# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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); ` `    ``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 `

Output:

```0.5
```

Time Complexity: O(N)

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