# Largest Ratio Contiguous subarray

Given an array arr[] of N numbers, the task is to find the largest ratio of contiguous subarray from the given array.

Examples:

Input: arr = { -1, 10, 0.1, -8, -2 }
Output: 100
Explanation:
The subarray {10, 0.1} gives 10 / 0.1 = 100 which is the largest ratio.

Input: arr = { 2, 2, 4, -0.2, -1 }
Output: 20
Explanation:
The subarray {4, -0.2, -1} has the largest ratio as 20.

Approach: The idea is to generate all the subarrays of the array and for each subarray, find the ratio of the subarray as arr[i] / arr[i+1] / arr[i+2] and so on. Keep track of the maximum ratio and return it at the end.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return maximum ` `// of two double values ` `double` `maximum(``double` `a, ``double` `b) ` `{ ` `    ``// Check if a is greater ` `    ``// than b then return a ` `    ``if` `(a > b) ` `        ``return` `a; ` ` `  `    ``return` `b; ` `} ` ` `  `// Function that returns the ` `// Ratio of max Ratio subarray ` `double` `maxSubarrayRatio( ` `  ``double` `arr[], ``int` `n) ` `{ ` `   `  `    ``// Variable to store ` `    ``// the maximum ratio ` `    ``double` `maxRatio = INT_MIN; ` ` `  `    ``// Compute the product while ` `    ``// traversing for subarrays ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``for` `(``int` `j = i; j < n; j++) { ` `           `  `            ``double` `ratio = arr[i]; ` `           `  `            ``for` `(``int` `k = i + 1; k <= j; k++) { ` `               `  `                ``// Calculate the ratio ` `                ``ratio = ratio / arr[k]; ` `            ``} ` `           `  `            ``// Update max ratio ` `            ``maxRatio = maximum(maxRatio, ratio); ` `        ``} ` `    ``} ` ` `  `    ``// Print the answer ` `    ``return` `maxRatio; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``double` `arr[] = { 2, 2, 4, -0.2, -1 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``cout << maxSubarrayRatio(arr, n); ` `    ``return` `0; ` `}`

## Java

 `// Java program for the above approach ` `class` `GFG{ ` `     `  `// Function to return maximum ` `// of two double values ` `static` `double` `maximum(``double` `a, ``double` `b) ` `{ ` `     `  `    ``// Check if a is greater ` `    ``// than b then return a ` `    ``if` `(a > b) ` `        ``return` `a; ` ` `  `    ``return` `b; ` `} ` ` `  `// Function that returns the ` `// Ratio of max Ratio subarray ` `static` `double` `maxSubarrayRatio(``double` `arr[], ` `                               ``int` `n) ` `{ ` `     `  `    ``// Variable to store ` `    ``// the maximum ratio ` `    ``double` `maxRatio = Integer.MIN_VALUE; ` ` `  `    ``// Compute the product while ` `    ``// traversing for subarrays ` `    ``for``(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` `        ``for``(``int` `j = i; j < n; j++)  ` `        ``{ ` `            ``double` `ratio = arr[i]; ` `             `  `            ``for``(``int` `k = i + ``1``; k <= j; k++) ` `            ``{ ` `                 `  `                ``// Calculate the ratio ` `                ``ratio = ratio / arr[k]; ` `            ``} ` `             `  `            ``// Update max ratio ` `            ``maxRatio = maximum(maxRatio, ratio); ` `        ``} ` `    ``} ` ` `  `    ``// Print the answer ` `    ``return` `maxRatio; ` `} ` `     `  `// Driver code     ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``double` `arr[] = { ``2``, ``2``, ``4``, -``0.2``, -``1` `}; ` `    ``int` `n = arr.length; ` `     `  `    ``System.out.println(maxSubarrayRatio(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by rutvik_56 `

## Python3

 `# Python3 program for the above approach ` `import` `sys ` ` `  `# Function to return maximum ` `# of two double values ` `def` `maximum(a, b): ` ` `  `    ``# Check if a is greater ` `    ``# than b then return a ` `    ``if` `(a > b): ` `        ``return` `a ` ` `  `    ``return` `b ` ` `  `# Function that returns the ` `# Ratio of max Ratio subarray ` `def` `maxSubarrayRatio(arr, n): ` ` `  `    ``# Variable to store ` `    ``# the maximum ratio ` `    ``maxRatio ``=` `-``sys.maxsize ``-` `1` ` `  `    ``# Compute the product while ` `    ``# traversing for subarrays ` `    ``for` `i ``in` `range``(n): ` `        ``for` `j ``in` `range``(i, n): ` `            ``ratio ``=` `arr[i] ` `         `  `            ``for` `k ``in` `range``(i ``+` `1``, j ``+` `1``): ` `             `  `                ``# Calculate the ratio ` `                ``ratio ``=` `ratio ``/``/` `arr[k] ` `         `  `            ``# Update max ratio ` `            ``maxRatio ``=` `maximum(maxRatio, ratio) ` `         `  `    ``# Print the answer ` `    ``return` `int``(maxRatio) ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` `     `  `    ``arr ``=` `[ ``2``, ``2``, ``4``, ``-``0.2``, ``-``1` `] ` `    ``n ``=` `len``(arr) ` `     `  `    ``print``(maxSubarrayRatio(arr, n)) ` ` `  `# This code is contributed by chitranayal `

## C#

 `// C# program for the above approach ` `using` `System; ` ` `  `class` `GFG{ ` `     `  `// Function to return maximum ` `// of two double values ` `static` `double` `maximum(``double` `a, ``double` `b) ` `{ ` `     `  `    ``// Check if a is greater ` `    ``// than b then return a ` `    ``if` `(a > b) ` `        ``return` `a; ` ` `  `    ``return` `b; ` `} ` ` `  `// Function that returns the ` `// Ratio of max Ratio subarray ` `static` `double` `maxSubarrayRatio(``double` `[]arr, ` `                               ``int` `n) ` `{ ` `     `  `    ``// Variable to store ` `    ``// the maximum ratio ` `    ``double` `maxRatio = ``int``.MinValue; ` ` `  `    ``// Compute the product while ` `    ``// traversing for subarrays ` `    ``for``(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        ``for``(``int` `j = i; j < n; j++)  ` `        ``{ ` `            ``double` `ratio = arr[i]; ` `             `  `            ``for``(``int` `k = i + 1; k <= j; k++) ` `            ``{ ` `                 `  `                ``// Calculate the ratio ` `                ``ratio = ratio / arr[k]; ` `            ``} ` `             `  `            ``// Update max ratio ` `            ``maxRatio = maximum(maxRatio, ratio); ` `        ``} ` `    ``} ` ` `  `    ``// Print the answer ` `    ``return` `maxRatio; ` `} ` `     `  `// Driver code  ` `public` `static` `void` `Main(String[] args)  ` `{ ` `    ``double` `[]arr = { 2, 2, 4, -0.2, -1 }; ` `    ``int` `n = arr.Length; ` `     `  `    ``Console.WriteLine(maxSubarrayRatio(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar  `

Output

```20
```

Time Complexity: (N3)
Auxiliary Space: O(1)

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