Dada una array que contiene enteros positivos y negativos, encuentre el producto del subarreglo de producto máximo. La complejidad del tiempo esperado es O(n) y solo se puede usar O(1) espacio extra.
Ejemplos:
Input: arr[] = {6, -3, -10, 0, 2} Output: 180 // The subarray is {6, -3, -10} Input: arr[] = {-1, -3, -10, 0, 60} Output: 60 // The subarray is {60} Input: arr[] = {-1, -2, -3, 4} Output: 24 // The subarray is {-2, -3, 4} Input: arr[] = {-10} Output: 0 // An empty array is also subarray // and product of empty subarray is // considered as 0.
Hemos discutido una solución a este problema aquí .
En este post se discute una solución interesante. La idea se basa en el hecho de que el producto máximo general es el máximo de los dos siguientes:
- Producto máximo en recorrido de izquierda a derecha.
- Producto máximo en recorrido de derecha a izquierda
Por ejemplo, considere la entrada de la tercera muestra anterior {-1, -2, -3, 4}. Si recorremos la array solo hacia adelante (considerando -1 como parte de la salida), el producto máximo será 2. Si recorremos la array hacia atrás (considerando 4 como parte de la salida), el producto máximo será 24, es decir; {-2, -3, 4}.
Una cosa importante es manejar 0’s. Necesitamos calcular una nueva suma hacia adelante (o hacia atrás) cada vez que veamos 0.
A continuación se muestra la implementación de la idea anterior:
C++
// C++ program to find maximum product subarray #include<bits/stdc++.h> using namespace std; // Function for maximum product int max_product(int arr[], int n) { // Initialize maximum products in forward and // backward directions int max_fwd = INT_MIN, max_bkd = INT_MIN; // Initialize current product int max_till_now = 1; //check if zero is present in an array or not bool isZero=false; // max_fwd for maximum contiguous product in // forward direction // max_bkd for maximum contiguous product in // backward direction // iterating within forward direction in array for (int i=0; i<n; i++) { // if arr[i]==0, it is breaking condition // for contiguous subarray max_till_now = max_till_now*arr[i]; if (max_till_now == 0) { isZero=true; max_till_now = 1; continue; } if (max_fwd < max_till_now) // update max_fwd max_fwd = max_till_now; } max_till_now = 1; // iterating within backward direction in array for (int i=n-1; i>=0; i--) { max_till_now = max_till_now * arr[i]; if (max_till_now == 0) { isZero=true; max_till_now = 1; continue; } // update max_bkd if (max_bkd < max_till_now) max_bkd = max_till_now; } // return max of max_fwd and max_bkd int res = max(max_fwd, max_bkd); // Product should not be negative. // (Product of an empty subarray is // considered as 0) if(isZero) return max(res, 0); return res; } // Driver Program to test above function int main() { int arr[] = {-1, -2, -3, 4}; int n = sizeof(arr)/sizeof(arr[0]); cout << max_product(arr, n) << endl; return 0; }
Java
// Java program to find // maximum product subarray import java.io.*; class GFG { // Function for maximum product static int max_product(int arr[], int n) { // Initialize maximum products in // forward and backward directions int max_fwd = Integer.MIN_VALUE, max_bkd = Integer.MIN_VALUE; //check if zero is present in an array or not boolean isZero=false; // Initialize current product int max_till_now = 1; // max_fwd for maximum contiguous // product in forward direction // max_bkd for maximum contiguous // product in backward direction // iterating within forward // direction in array for (int i = 0; i < n; i++) { // if arr[i]==0, it is breaking // condition for contiguous subarray max_till_now = max_till_now * arr[i]; if (max_till_now == 0) { isZero=true; max_till_now = 1; continue; } // update max_fwd if (max_fwd < max_till_now) max_fwd = max_till_now; } max_till_now = 1; // iterating within backward // direction in array for (int i = n - 1; i >= 0; i--) { max_till_now = max_till_now * arr[i]; if (max_till_now == 0) { isZero=true; max_till_now = 1; continue; } // update max_bkd if (max_bkd < max_till_now) max_bkd = max_till_now; } // return max of max_fwd and max_bkd int res = Math. max(max_fwd, max_bkd); // Product should not be negative. // (Product of an empty subarray is // considered as 0) if(isZero) return Math.max(res, 0); return res; } // Driver Code public static void main (String[] args) { int arr[] = {-1, -2, -3, 4}; int n = arr.length; System.out.println( max_product(arr, n) ); } } // This code is contributed by anuj_67.
