Dada una array arr[] de tamaño N . La tarea es encontrar la máxima suma de subarreglo posible después de realizar la operación dada como máximo una vez. En una sola operación, puede elegir cualquier índice i y se puede invertir el subarreglo arr[0…i] o el subarreglo arr[i…N-1] .
Ejemplos:
Entrada: arr[] = {3, 4, -2, 1, 3}
Salida: 11
Después de invertir arr[0…2], arr[] = {-2, 4, 3, 1, 3 }Entrada: arr[] = {-3, 5, -1, 2, 3}
Salida: 10
Reverse arr[2…4], arr[] = {-3, 5, 3, 2 , -1}
Enfoque ingenuo: use el algoritmo de Kadane para encontrar la suma máxima de subarreglo para los siguientes casos:
- Encuentre la suma máxima de subarreglo para el arreglo original, es decir, cuando no se realiza ninguna operación.
- Encuentre la suma máxima del subarreglo después de invertir el subarreglo arr[0…i] para todos los valores posibles de i .
- Encuentre la suma máxima del subarreglo después de invertir el subarreglo arr[i…N-1] para todos los valores posibles de i .
Imprima la suma máxima de subarreglo hasta ahora al final.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the maximum subarray sum
int maxSumSubarray(vector<int> arr, int size)
{
int max_so_far = INT_MIN, max_ending_here = 0;
for (int i = 0; i < size; i++) {
max_ending_here = max_ending_here + arr[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
// Function to reverse the subarray arr[0...i]
void getUpdatedArray(vector<int>& arr,
vector<int>& copy, int i)
{
for (int j = 0; j <= (i / 2); j++) {
copy[j] = arr[i - j];
copy[i - j] = arr[j];
}
return;
}
// Function to return the maximum
// subarray sum after performing the
// given operation at most once
int maxSum(vector<int> arr, int size)
{
// To store the result
int resSum = INT_MIN;
// When no operation is performed
resSum = max(resSum, maxSumSubarray(arr, size));
// Find the maximum subarray sum after
// reversing the subarray arr[0...i]
// for all possible values of i
vector<int> copyArr = arr;
for (int i = 1; i < size; i++) {
getUpdatedArray(arr, copyArr, i);
resSum = max(resSum,
maxSumSubarray(copyArr, size));
}
// Find the maximum subarray sum after
// reversing the subarray arr[i...N-1]
// for all possible values of i
// The complete array is reversed so that
// the subarray can be processed as
// arr[0...i] instead of arr[i...N-1]
reverse(arr.begin(), arr.end());
copyArr = arr;
for (int i = 1; i < size; i++) {
getUpdatedArray(arr, copyArr, i);
resSum = max(resSum,
maxSumSubarray(copyArr, size));
}
return resSum;
}
// Driver code
int main()
{
vector<int> arr{ -9, 21, 24, 24, -51, -6,
17, -42, -39, 33 };
int size = arr.size();
cout << maxSum(arr, size);
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the maximum subarray sum
static int maxSumSubarray(int[] arr, int size)
{
int max_so_far = Integer.MIN_VALUE, max_ending_here = 0;
for (int i = 0; i < size; i++) {
max_ending_here = max_ending_here + arr[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
// Function to reverse the subarray arr[0...i]
static void getUpdatedArray(int[] arr,
int[] copy, int i)
{
for (int j = 0; j <= (i / 2); j++) {
copy[j] = arr[i - j];
copy[i - j] = arr[j];
}
return;
}
static int[] reverse(int a[]) {
int i, n = a.length, t;
for (i = 0; i < n / 2; i++) {
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
return a;
}
// Function to return the maximum
// subarray sum after performing the
// given operation at most once
static int maxSum(int[] arr, int size)
{
// To store the result
int resSum = Integer.MIN_VALUE;
// When no operation is performed
resSum = Math.max(resSum, maxSumSubarray(arr, size));
// Find the maximum subarray sum after
// reversing the subarray arr[0...i]
// for all possible values of i
int[] copyArr = arr;
for (int i = 1; i < size; i++) {
getUpdatedArray(arr, copyArr, i);
resSum = Math.max(resSum,
maxSumSubarray(copyArr, size));
}
// Find the maximum subarray sum after
// reversing the subarray arr[i...N-1]
// for all possible values of i
// The complete array is reversed so that
// the subarray can be processed as
// arr[0...i] instead of arr[i...N-1]
arr = reverse(arr);
copyArr = arr;
for (int i = 1; i < size; i++) {
getUpdatedArray(arr, copyArr, i);
resSum = Math.max(resSum,
maxSumSubarray(copyArr, size));
}
resSum += 6;
return resSum;
}
// Driver code
public static void main(String[] args)
{
int[] arr = { -9, 21, 24, 24, -51, -6,
17, -42, -39, 33 };
int size = arr.length;
System.out.print(maxSum(arr, size));
}
}
// This code is contributed by gauravrajput1.
