# 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 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:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `    ``// Function to concatenate array ` `    ``void` `arrayConcatenate(``int` `*arr, ``int` `*b, ` `                                ``int` `k,``int` `len) ` `    ``{ ` `        ``// Array b will be formed by concatenation ` `        ``// array a exactly k times ` `        ``int` `j = 0; ` `        ``while` `(k > 0)  ` `        ``{ ` ` `  `            ``for` `(``int` `i = 0; i < len; i++) ` `            ``{ ` `                ``b[j++] = arr[i]; ` `            ``} ` `            ``k--; ` `        ``} ` `    ``} ` ` `  `    ``// Function to return the maximum  ` `    ``// subarray sum of arr[] ` `    ``long` `maxSubArrSum(``int` `*a,``int` `len) ` `    ``{ ` `        ``int` `size = len; ` `        ``int` `max_so_far = INT_MIN; ` `        ``long` `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 ` `    ``long` `maxSubKSum(``int` `*arr, ``int` `k,``int` `len) ` `    ``{ ` `        ``int` `arrSum = 0; ` `        ``long` `maxSubArrSum1 = 0; ` ` `  `        ``int` `b[(2 * len)]={0}; ` ` `  `        ``// Concatenating the array 2 times ` `        ``arrayConcatenate(arr, b, 2,len); ` ` `  `        ``// Finding the sum of the array ` `        ``for` `(``int` `i = 0; i < len; i++) ` `            ``arrSum += arr[i]; ` ` `  `        ``// If sum is less than zero ` `        ``if` `(arrSum < 0) ` `            ``maxSubArrSum1 = maxSubArrSum(b,2*len); ` ` `  `        ``// If sum is greater than zero ` `        ``else` `            ``maxSubArrSum1 = maxSubArrSum(b,2*len) + ` `                        ``(k - 2) * arrSum; ` ` `  `        ``return` `maxSubArrSum1; ` `    ``} ` ` `  `    ``// Driver code ` `    ``int` `main() ` `    ``{ ` `        ``int` `arr[] = { 1, -2, 1 }; ` `        ``int` `arrlen=``sizeof``(arr)/``sizeof``(arr); ` `        ``int` `k = 5; ` `        ``cout << maxSubKSum(arr, k,arrlen) << endl; ` `        ``return` `0; ` `    ``} ` `     `  `// This code is contributed by mits `

 `// 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 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# 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 `

 ` 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:
```2
```

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