Maximum subarray sum in array formed by repeating the given array k times

Given an integer k and an integer array arr[] of n elements, the task is to find the largest sub-array sum in the modified array (formed by repeating the given array k times). For example, if arr[] = {1, 2} and k = 3 then modified array will be {1, 2, 1, 2, 1, 2}.

Examples:

Input: arr[] = {1, 2}, k = 3
Output: 9
Modified array will be {1, 2, 1, 2, 1, 2}
And the maximum sub-array sum will be 1 + 2 + 1 + 2 + 1 + 2 = 9

Input: arr[] = {1, -2, 1}, k = 5
Output: 2

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

A simple solution is to create an array of size n * k, then run Kadane’s algorithm to find the maximum sub-array sum. Time complexity would be O(n * k) with auxiliary space O(n * k).

A better solution is to calculate the sum of the array arr[] and store it in sum.

If sum < 0 then calculate the maximum sub-array sum of an array formed by concatenating the the array two times irrespective of the K. For example, take arr[] = {1, -4, 1} and k = 5. The sum of the array is less than 0. So, the maximum sub-array sum of the array can be found after concatenating the array two times only irrespective of the value of K i.e. b[] = {1, -4, 1, 1, -4, 1} and the maximum sub-array sum = 1 + 1 = 2

If sum > 0 then maximum sub-array will include the maximum sum as calculated in the previous step (where the array was concatenated twice) + the rest (k – 2) repetitions of the array can also be included as their sum is greater than 0 that will contribute to the maximum sum.

Below is the implementation of the above approach:

Java

// Java implementation of the approach 
public class GFG { 
  
    // Function to concatenate array 
    static void arrayConcatenate(int arr[], int b[],  
                                             int k) 
    { 
        // Array b will be formed by concatenation 
        // array a exactly k times 
        int j = 0; 
        while (k > 0) { 
  
            for (int i = 0; i < arr.length; i++) { 
                b[j++] = arr[i]; 
            } 
            k--; 
        } 
    } 
  
    // Function to return the maximum  
    // subarray sum of arr[] 
    static int maxSubArrSum(int a[]) 
    { 
        int size = a.length; 
        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 maximum sub-array  
    // sum of the modified array 
    static long maxSubKSum(int arr[], int k) 
    { 
        int arrSum = 0; 
        long maxSubArrSum = 0; 
  
        int b[] = new int[(2 * arr.length)]; 
  
        // Concatenating the array 2 times 
        arrayConcatenate(arr, b, 2); 
  
        // Finding the sum of the array 
        for (int i = 0; i < arr.length; i++) 
            arrSum += arr[i]; 
  
        // If sum is less than zero 
        if (arrSum < 0) 
            maxSubArrSum = maxSubArrSum(b); 
  
        // If sum is greater than zero 
        else
            maxSubArrSum = maxSubArrSum(b) + 
                          (k - 2) * arrSum; 
  
        return maxSubArrSum; 
    } 
  
    // Driver code 
    public static void main(String[] args) 
    { 
        int arr[] = { 1, -2, 1 }; 
        int k = 5; 
        System.out.println(maxSubKSum(arr, k)); 
    } 
} 

Below is the implementation of the above approach:

Python

# Python approach to this problem 
  
# A python module where element  
# are added to list k times 
def MaxsumArrKtimes(c, ktimes): 
      
    # Store element in list d k times 
    d = c * ktimes 
      
    # two variable which can keep  
    # track of maximum sum seen 
    # so far and maximum sum ended. 
    maxsofar = -99999
    maxending = 0
      
    for i in d: 
        maxending = maxending + i 
        if maxsofar < maxending: 
            maxsofar = maxending 
        if maxending < 0: 
            maxending = 0
    return maxsofar 
      
# Get the Maximum sum of element 
print(MaxsumArrKtimes([1, -2, 1], 5)) 
      
# This code is contributed by AnupGaurav. 
        

C#

// C# implementation of the approach  
using System; 
  
class GFG  
{  
  
// Function to concatenate array  
static void arrayConcatenate(int []arr,  
                             int []b, int k)  
{  
    // Array b will be formed by concatenation  
    // array a exactly k times  
    int j = 0;  
    while (k > 0) 
    {  
        for (int i = 0; i < arr.Length; i++)  
        {  
            b[j++] = arr[i];  
        }  
        k--;  
    }  
}  
  
// Function to return the maximum  
// subarray sum of arr[]  
static int maxSubArrSum(int []a)  
{  
    int size = a.Length;  
    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 maximum sub-array  
// sum of the modified array  
static long maxSubKSum(int []arr, int k)  
{  
    int arrSum = 0;  
    long maxSubArrsum = 0;  
  
    int []b = new int[(2 * arr.Length)];  
  
    // Concatenating the array 2 times  
    arrayConcatenate(arr, b, 2);  
  
    // Finding the sum of the array  
    for (int i = 0; i < arr.Length; i++)  
        arrSum += arr[i];  
  
    // If sum is less than zero  
    if (arrSum < 0)  
        maxSubArrsum = maxSubArrSum(b);  
  
    // If sum is greater than zero  
    else
        maxSubArrsum = maxSubArrSum(b) +  
                       (k - 2) * arrSum;  
  
    return maxSubArrsum;  
}  
  
// Driver code  
public static void Main()  
{  
    int []arr = { 1, -2, 1 };  
    int k = 5;  
    Console.WriteLine(maxSubKSum(arr, k));  
}  
}  
  
// This code is contributed by Ryuga 

PHP

<?php  
  
// PHP implementation of the approach 
  
// Function to concatenate array 
function arrayConcatenate(&$arr, &$b, $k) 
{ 
    // Array b will be formed by concatenation 
    // array a exactly k times 
    $j = 0; 
    while ($k > 0)  
    { 
  
        for ($i = 0; $i < sizeof($arr); $i++)  
        { 
            $b[$j++] = $arr[$i]; 
        } 
        $k--; 
    } 
} 
  
// Function to return the maximum  
// subarray sum of arr[] 
function maxSubArrSum(&$a) 
{ 
    $size = sizeof($a); 
    $max_so_far = 0; 
    $max_ending_here = 0; 
  
    for ($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 maximum sub-array  
// sum of the modified array 
function maxSubKSum(&$arr,$k) 
{ 
    $arrSum = 0; 
    $maxSubArrSum = 0; 
  
    $b = array_fill(0,(2 * sizeof($arr)),NULL); 
  
    // Concatenating the array 2 times 
    arrayConcatenate($arr, $b, 2); 
  
    // Finding the sum of the array 
    for ($i = 0; $i < sizeof($arr); $i++) 
        $arrSum += $arr[$i]; 
  
    // If sum is less than zero 
    if ($arrSum < 0) 
        $maxSubArrSum = maxSubArrSum($b); 
  
    // If sum is greater than zero 
    else
        $maxSubArrSum = maxSubArrSum($b) + 
                    ($k - 2) * $arrSum; 
  
    return $maxSubArrSum; 
} 
  
    // Driver code 
    $arr = array(1, -2, 1 ); 
    $k = 5; 
    echo maxSubKSum($arr, $k); 
      
// This code is contributed by Ita_c.    
?> 

Output:

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