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# Ratio of all subarrays of size K

• Last Updated : 17 Mar, 2021

Given an array arr[] and an integer K, the task is to calculate the ratio of all subarrays of size K.

Examples:

Input: arr[] = {24, 3, 2, 1}, K = 3
Output: 4 1.5
Explanation:
All subarrays of size K and their ratio:
Subarray 1: {24, 3, 2} = 24 / 3 / 2 = 4
Subarray 2: {3, 2, 1} = 3 / 2 / 1= 1.5

Input: arr[] = {1, -2, 3, -4, 5, 6}, K = 2
Output: -0.5 -0.666667 -0.75 -0.8 0.833333

Approach: The idea is to iterate over every subarray of size K present in the given array and follow the steps below to solve the problem:

1. Initialize a variable, say ratio with the first element of the subarray.
2. Iterate over the remaining subarray and keep on dividing ratio by the encountered elements one by one.
3. Finally, print the final value of ratio for that subarray.
4. Repeat the above steps for all subarrays.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach` `#include ``using` `namespace` `std;` `// Function to find the ratio of``// all subarrays of size K``int` `calcRatio(``double` `arr[], ``int` `n, ``int` `k)``{` `    ``// Traverse every subarray of size K``    ``for` `(``int` `i = 0; i <= n - k; i++) {` `        ``// Initialize ratio``        ``double` `ratio = arr[i];` `        ``// Calculate ratio of the``        ``// current subarray``        ``for` `(``int` `j = i + 1; j < k + i; j++)``            ``ratio /= arr[j];` `        ``// Print ratio of the subarray``        ``cout << ratio << ``" "``;``    ``}``}` `// Driver Code``int` `main()``{``    ``// Given array``    ``double` `arr[] = { 24, 3, 2, 1 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``int` `k = 3;` `    ``// Function Call``    ``calcRatio(arr, n, k);` `    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``import` `java.util.*;` `class` `GFG{``  ` `// Function to find the ratio of``// all subarrays of size K``static` `void` `calcRatio(``double` `arr[], ``int` `n,``                                    ``int` `k)``{``    ` `    ``// Traverse every subarray of size K``    ``for``(``int` `i = ``0``; i <= n - k; i++)``    ``{``  ` `        ``// Initialize ratio``        ``double` `ratio = arr[i];``  ` `        ``// Calculate ratio of the``        ``// current subarray``        ``for``(``int` `j = i + ``1``; j < k + i; j++)``            ``ratio /= arr[j];``  ` `        ``// Print ratio of the subarray``        ``System.out.print(ratio + ``" "``);``    ``}``}`` ` `// Driver code``public` `static` `void` `main (String[] args)``{``    ` `    ``// Given array``    ``double` `arr[] = { ``24``, ``3``, ``2``, ``1` `};``    ``int` `n = arr.length;``    ``int` `k = ``3``;``    ` `    ``// Function call``    ``calcRatio(arr, n, k);``}``}` `// This code is contributed by offbeat`

## Python3

 `# Python3 implementation``# of the above approach` `# Function to find the ratio of``# all subarrays of size K``def` `calcRatio(arr, n, k):` `    ``# Traverse every subarray``    ``# of size K``    ``for` `i ``in` `range``(n ``-` `k ``+` `1``):` `        ``# Initialize ratio``        ``ratio ``=` `arr[i]` `        ``# Calculate ratio of the``        ``# current subarray``        ``for` `j ``in` `range``(i ``+` `1``, k ``+` `i):``            ``ratio ``=` `ratio ``/` `arr[j]` `        ``# Print ratio of the subarray``        ``print``(ratio, end ``=` `" "``)` `# Given array``arr ``=` `[``24``, ``3``, ``2``, ``1``]``n ``=` `len``(arr)``k ``=` `3` `# Function Call``calcRatio(arr, n, k)` `# This code is contributed by divyeshrabadiya07`

## C#

 `// C# implementation of the above approach``using` `System;`` ` `class` `GFG{``   ` `// Function to find the ratio of``// all subarrays of size K``static` `void` `calcRatio(``double` `[]arr, ``int` `n,``                                    ``int` `k)``{``  ` `    ``// Traverse every subarray of size K``    ``for``(``int` `i = 0; i <= n - k; i++)``    ``{``   ` `        ``// Initialize ratio``        ``double` `ratio = arr[i];``   ` `        ``// Calculate ratio of the``        ``// current subarray``        ``for``(``int` `j = i + 1; j < k + i; j++)``            ``ratio /= arr[j];``   ` `        ``// Print ratio of the subarray``        ``Console.Write(ratio + ``" "``);``    ``}``}``  ` `// Driver code``public` `static` `void` `Main(``string``[] args)``{``     ` `    ``// Given array``    ``double` `[]arr = { 24, 3, 2, 1 };``    ``int` `n = arr.Length;``    ``int` `k = 3;``     ` `    ``// Function call``    ``calcRatio(arr, n, k);``}``}` `// This code is contributed by rutvik_56`

## Javascript

 ``
Output
`4 1.5 `

Time Complexity: O(N2
Auxiliary Space: O(1)

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