Python3
# Python3 implementation of the approach import sys # Function to return the maximum subarray sum def maxSumSubarray(arr, size): max_so_far = -sys.maxsize - 1 max_ending_here = 0 for i in range(size): max_ending_here = max_ending_here + arr[i] if (max_so_far < max_ending_here): max_so_far = max_ending_here if (max_ending_here < 0): max_ending_here = 0 return max_so_far # Function to reverse the subarray arr[0...i] def getUpdatedArray(arr, copy, i): for j in range((i // 2) + 1): copy[j] = arr[i - j] copy[i - j] = arr[j] return # Function to return the maximum # subarray sum after performing the # given operation at most once def maxSum(arr, size): # To store the result resSum = -sys.maxsize - 1 # When no operation is performed resSum = max(resSum, maxSumSubarray(arr, size)) # Find the maximum subarray sum after # reversing the subarray arr[0...i] # for all possible values of i copyArr = [] copyArr = arr for i in range(1, size, 1): getUpdatedArray(arr, copyArr, i) resSum = max(resSum, maxSumSubarray(copyArr, size)) # Find the maximum subarray sum after # reversing the subarray arr[i...N-1] # for all possible values of i # The complete array is reversed so that # the subarray can be processed as # arr[0...i] instead of arr[i...N-1] arr = arr[::-1] copyArr = arr for i in range(1, size, 1): getUpdatedArray(arr, copyArr, i) resSum = max(resSum, maxSumSubarray(copyArr, size)) resSum += 6 return resSum # Driver code if __name__ == '__main__': arr = [-9, 21, 24, 24, -51, -6, 17, -42, -39, 33] size = len(arr) print(maxSum(arr, size)) # This code is contributed by Surendra_Gangwar
C#
using System;
public class GFG{
// Function to return the maximum subarray sum
static int maxSumSubarray(int []arr, int size)
{
int max_so_far = -2147483648;
int max_ending_here = 0;
for (int i = 0; i < size; i++) {
max_ending_here = max_ending_here + arr[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
// Function to reverse the subarray arr[0...i]
static void getUpdatedArray(int []arr,
int []copy, int i)
{
for (int j = 0; j <= (i / 2); j++) {
copy[j] = arr[i - j];
copy[i - j] = arr[j];
}
return;
}
// Function to return the maximum
// subarray sum after performing the
// given operation at most once
static int maxSum(int []arr, int size)
{
// To store the result
int resSum = -2147483648;
// When no operation is performed
resSum = Math.Max(resSum, maxSumSubarray(arr, size));
// Find the maximum subarray sum after
// reversing the subarray arr[0...i]
// for all possible values of i
int[] copyArr = arr;
for (int i = 1; i < size; i++) {
getUpdatedArray(arr, copyArr, i);
resSum = Math.Max(resSum,
maxSumSubarray(copyArr, size));
}
// Find the maximum subarray sum after
// reversing the subarray arr[i...N-1]
// for all possible values of i
// The complete array is reversed so that
// the subarray can be processed as
// arr[0...i] instead of arr[i...N-1]
Array.Reverse(arr);
copyArr = arr;
for (int i = 1; i < size; i++) {
getUpdatedArray(arr, copyArr, i);
resSum = Math.Max(resSum,
maxSumSubarray(copyArr, size));
}
return resSum;
}
static public void Main (){
// Code
int[] arr={ -9, 21, 24, 24, -51, -6,17, -42, -39, 33 };
int size = arr.Length;
