``// in the array up to four decimal places. ` `    ``cout << fixed << setprecision(4) << (positiveCount / len)<<``" "``; ` `    ``cout << fixed << setprecision(4) << (negativeCount / len)<<``" "``; ` `    ``cout << fixed << setprecision(4) << (zeroCount / len); ` `    ``cout << endl; ` `} ` ` `  `// Driver Code. ` `int` `main() ` `{ ` `    ``// Test Case 1: ` `    ``int` `a1[] = { 2, -1, 5, 6, 0, -3 }; ` `    ``int` `len=``sizeof``(a1)/``sizeof``(a1); ` `    ``positiveNegativeZero(a1,len); ` ` `  `    ``// Test Case 2: ` `    ``int` `a2[] = { 4, 0, -2, -9, -7, 1 }; ` `    ``len=``sizeof``(a2)/``sizeof``(a2); ` `    ``positiveNegativeZero(a2,len); ` `} ` ` `  `// This code is contributed by chitranayal `

 `// Java program to find the ratio of positive, ` `// negative, and zero elements in the array. ` ` `  `class` `GFG { ` ` `  `    ``// Function to find the ratio of ` `    ``// positive, negative, and zero elements ` `    ``static` `void` `positiveNegativeZero(``int``[] arr) ` `    ``{ ` ` `  `        ``// Store the array length into the variable len. ` `        ``int` `len = arr.length; ` ` `  `        ``// Initialize the postiveCount, negativeCount, and ` `        ``// zeroCountby 0 which will count the total number ` `        ``// of positive, negative and zero elements ` `        ``float` `positiveCount = ``0``; ` `        ``float` `negativeCount = ``0``; ` `        ``float` `zeroCount = ``0``; ` ` `  `        ``// Traverse the array and count the total number of ` `        ``// positive, negative, and zero elements. ` `        ``for` `(``int` `i = ``0``; i < len; i++) { ` `            ``if` `(arr[i] > ``0``) { ` `                ``positiveCount++; ` `            ``} ` `            ``else` `if` `(arr[i] < ``0``) { ` `                ``negativeCount++; ` `            ``} ` `            ``else` `if` `(arr[i] == ``0``) { ` `                ``zeroCount++; ` `            ``} ` `        ``} ` ` `  `        ``// Print the ratio of positive, ` `        ``// negative, and zero elements ` `        ``// in the array up to four decimal places. ` `        ``System.out.printf(``"%1.4f "``, positiveCount / len); ` `        ``System.out.printf(``"%1.4f "``, negativeCount / len); ` `        ``System.out.printf(``"%1.4f "``, zeroCount / len); ` `        ``System.out.println(); ` `    ``} ` ` `  `    ``// Driver Code. ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` ` `  `        ``// Test Case 1: ` `        ``int``[] a1 = { ``2``, -``1``, ``5``, ``6``, ``0``, -``3` `}; ` `        ``positiveNegativeZero(a1); ` ` `  `        ``// Test Case 2: ` `        ``int``[] a2 = { ``4``, ``0``, -``2``, -``9``, -``7``, ``1` `}; ` `        ``positiveNegativeZero(a2); ` `    ``} ` `} `

 `# Python3 program to find the ratio of positive, ` `# negative, and zero elements in the array. ` ` `  `# Function to find the ratio of ` `# positive, negative, and zero elements ` `def` `positiveNegativeZero(arr): ` ` `  `    ``# Store the array length into the variable len. ` `    ``length ``=` `len``(arr); ` ` `  `    ``# Initialize the postiveCount, negativeCount, and ` `    ``# zeroCountby 0 which will count the total number ` `    ``# of positive, negative and zero elements ` `    ``positiveCount ``=` `0``; ` `    ``negativeCount ``=` `0``; ` `    ``zeroCount ``=` `0``; ` ` `  `    ``# Traverse the array and count the total number of ` `    ``# positive, negative, and zero elements. ` `    ``for` `i ``in` `range``(length): ` `        ``if` `(arr[i] > ``0``): ` `            ``positiveCount ``+``=` `1``; ` `        ``elif``(arr[i] < ``0``): ` `            ``negativeCount ``+``=` `1``; ` `        ``elif``(arr[i] ``=``=` `0``): ` `            ``zeroCount ``+``=` `1``; ` `         `  `    ``# Print the ratio of positive, ` `    ``# negative, and zero elements ` `    ``# in the array up to four decimal places. ` `    ``print``(``"{0:.4f}"``.``format``((positiveCount ``/` `length)), end``=``" "``); ` `    ``print``(``"%1.4f "``%``(negativeCount ``/` `length), end``=``" "``); ` `    ``print``(``"%1.4f "``%``(zeroCount ``/` `length), end``=``" "``); ` `    ``print``(); ` ` `  `# Driver Code. ` `if` `__name__ ``=``=` `'__main__'``: ` ` `  `    ``# Test Case 1: ` `    ``a1 ``=` `[ ``2``, ``-``1``, ``5``, ``6``, ``0``, ``-``3` `]; ` `    ``positiveNegativeZero(a1); ` ` `  `    ``# Test Case 2: ` `    ``a2 ``=` `[ ``4``, ``0``, ``-``2``, ``-``9``, ``-``7``, ``1` `]; ` `    ``positiveNegativeZero(a2); ` `     `  `# This code is contributed by Rajput-Ji `

 `// C# program to find the ratio of positive, ` `// negative, and zero elements in the array. ` `using` `System; ` ` `  `class` `GFG { ` `  `  `    ``// Function to find the ratio of ` `    ``// positive, negative, and zero elements ` `    ``static` `void` `positiveNegativeZero(``int``[] arr) ` `    ``{ ` `  `  `        ``// Store the array length into the variable len. ` `        ``int` `len = arr.Length; ` `  `  `        ``// Initialize the postiveCount, negativeCount, and ` `        ``// zeroCountby 0 which will count the total number ` `        ``// of positive, negative and zero elements ` `        ``float` `positiveCount = 0; ` `        ``float` `negativeCount = 0; ` `        ``float` `zeroCount = 0; ` `  `  `        ``// Traverse the array and count the total number of ` `        ``// positive, negative, and zero elements. ` `        ``for` `(``int` `i = 0; i < len; i++) { ` `            ``if` `(arr[i] > 0) { ` `                ``positiveCount++; ` `            ``} ` `            ``else` `if` `(arr[i] < 0) { ` `                ``negativeCount++; ` `            ``} ` `            ``else` `if` `(arr[i] == 0) { ` `                ``zeroCount++; ` `            ``} ` `        ``} ` `  `  `        ``// Print the ratio of positive, ` `        ``// negative, and zero elements ` `        ``// in the array up to four decimal places. ` `        ``Console.Write(``"{0:F4} "``, positiveCount / len); ` `        ``Console.Write(``"{0:F4} "``, negativeCount / len); ` `        ``Console.Write(``"{0:F4} "``, zeroCount / len); ` `        ``Console.WriteLine(); ` `    ``} ` `  `  `    ``// Driver Code. ` `    ``public` `static` `void` `Main(String []args) ` `    ``{ ` `  `  `        ``// Test Case 1: ` `        ``int``[] a1 = { 2, -1, 5, 6, 0, -3 }; ` `        ``positiveNegativeZero(a1); ` `  `  `        ``// Test Case 2: ` `        ``int``[] a2 = { 4, 0, -2, -9, -7, 1 }; ` `        ``positiveNegativeZero(a2); ` `    ``} ` `} ` ` `  `// This code is contributed by sapnasingh4991 `

Output:
```0.5000 0.3333 0.1667
0.3333 0.5000 0.1667
```

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