Maximum subarray sum in an array created after repeated concatenation

Last Updated : 02 Nov, 2024

Given an array and a number k, find the largest sum of contiguous array in the modified array which is formed by repeating the given array k times.

Examples :

Input : arr[] = {-1, 10, 20}, k = 2
Output : 59
After concatenating array twice, we get {-1, 10, 20, -1, 10, 20} which has maximum subarray sum as 59.

Input : arr[] = {-1, -2, -3}, k = 3
Output : -1

Naive Approach – O(nk) Time and O(nk) Space

A simple solution is to create an array of size n*k, then run Kadane’s algorithm for the whole array.

Expected Approach – O(nk) Time and O(1) Space

An efficient solution is to run a loop on same array and use modular arithmetic to move back beginning after end of array.

Below is the implementation:

C++

#include <bits/stdc++.h>
using namespace std;

// Returns sum of maximum sum subarray created
// after concatenating a[0..n-1] k times.
int maxSubArraySumRepeated(vector<int>& arr, int k) {
    int n = arr.size();
    int maxSoFar = INT_MIN, maxEndingHere = 0;

    for (int i = 0; i < n * k; i++) {
      
        // Use modular arithmetic to get the next element
        maxEndingHere += arr[i % n];

        maxSoFar = max(maxSoFar, maxEndingHere);

        if (maxEndingHere < 0)
            maxEndingHere = 0;
    }
    return maxSoFar;
}

/* Driver program to test maxSubArraySum */
int main() {
    vector<int> arr = {10, 20, -30, -1};
    int k = 3;
    cout << "Maximum contiguous sum is "
         << maxSubArraySumRepeated(arr, k);
    return 0;
}

// Returns sum of maximum sum subarray created
// after concatenating a[0..n-1] k times.
#include <stdio.h>
#include <limits.h>

int maxSubArraySumRepeated(int arr[], int n, int k) {
    int maxSoFar = INT_MIN, maxEndingHere = 0;

    for (int i = 0; i < n * k; i++) {
      
        // Use modular arithmetic to get the next element
        maxEndingHere += arr[i % n];

        if (maxEndingHere > maxSoFar) {
            maxSoFar = maxEndingHere;
        }

        if (maxEndingHere < 0) {
            maxEndingHere = 0;
        }
    }
    return maxSoFar;
}

// Driver program to test maxSubArraySumRepeated
int main() {
    int arr[] = {10, 20, -30, -1};
    int k = 3;
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("Maximum contiguous sum is %d\n", maxSubArraySumRepeated(arr, n, k));
    return 0;
}

Java

// Returns sum of maximum sum subarray created
// after concatenating a[0..n-1] k times.
public class GfG {
    public static int maxSubArraySumRepeated(int[] arr, int k) {
        int n = arr.length;
        int maxSoFar = Integer.MIN_VALUE, maxEndingHere = 0;

        for (int i = 0; i < n * k; i++) {
          
            // Use modular arithmetic to get the next element
            maxEndingHere += arr[i % n];

            maxSoFar = Math.max(maxSoFar, maxEndingHere);

            if (maxEndingHere < 0)
                maxEndingHere = 0;
        }
        return maxSoFar;
    }

    // Driver program to test maxSubArraySumRepeated
    public static void main(String[] args) {
        int[] arr = {10, 20, -30, -1};
        int k = 3;
        System.out.println("Maximum contiguous sum is " + maxSubArraySumRepeated(arr, k));
    }
}

Python

# Returns sum of maximum sum subarray created
# after concatenating a[0..n-1] k times.
def max_subarray_sum_repeated(arr, k):
    n = len(arr)
    max_so_far = float('-inf')
    max_ending_here = 0

    for i in range(n * k):
      
        # Use modular arithmetic to get the next element
        max_ending_here += arr[i % n]

        max_so_far = max(max_so_far, max_ending_here)

        if max_ending_here < 0:
            max_ending_here = 0
    return max_so_far

# Driver program to test max_subarray_sum_repeated
arr = [10, 20, -30, -1]
k = 3
print("Maximum contiguous sum is", max_subarray_sum_repeated(arr, k))

// Returns sum of maximum sum subarray created
// after concatenating a[0..n-1] k times.
using System;
using System.Linq;

class Program {
    public static int MaxSubArraySumRepeated(int[] arr, int k) {
        int n = arr.Length;
        int maxSoFar = int.MinValue, maxEndingHere = 0;

        for (int i = 0; i < n * k; i++) {
          
            // Use modular arithmetic to get the next element
            maxEndingHere += arr[i % n];

            maxSoFar = Math.Max(maxSoFar, maxEndingHere);

            if (maxEndingHere < 0)
                maxEndingHere = 0;
        }
        return maxSoFar;
    }

    // Driver program to test MaxSubArraySumRepeated
    static void Main() {
        int[] arr = {10, 20, -30, -1};
        int k = 3;
        Console.WriteLine("Maximum contiguous sum is " + MaxSubArraySumRepeated(arr, k));
    }
}

JavaScript

// Returns sum of maximum sum subarray created
// after concatenating a[0..n-1] k times.
function maxSubArraySumRepeated(arr, k) {
    let n = arr.length;
    let maxSoFar = Number.MIN_SAFE_INTEGER, maxEndingHere = 0;

    for (let i = 0; i < n * k; i++) {
    
        // Use modular arithmetic to get the next element
        maxEndingHere += arr[i % n];

        maxSoFar = Math.max(maxSoFar, maxEndingHere);

        if (maxEndingHere < 0)
            maxEndingHere = 0;
    }
    return maxSoFar;
}

// Driver program to test maxSubArraySumRepeated
let arr = [10, 20, -30, -1];
let k = 3;
console.log("Maximum contiguous sum is " + maxSubArraySumRepeated(arr, k));

Output

Maximum contiguous sum is 30

JavaScript Number toString() Method

SandeepC

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Practice Tags :

Maximum subarray sum in an array created after repeated concatenation

Naive Approach – O(nk) Time and O(nk) Space

Expected Approach – O(nk) Time and O(1) Space

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