Minimum possible sum of array elements after performing the given operation

Given an array arr[] of size N and a number X. If any sub array of the array(possibly empty) arr[i], arr[i+1], … can be replaced with arr[i]/x, arr[i+1]/x, …. The task is to find the minimum possible sum of the array which can be obtained.
Note: The given operation can only be performed once.

Examples:

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.

Naive Approach: A naive approach is to replace all possible positive sub-arrays with the values which we get by dividing it by X and compute the sum. But the total number of sub-arrays for any given array is (N * (N + 1))/2 where N is the size of the array. Therefore, the running time of this algorithm is O(N2).



Efficient Approach: This problem can be solved efficiently in O(N) time. On observing carefully, it can be deduced that sum can be made minimum if the subarray to be replaced satisfies the following conditions:

  1. The subarray should be positive.
  2. The sum of the subarray should be maximum.

Therefore, by reducing the subarray satisfying the above conditions, the sum can be made minimum. The subarray which satisfies the above conditions can be found by using a slight variation of Kadane’s algorithm.

After finding the maximum sum of the subarray, this can be subtracted from the sum of the array to get the minimum sum. Since the maximum sum of the sub-array is being replaced with the maximum sum / X, this factor can be added to the sum which is found. That is:

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.

Below is the implementation of the above approach:

CPP

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// C++ program to find the minimum possible sum
// of the array elements after performing
// the given operation
#include <bits/stdc++.h>
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);
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

0.5

Time Complexity: O(N)

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