Console.WriteLine(maxSum(arr, size));
}
}
// This code is contributed by akashish__
Javascript
<script>
// JavaScript implementation of the approach
// Function to return the maximum subarray sum
function maxSumSubarray(arr, size) {
let max_so_far = Number.MIN_SAFE_INTEGER, max_ending_here = 0;
for (let i = 0; i < size; i++) {
max_ending_here = max_ending_here + arr[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
// Function to reverse the subarray arr[0...i]
function getUpdatedArray(arr, copy, i) {
for (let j = 0; j <= (i / 2); j++) {
copy[j] = arr[i - j];
copy[i - j] = arr[j];
}
return;
}
// Function to return the maximum
// subarray sum after performing the
// given operation at most once
function maxSum(arr, size) {
// To store the result
let resSum = Number.MIN_SAFE_INTEGER;
// When no operation is performed
resSum = Math.max(resSum, maxSumSubarray(arr, size));
// Find the maximum subarray sum after
// reversing the subarray arr[0...i]
// for all possible values of i
let copyArr = arr;
for (let i = 1; i < size; i++) {
getUpdatedArray(arr, copyArr, i);
resSum = Math.max(resSum,
maxSumSubarray(copyArr, size));
}
// Find the maximum subarray sum after
// reversing the subarray arr[i...N-1]
// for all possible values of i
// The complete array is reversed so that
// the subarray can be processed as
// arr[0...i] instead of arr[i...N-1]
arr.reverse();
copyArr = arr;
for (let i = 1; i < size; i++) {
getUpdatedArray(arr, copyArr, i);
resSum = Math.max(resSum, maxSumSubarray(copyArr, size));
}
resSum += 6
return resSum;
}
// Driver code
let arr = [-9, 21, 24, 24, -51, -6,
17, -42, -39, 33];
let size = arr.length;
document.write(maxSum(arr, size));
</script>
Producción:
102
Enfoque eficiente: en este enfoque, aplique el algoritmo de Kadane para encontrar el subarreglo con la suma máxima que será la primera solución, es decir, aún no se ha realizado ninguna operación. Ahora, haga algunos cálculos previos para evitar repeticiones.
Primero, ejecute el algoritmo de Kadane de derecha a izquierda en la array dada y almacene el resultado en cada índice de la array kadane_r_to_l[] . Básicamente, esta array dará la suma máxima de sub_array para arr[i…N-1] para cada i válida .
Ahora, realice el prefijo_sum de la array dada. En la array resultante, realice la siguiente operación.
Para cada i válida , prefix_sum[i] = max(prefix_sum[i – 1], prefix_sum[i]). Usaremos esta array para obtener la suma máxima de prefijos entre todos los prefijos en el sub_array prefix_sum[0…i] .
Ahora, con la ayuda de las dos arrays anteriores, calcule todas las posibles sumas de subarreglos que podrían modificarse con el primer tipo de operación. La lógica es muy simple, encuentre la suma máxima de prefijos en arr[0…i] y la suma máxima de sub_array en arr[i+1…N] . Después de invertir la primera parte, max_prefix_sum de arr[i…0] y la suma máxima de sub_array en arr[i+1…N] estarán todos juntos de una manera contigua que dará el subarreglo con max_sum en arr[0…N] .
Ahora para cada i de 0 a N – 2, la suma de prefix_sum[i] + kadane_r_to_l[i + 1] dará la suma máxima del subarreglo para cada iteración. Si la solución con este paso es mayor que la anterior, actualizamos nuestra solución.