Python3
# Python3 program to find # maximum product subarray import sys # Function for maximum product def max_product(arr, n): # Initialize maximum products # in forward and backward directions max_fwd = -sys.maxsize - 1 max_bkd = -sys.maxsize - 1 #check if zero is present in an array or not isZero=False; # Initialize current product max_till_now = 1 # max_fwd for maximum contiguous # product in forward direction # max_bkd for maximum contiguous # product in backward direction # iterating within forward # direction in array for i in range(n): # if arr[i]==0, it is breaking # condition for contiguous subarray max_till_now = max_till_now * arr[i] if (max_till_now == 0): isZero=True max_till_now = 1; continue if (max_fwd < max_till_now): #update max_fwd max_fwd = max_till_now max_till_now = 1 # iterating within backward # direction in array for i in range(n - 1, -1, -1): max_till_now = max_till_now * arr[i] if (max_till_now == 0): isZero=True max_till_now = 1 continue # update max_bkd if (max_bkd < max_till_now) : max_bkd = max_till_now # return max of max_fwd and max_bkd res = max(max_fwd, max_bkd) # Product should not be negative. # (Product of an empty subarray is # considered as 0) if isZero==True : return max(res, 0) return res # Driver Code arr = [-1, -2, -3, 4] n = len(arr) print(max_product(arr, n)) # This code is contributed # by Yatin Gupta
C#
// C# program to find maximum product // subarray using System; class GFG { // Function for maximum product static int max_product(int []arr, int n) { // Initialize maximum products in // forward and backward directions int max_fwd = int.MinValue, max_bkd = int.MinValue; // Initialize current product int max_till_now = 1; // max_fwd for maximum contiguous // product in forward direction // max_bkd for maximum contiguous // product in backward direction // iterating within forward // direction in array for (int i = 0; i < n; i++) { // if arr[i]==0, it is breaking // condition for contiguous subarray max_till_now = max_till_now * arr[i]; if (max_till_now == 0) { max_till_now = 1; continue; } // update max_fwd if (max_fwd < max_till_now) max_fwd = max_till_now; } max_till_now = 1; // iterating within backward // direction in array for (int i = n - 1; i >= 0; i--) { max_till_now = max_till_now * arr[i]; if (max_till_now == 0) { max_till_now = 1; continue; } // update max_bkd if (max_bkd < max_till_now) max_bkd = max_till_now; } // return max of max_fwd and max_bkd int res = Math. Max(max_fwd, max_bkd); // Product should not be negative. // (Product of an empty subarray is // considered as 0) return Math.Max(res, 0); } // Driver Code public static void Main () { int []arr = {-1, -2, -3, 4}; int n = arr.Length; Console.Write( max_product(arr, n) ); } } // This code is contributed by nitin mittal.
PHP
<?php // PHP program to find maximum // product subarray // Function for maximum product function max_product( $arr, $n) { // Initialize maximum products // in forward and backward // directions $max_fwd = PHP_INT_MIN; $max_bkd = PHP_INT_MIN; // Initialize current product $max_till_now = 1; // max_fwd for maximum contiguous // product in forward direction // max_bkd for maximum contiguous // product in backward direction // iterating within forward direction // in array for ($i = 0; $i < $n; $i++) { // if arr[i]==0, it is // breaking condition // for contiguous subarray $max_till_now = $max_till_now * $arr[$i]; if ($max_till_now == 0) { $max_till_now = 1; continue; } // update max_fwd if ($max_fwd < $max_till_now) $max_fwd = $max_till_now; } $max_till_now = 1; // iterating within backward // direction in array for($i = $n - 1; $i >= 0; $i--) { $max_till_now = $max_till_now * $arr[$i]; if ($max_till_now == 0) { $max_till_now = 1; continue; } // update max_bkd if ($max_bkd < $max_till_now) $max_bkd = $max_till_now; } // return max of max_fwd // and max_bkd $res = max($max_fwd, $max_bkd); // Product should not be negative. // (Product of an empty subarray is // considered as 0) return max($res, 0); } // Driver Code $arr = array(-1, -2, -3, 4); $n = count($arr); echo max_product($arr, $n); // This code is contributed by anuj_67. ?>
Javascript
<script> // JavaScript program to find maximum product // subarray // Function for maximum product function max_product(arr, n) { // Initialize maximum products in // forward and backward directions let max_fwd = Number.MIN_VALUE, max_bkd = Number.MIN_VALUE; // Initialize current product let max_till_now = 1; // max_fwd for maximum contiguous // product in forward direction // max_bkd for maximum contiguous // product in backward direction // iterating within forward // direction in array for (let i = 0; i < n; i++) { // if arr[i]==0, it is breaking // condition for contiguous subarray max_till_now = max_till_now * arr[i]; if (max_till_now == 0) { max_till_now = 1; continue; } // update max_fwd if (max_fwd < max_till_now) max_fwd = max_till_now; } max_till_now = 1; // iterating within backward // direction in array for (let i = n - 1; i >= 0; i--) { max_till_now = max_till_now * arr[i]; if (max_till_now == 0) { max_till_now = 1; continue; } // update max_bkd if (max_bkd < max_till_now) max_bkd = max_till_now; } // return max of max_fwd and max_bkd let res = Math.max(max_fwd, max_bkd); // Product should not be negative. // (Product of an empty subarray is // considered as 0) return Math.max(res, 0); } let arr = [-1, -2, -3, 4]; let n = arr.length; document.write(max_product(arr, n) ); </script>
24
Tiempo Complejidad : O(n)
Espacio Auxiliar : O(1)
Tenga en cuenta que la solución anterior requiere dos recorridos de una array, mientras que la solución anterior requiere solo un recorrido.
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA