Minimize sum by dividing all elements of a subarray by K

Given an array arr[] of N integers and a positive integer K, the task is to minimize the sum of the array elements after performing the given operation atmost one time. The operation is to choose a subarray and divide all elements of the subarray by K. Find and print the minimum possible sum.
Examples:

Input: arr[] = {1, -2, 3}, K = 2 
Output: 0.5 
Choose the subarray {3} and divide them by K 
The array becomes {1, -2, 1.5} where 1 – 2 + 1.5 = 0.5
Input: arr[] = {-1, -2, -3, -5}, K = 4 
Output: -11 
There is no need to perform the operation as the 
sum of the array elements is already minimum. 

Approach:

• Find the maximum sum subarray using Kadane’s Algorithm say maxSum as it will be the subarray which will be contributing in maximizing the sum of the array.
• Now there are two cases: 
  1. maxSum > 0: Divide every element of the found subarray with K and the sum of the resultant array will be minimum possible.
  2. maxSum ? 0: No need to perform the operation as the sum of the array is already minimum.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the maximum subarray sum
int maxSubArraySum(int a[], 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 + a[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 return the minimized sum
// of the array elements after performing
// the given operation
double minimizedSum(int a[], int n, int K)
{

    // Find maximum subarray sum
    int sum = maxSubArraySum(a, n);
    double totalSum = 0;

    // Find total sum of the array
    for (int i = 0; i < n; i++)
        totalSum += a[i];

    // Maximum subarray sum is already negative
    if (sum < 0)
        return totalSum;

    // Choose the subarray whose sum is
    // maximum and divide all elements by K
    totalSum = totalSum - sum + (double)sum / (double)K;
    return totalSum;
}

// Driver code
int main()
{

    int a[] = { 1, -2, 3 };
    int n = sizeof(a) / sizeof(a[0]);
    int K = 2;

    cout << minimizedSum(a, n, K);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
class GFG
{

// Function to return the maximum subarray sum
static int maxSubArraySum(int a[], 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 + a[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 return the minimized sum
// of the array elements after performing
// the given operation
static double minimizedSum(int a[], int n, int K)
{

    // Find maximum subarray sum
    int sum = maxSubArraySum(a, n);
    double totalSum = 0;

    // Find total sum of the array
    for (int i = 0; i < n; i++)
        totalSum += a[i];

    // Maximum subarray sum is already negative
    if (sum < 0)
        return totalSum;

    // Choose the subarray whose sum is
    // maximum and divide all elements by K
    totalSum = totalSum - sum + (double)sum /
                                (double)K;
    return totalSum;
}

// Driver code
public static void main(String []args)
{
    int a[] = { 1, -2, 3 };
    int n = a.length;
    int K = 2;

    System.out.println(minimizedSum(a, n, K));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
import sys

# Function to return the maximum subarray sum
def maxSubArraySum(a, size) :

    max_so_far = -(sys.maxsize - 1);
    max_ending_here = 0;

    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 (max_ending_here < 0) :
            max_ending_here = 0;

    return max_so_far;

# Function to return the minimized sum
# of the array elements after performing
# the given operation
def minimizedSum(a, n, K) :

    # Find maximum subarray sum
    sum = maxSubArraySum(a, n);
    totalSum = 0;

    # Find total sum of the array
    for i in range(n) :
        totalSum += a[i];

    # Maximum subarray sum is already negative
    if (sum < 0) :
        return totalSum;

    # Choose the subarray whose sum is
    # maximum and divide all elements by K
    totalSum = totalSum - sum + sum / K;

    return totalSum;

# Driver code
if __name__ == "__main__" :

    a = [ 1, -2, 3 ];
    n = len(a);
    K = 2;

    print(minimizedSum(a, n, K));

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the maximum subarray sum
static int maxSubArraySum(int []a, int size)
{
    int max_so_far = int.MinValue,
        max_ending_here = 0;

    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 (max_ending_here < 0)
            max_ending_here = 0;
    }
    return max_so_far;
}

// Function to return the minimized sum
// of the array elements after performing
// the given operation
static double minimizedSum(int []a, int n, int K)
{

    // Find maximum subarray sum
    int sum = maxSubArraySum(a, n);
    double totalSum = 0;

    // Find total sum of the array
    for (int i = 0; i < n; i++)
        totalSum += a[i];

    // Maximum subarray sum is already negative
    if (sum < 0)
        return totalSum;

    // Choose the subarray whose sum is
    // maximum and divide all elements by K
    totalSum = totalSum - sum + (double)sum /
                                (double)K;
    return totalSum;
}

// Driver code
public static void Main(String []args)
{
    int []a = { 1, -2, 3 };
    int n = a.Length;
    int K = 2;

    Console.WriteLine(minimizedSum(a, n, K));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript implementation of the approach

// Function to return the maximum subarray sum
function maxSubArraySum(a, size)
{
    var max_so_far = -1000000000, max_ending_here = 0;

    for (var 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 (max_ending_here < 0)
            max_ending_here = 0;
    }
    return max_so_far;
}

// Function to return the minimized sum
// of the array elements after performing
// the given operation
function minimizedSum(a, n, K)
{

    // Find maximum subarray sum
    var sum = maxSubArraySum(a, n);
    var totalSum = 0;

    // Find total sum of the array
    for (var i = 0; i < n; i++)
        totalSum += a[i];

    // Maximum subarray sum is already negative
    if (sum < 0)
        return totalSum;

    // Choose the subarray whose sum is
    // maximum and divide all elements by K
    totalSum = totalSum - sum + sum / K;
    return totalSum;
}

// Driver code
var a = [1, -2, 3];
var n = a.length;
var K = 2;
document.write( minimizedSum(a, n, K));

// This code is contributed by rrrtnx.
</script>

Output:

0.5

Time Complexity: O(N)

Auxiliary Space: O(1)