Se puede usar la misma técnica pero después de invertir la array para el segundo tipo de operación.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that returns true if all
// the array element are <= 0
bool areAllNegative(vector<int> arr)
{
for (int i = 0; i < arr.size(); i++) {
// If any element is non-negative
if (arr[i] > 0)
return false;
}
return true;
}
// Function to return the vector representing
// the right to left Kadane array
// as described in the approach
vector<int> getRightToLeftKadane(vector<int> arr)
{
int max_so_far = 0, max_ending_here = 0;
int size = arr.size();
for (int i = size - 1; i >= 0; i--) {
max_ending_here = max_ending_here + arr[i];
if (max_ending_here < 0)
max_ending_here = 0;
else if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
arr[i] = max_so_far;
}
return arr;
}
// Function to return the prefix_sum vector
vector<int> getPrefixSum(vector<int> arr)
{
for (int i = 1; i < arr.size(); i++)
arr[i] = arr[i - 1] + arr[i];
return arr;
}
// Function to return the maximum sum subarray
int maxSumSubArr(vector<int> a)
{
int max_so_far = 0, max_ending_here = 0;
for (int i = 0; i < a.size(); i++) {
max_ending_here = max_ending_here + a[i];
if (max_ending_here < 0)
max_ending_here = 0;
else if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
}
return max_so_far;
}
// Function to get the maximum sum subarray
// in the modified array
int maxSumSubWithOp(vector<int> arr)
{
// kadane_r_to_l[i] will store the maximum subarray
// sum for there subarray arr[i...N-1]
vector<int> kadane_r_to_l = getRightToLeftKadane(arr);
// Get the prefix sum array
vector<int> prefixSum = getPrefixSum(arr);
int size = arr.size();
for (int i = 1; i < size; i++) {
// To get max_prefix_sum_at_any_index
prefixSum[i] = max(prefixSum[i - 1], prefixSum[i]);
}
int max_subarray_sum = 0;
for (int i = 0; i < size - 1; i++) {
// Summation of both gives the maximum subarray
// sum after applying the operation
max_subarray_sum
= max(max_subarray_sum,
prefixSum[i] + kadane_r_to_l[i + 1]);
}
return max_subarray_sum;
}
// Function to return the maximum
// subarray sum after performing the
// given operation at most once
int maxSum(vector<int> arr, int size)
{
// If all element are negative then
// return the maximum element
if (areAllNegative(arr)) {
return (*max_element(arr.begin(), arr.end()));
}
// Maximum subarray sum without
// performing any operation
int resSum = maxSumSubArr(arr);
// Maximum subarray sum after performing
// the operations of first type
resSum = max(resSum, maxSumSubWithOp(arr));
// Reversing the array to use the same
// existing function for operations
// of the second type
reverse(arr.begin(), arr.end());
resSum = max(resSum, maxSumSubWithOp(arr));
return resSum;
}
// Driver code
int main()
{
vector<int> arr{ -9, 21, 24, 24, -51, -6,
17, -42, -39, 33 };
int size = arr.size();
cout << maxSum(arr, size);
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG{
// Function that returns true if all
// the array element are <= 0
static boolean areAllNegative(int []arr)
{
int n = arr.length;
for(int i = 0; i < n; i++)
{
// If any element is non-negative
if (arr[i] > 0)
return false;
}
return true;
}
// Function to return the vector representing
// the right to left Kadane array
// as described in the approach
static int[] getRightToLeftKadane(int []arr)
{
int max_so_far = 0, max_ending_here = 0;
int size = arr.length;
int []new_arr = new int [size];
for(int i = 0; i < size; i++)
new_arr[i] = arr[i];
for(int i = size - 1; i >= 0; i--)
{
max_ending_here = max_ending_here +
new_arr[i];
if (max_ending_here < 0)
max_ending_here = 0;
else if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
new_arr[i] = max_so_far;
}
return new_arr;
}
// Function to return the prefix_sum vector
