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# 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)

Minimize sum by dividing all elements of a subarray by K

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