Dada una array arr[] de N números, la tarea es encontrar la proporción más grande de subarreglo contiguo de la array dada.
Ejemplos:
Entrada: arr = { -1, 10, 0.1, -8, -2 }
Salida: 100
Explicación:
El subarreglo {10, 0.1} da 10 / 0.1 = 100 que es la relación más grande.Entrada: arr = { 2, 2, 4, -0.2, -1 }
Salida: 20
Explicación:
El subarreglo {4, -0.2, -1} tiene la relación más grande como 20.
Enfoque: La idea es generar todos los subarreglos del arreglo y para cada subarreglo, encontrar la razón del subarreglo como arr[i] / arr[i+1] / arr[i+2] y así sucesivamente. Lleve un registro de la relación máxima y devuélvala al final.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
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[0]);
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
Javascript
<script>
// Javascript program for the above approach
// Function to return maximum
// of two double values
function 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
function maxSubarrayRatio(arr, n)
{
// Variable to store
// the maximum ratio
var maxRatio = -1000000000;
// Compute the product while
// traversing for subarrays
for (var i = 0; i < n; i++)
{
for (var j = i; j < n; j++)
{
var ratio = arr[i];
for (var 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
var arr = [ 2, 2, 4, -0.2, -1 ];
var n = arr.length;
document.write( maxSubarrayRatio(arr, n));
// This code is contributed by rrrtnx.
</script>
Producción
20
Complejidad Temporal: (N 3 )
Espacio Auxiliar: O(1)