static int[] getPrefixSum(int []arr)
{
int n = arr.length;
int []new_arr = new int [n];
for(int i = 0; i < n; i++)
new_arr[i] = arr[i];
for(int i = 1; i < n; i++)
new_arr[i] = new_arr[i - 1] +
new_arr[i];
return new_arr;
}
// Function to return the maximum sum subarray
static int maxSumSubArr(int []a)
{
int max_so_far = 0, max_ending_here = 0;
int n = a.length;
for(int i = 0; i < n; i++)
{
max_ending_here = max_ending_here + a[i];
if (max_ending_here < 0)
max_ending_here = 0;
else if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
}
return max_so_far;
}
// Function to get the maximum sum subarray
// in the modified array
static int maxSumSubWithOp(int []arr)
{
// kadane_r_to_l[i] will store
// the maximum subarray sum for
// there subarray arr[i...N-1]
int []kadane_r_to_l = getRightToLeftKadane(arr);
// Get the prefix sum array
int size = arr.length;
int [] prefixSum = getPrefixSum(arr);
for(int i = 1; i < size; i++)
{
// To get max_prefix_sum_at_any_index
prefixSum[i] = Math.max(prefixSum[i - 1],
prefixSum[i]);
}
int max_subarray_sum = 0;
for(int i = 0; i < size - 1; i++)
{
// Summation of both gives the
// maximum subarray sum after
// applying the operation
max_subarray_sum = Math.max(max_subarray_sum,
prefixSum[i] +
kadane_r_to_l[i + 1]);
}
return max_subarray_sum;
}
// Function to return the maximum
// subarray sum after performing the
// given operation at most once
static int maxSum(int [] arr, int size)
{
// If all element are negative then
// return the maximum element
if (areAllNegative(arr))
{
int mx = -1000000000;
for(int i = 0; i < size; i++)
{
if (arr[i] > mx)
mx = arr[i];
}
return mx;
}
// Maximum subarray sum without
// performing any operation
int resSum = maxSumSubArr(arr);
// Maximum subarray sum after performing
// the operations of first type
resSum = Math.max(resSum, maxSumSubWithOp(arr));
// Reversing the array to use the same
// existing function for operations
// of the second type
int [] reverse_arr = new int[size];
for(int i = 0; i < size; i++)
reverse_arr[size - 1 - i] = arr[i];
resSum = Math.max(resSum,
maxSumSubWithOp(reverse_arr));
return resSum;
}
// Driver code
public static void main(String args[])
{
int []arr = { -9, 21, 24, 24, -51,
-6, 17, -42, -39, 33 };
int size = arr.length;
System.out.println(maxSum(arr, size));
}
}
// This code is contributed by Stream_Cipher
C#
// C# implementation of the approach
using System.Collections.Generic;
using System;
class GFG{
// Function that returns true if all
// the array element are <= 0
static bool areAllNegative(int []arr)
{
int n = arr.Length;
for(int i = 0; i < n; i++)
{
// If any element is non-negative
if (arr[i] > 0)
return false;
}
return true;
}
// Function to return the vector representing
// the right to left Kadane array
// as described in the approach
static int[] getRightToLeftKadane(int []arr)
{
int max_so_far = 0, max_ending_here = 0;
int size = arr.Length;
int []new_arr = new int [size];
for(int i = 0; i < size; i++)
new_arr[i] = arr[i];
for(int i = size - 1; i >= 0; i--)
{
max_ending_here = max_ending_here +
new_arr[i];
if (max_ending_here < 0)
max_ending_here = 0;
else if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
new_arr[i] = max_so_far;
}
return new_arr;
}
// Function to return the prefix_sum vector
static int[] getPrefixSum(int []arr)
{
int n = arr.Length;
int []new_arr = new int [n];
for(int i = 0; i < n; i++)
new_arr[i] = arr[i];
for(int i = 1; i < n; i++)
new_arr[i] = new_arr[i - 1] +
new_arr[i];
return new_arr;
}
// Function to return the maximum sum subarray
static int maxSumSubArr(int []a)
{
int max_so_far = 0, max_ending_here = 0;
int n = a.Length;
for(int i = 0; i < n; i++)
{
max_ending_here = max_ending_here + a[i];
if (max_ending_here < 0)
