给定大小为N的数组arr []和数字X。如果数组的任何子数组(可能为空)arr [i],arr [i + 1],…都可以替换为arr [i] / x,arr [i + 1] / x,…。任务是找到可以获取的数组的最小和。
注意:给定的操作只能执行一次。
例子:
Input: N = 3, X = 2, arr[] = {1, -2, 3}
Output: 0.5
Explanation:
On selecting subarray {3} and replacing it with {1.5}, the array becomes {1, -2, 1.5}, which gives sum = 0.5
Input: N = 5, X = 5, arr[] = {5, 5, 5, 5, 5}
Output: 5
Explanation:
On selecting subarray {5, 5, 5, 5, 5} and replacing it with {1, 1, 1, 1, 1}, the sum of the array becomes 5.
天真的方法:天真的方法是将所有可能的正子数组替换为我们得到的值,方法是将其除以X并计算总和。但是任何给定数组的子数组总数为(N *(N + 1))/ 2 ,其中N是数组的大小。因此,该算法的运行时间为O(N 2 ) 。
高效的方法:此问题可以在O(N)时间内有效解决。仔细观察,可以推断出,如果要替换的子数组满足以下条件,则总和可以为最小值:
- 子数组应为正数。
- 子数组的总和应为最大值。
因此,通过减少满足上述条件的子阵列,可以使总和最小。通过使用Kadane算法的细微变化,可以找到满足上述条件的子阵列。
在找到子数组的最大和之后,可以将其从数组的和中减去以获得最小和。由于子数组的最大和被替换为最大和/ X,因此可以将此因子添加到找到的和中。那是:
Minimum sum = (sumOfArr - subSum) + (subSum/ X)
where sumOfArr is the sum of all elements of the array
and subSum is the maximum sum of the subarray.
下面是上述方法的实现:
CPP
// C++ program to find the minimum possible sum
// of the array elements after performing
// the given operation
#include
using namespace std;
// Function to find the maximum sum
// of the sub array
double maxSubArraySum(double a[], int size)
{
// max_so_far represents the maximum sum
// found till now and max_ending_here
// represents the maximum sum ending at
// a specific index
double max_so_far = INT_MIN,
max_ending_here = 0;
// Iterating through the array to find
// the maximum sum of the subarray
for (int i = 0; i < size; i++) {
max_ending_here = max_ending_here + a[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
// If the maximum sum ending at a
// specific index becomes less than 0,
// then making it equal to 0.
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
// Function to find the minimum possible sum
// of the array elements after performing
// the given operation
double minPossibleSum(double a[], int n, double x)
{
double mxSum = maxSubArraySum(a, n);
double sum = 0;
// Finding the sum of the array
for (int i = 0; i < n; i++) {
sum += a[i];
}
// Computing the minimum sum of the array
sum = sum - mxSum + mxSum / x;
cout << setprecision(2) << sum << endl;
}
// Driver code
int main()
{
int N = 3;
double X = 2;
double A[N] = { 1, -2, 3 };
minPossibleSum(A, N, X);
}
Java
// Java program to find the minimum possible sum
// of the array elements after performing
// the given operation
class GFG{
// Function to find the maximum sum
// of the sub array
static double maxSubArraySum(double a[], int size)
{
// max_so_far represents the maximum sum
// found till now and max_ending_here
// represents the maximum sum ending at
// a specific index
double max_so_far = Integer.MIN_VALUE,
max_ending_here = 0;
// Iterating through the array to find
// the maximum sum of the subarray
for (int i = 0; i < size; i++) {
max_ending_here = max_ending_here + a[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
// If the maximum sum ending at a
// specific index becomes less than 0,
// then making it equal to 0.
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
// Function to find the minimum possible sum
// of the array elements after performing
// the given operation
static void minPossibleSum(double a[], int n, double x)
{
double mxSum = maxSubArraySum(a, n);
double sum = 0;
// Finding the sum of the array
for (int i = 0; i < n; i++) {
sum += a[i];
}
// Computing the minimum sum of the array
sum = sum - mxSum + mxSum / x;
System.out.print(sum +"\n");
}
// Driver code
public static void main(String[] args)
{
int N = 3;
double X = 2;
double A[] = { 1, -2, 3 };
minPossibleSum(A, N, X);
}
}
// This code is contributed by Princi Singh
Python3
# Python3 program to find the minimum possible sum
# of the array elements after performing
# the given operation
# Function to find the maximum sum
# of the sub array
def maxSubArraySum(a, size):
# max_so_far represents the maximum sum
# found till now and max_ending_here
# represents the maximum sum ending at
# a specific index
max_so_far = -10**9
max_ending_here = 0
# Iterating through the array to find
# the maximum sum of the subarray
for i in range(size):
max_ending_here = max_ending_here + a[i]
if (max_so_far < max_ending_here):
max_so_far = max_ending_here
# If the maximum sum ending at a
# specific index becomes less than 0,
# then making it equal to 0.
if (max_ending_here < 0):
max_ending_here = 0
return max_so_far
# Function to find the minimum possible sum
# of the array elements after performing
# the given operation
def minPossibleSum(a,n, x):
mxSum = maxSubArraySum(a, n)
sum = 0
# Finding the sum of the array
for i in range(n):
sum += a[i]
# Computing the minimum sum of the array
sum = sum - mxSum + mxSum / x
print(round(sum,2))
# Driver code
if __name__ == '__main__':
N = 3
X = 2
A=[1, -2, 3]
minPossibleSum(A, N, X)
# This code is contributed by mohit kumar 29
C#
// C# program to find the minimum possible sum
// of the array elements after performing
// the given operation
using System;
class GFG{
// Function to find the maximum sum
// of the sub array
static double maxSubArraySum(double[] a, int size)
{
// max_so_far represents the maximum sum
// found till now and max_ending_here
// represents the maximum sum ending at
// a specific index
double max_so_far = Int32.MinValue,
max_ending_here = 0;
// Iterating through the array to find
// the maximum sum of the subarray
for (int i = 0; i < size; i++) {
max_ending_here = max_ending_here + a[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
// If the maximum sum ending at a
// specific index becomes less than 0,
// then making it equal to 0.
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
// Function to find the minimum possible sum
// of the array elements after performing
// the given operation
static void minPossibleSum(double[] a, int n, double x)
{
double mxSum = maxSubArraySum(a, n);
double sum = 0;
// Finding the sum of the array
for (int i = 0; i < n; i++) {
sum += a[i];
}
// Computing the minimum sum of the array
sum = sum - mxSum + mxSum / x;
Console.Write(sum +"\n");
}
// Driver code
public static void Main()
{
int N = 3;
double X = 2;
double[] A = { 1, -2, 3 };
minPossibleSum(A, N, X);
}
}
// This code is contributed by chitranayal
Javascript
0.5
时间复杂度: O(N)