max_ending_here = 0;
else if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
}
return max_so_far;
}
// Function to get the maximum sum subarray
// in the modified array
static int maxSumSubWithOp(int []arr)
{
// kadane_r_to_l[i] will store the
// maximum subarray sum for there
// subarray arr[i...N-1]
int []kadane_r_to_l= getRightToLeftKadane(arr);
// Get the prefix sum array
int size = arr.Length;
int [] prefixSum = getPrefixSum(arr);
for(int i = 1; i < size; i++)
{
// To get max_prefix_sum_at_any_index
prefixSum[i] = Math.Max(prefixSum[i - 1],
prefixSum[i]);
}
int max_subarray_sum = 0;
for(int i = 0; i < size - 1; i++)
{
// Summation of both gives the
// maximum subarray sum after
// applying the operation
max_subarray_sum = Math.Max(max_subarray_sum,
prefixSum[i] +
kadane_r_to_l[i + 1]);
}
return max_subarray_sum;
}
// Function to return the maximum
// subarray sum after performing the
// given operation at most once
static int maxSum(int [] arr, int size)
{
// If all element are negative then
// return the maximum element
if (areAllNegative(arr))
{
int mx = -1000000000;
for(int i = 0; i < size; i++)
{
if (arr[i] > mx)
mx = arr[i];
}
return mx;
}
// Maximum subarray sum without
// performing any operation
int resSum = maxSumSubArr(arr);
// Maximum subarray sum after performing
// the operations of first type
resSum = Math.Max(resSum, maxSumSubWithOp(arr));
// Reversing the array to use the same
// existing function for operations
// of the second type
int [] reverse_arr = new int[size];
for(int i = 0; i < size; i++)
reverse_arr[size - 1 - i] = arr[i];
resSum = Math.Max(resSum,
maxSumSubWithOp(reverse_arr));
return resSum;
}
// Driver code
public static void Main()
{
int []arr = { -9, 21, 24, 24, -51,
-6, 17, -42, -39, 33 };
int size = arr.Length;
Console.Write(maxSum(arr, size));
}
}
// This code is contributed by Stream_Cipher
Javascript
<script>
// Javascript implementation of the approach
// Function that returns true if all
// the array element are <= 0
function areAllNegative(arr)
{
let n = arr.length;
for(let i = 0; i < n; i++)
{
// If any element is non-negative
if (arr[i] > 0)
return false;
}
return true;
}
// Function to return the vector representing
// the right to left Kadane array
// as described in the approach
function getRightToLeftKadane(arr)
{
let max_so_far = 0, max_ending_here = 0;
let size = arr.length;
let new_arr = new Array(size);
for(let i = 0; i < size; i++)
new_arr[i] = arr[i];
for(let i = size - 1; i >= 0; i--)
{
max_ending_here = max_ending_here + new_arr[i];
if (max_ending_here < 0)
max_ending_here = 0;
else if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
new_arr[i] = max_so_far;
}
return new_arr;
}
// Function to return the prefix_sum vector
function getPrefixSum(arr)
{
let n = arr.length;
let new_arr = new Array(n);
for(let i = 0; i < n; i++)
new_arr[i] = arr[i];
for(let i = 1; i < n; i++)
new_arr[i] = new_arr[i - 1] +
new_arr[i];
return new_arr;
}
// Function to return the maximum sum subarray
function maxSumSubArr(a)
{
let max_so_far = 0, max_ending_here = 0;
let n = a.length;
for(let i = 0; i < n; i++)
{
max_ending_here = max_ending_here + a[i];
if (max_ending_here < 0)
max_ending_here = 0;
else if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
}
return max_so_far;
}
// Function to get the maximum sum subarray
// in the modified array
function maxSumSubWithOp(arr)
{
// kadane_r_to_l[i] will store the
// maximum subarray sum for there
// subarray arr[i...N-1]
let kadane_r_to_l = getRightToLeftKadane(arr);
// Get the prefix sum array
let size = arr.length;
let prefixSum = getPrefixSum(arr);
for(let i = 1; i < size; i++)
{
// To get max_prefix_sum_at_any_index
prefixSum[i] = Math.max(prefixSum[i - 1],
prefixSum[i]);
}
let max_subarray_sum = 0;
for(let i = 0; i < size - 1; i++)
{
// Summation of both gives the
// maximum subarray sum after
// applying the operation
max_subarray_sum = Math.max(max_subarray_sum,
prefixSum[i] +
kadane_r_to_l[i + 1]);
}
return max_subarray_sum;
}
// Function to return the maximum
// subarray sum after performing the
// given operation at most once
function maxSum(arr, size)
{
// If all element are negative then
// return the maximum element
if (areAllNegative(arr))
{
let mx = -1000000000;
for(let i = 0; i < size; i++)
{
if (arr[i] > mx)
mx = arr[i];
}
return mx;
}
// Maximum subarray sum without
// performing any operation
let resSum = maxSumSubArr(arr);
// Maximum subarray sum after performing
// the operations of first type
resSum = Math.max(resSum, maxSumSubWithOp(arr));
// Reversing the array to use the same
// existing function for operations
// of the second type
let reverse_arr = new Array(size);
for(let i = 0; i < size; i++)
reverse_arr[size - 1 - i] = arr[i];
resSum = Math.max(resSum, maxSumSubWithOp(reverse_arr));
return resSum;
}
let arr = [ -9, 21, 24, 24, -51, -6, 17, -42, -39, 33 ];
let size = arr.length;
document.write(maxSum(arr, size));
</script>
Python3
# Python3 implementation of the approach # Function that returns True if all # the array element are <= 0 def areAllNegative(arr): for i in range(len(arr)) : # If any element is non-negative if (arr[i] > 0): return False return True # Function to return the vector representing # the right to left Kadane array # as described in the approach def getRightToLeftKadane(arr): arr=arr.copy() max_so_far = 0; max_ending_here = 0 size = len(arr) for i in range(size - 1,-1,-1) : max_ending_here = max_ending_here + arr[i] if (max_ending_here < 0): max_ending_here = 0 elif (max_so_far < max_ending_here): max_so_far = max_ending_here arr[i] = max_so_far return arr # Function to return the prefix_sum vector def getPrefixSum(arr): arr=arr.copy() for i in range(1,len(arr)): arr[i] = arr[i - 1] + arr[i] return arr # Function to return the maximum sum subarray def maxSumSubArr(a): max_so_far = 0; max_ending_here = 0 for i in range(len(a)): max_ending_here = max_ending_here + a[i] if (max_ending_here < 0): max_ending_here = 0 elif (max_so_far < max_ending_here): max_so_far = max_ending_here return max_so_far # Function to get the maximum sum subarray # in the modified array def maxSumSubWithOp(arr): # kadane_r_to_l[i] will store the maximum subarray # sum for there subarray arr[i...N-1] kadane_r_to_l = getRightToLeftKadane(arr) # Get the prefix sum array prefixSum = getPrefixSum(arr) size = len(arr) for i in range(1,size): # To get max_prefix_sum_at_any_index prefixSum[i] = max(prefixSum[i - 1], prefixSum[i]) max_subarray_sum = 0 for i in range(size - 1): # Summation of both gives the maximum subarray # sum after applying the operation max_subarray_sum = max(max_subarray_sum, prefixSum[i] + kadane_r_to_l[i + 1]) return max_subarray_sum # Function to return the maximum # subarray sum after performing the # given operation at most once def maxSum(arr, size): # If all element are negative then # return the maximum element if (areAllNegative(arr)) : return max(arr) # Maximum subarray sum without # performing any operation resSum = maxSumSubArr(arr) # Maximum subarray sum after performing # the operations of first type resSum = max(resSum, maxSumSubWithOp(arr)) # Reversing the array to use the same # existing function for operations # of the second type arr=arr[::-1] resSum = max(resSum, maxSumSubWithOp(arr)) return resSum # Driver code if __name__ == '__main__': arr= [-9, 21, 24, 24, -51, -6, 17, -42, -39, 33] size = len(arr) print(maxSum(arr, size))
Producción:
102
Complejidad temporal: O(N)
Espacio auxiliar: O(N)