1191. K-Concatenation Maximum Sum

Problem Description

The problem deals with finding the maximum sum of a sub-array within a modified array, which is the result of repeating the original integer array arr a total of k times. The challenge involves not just the repetition of the array but computing the maximum possible sum of a contiguous sub-array. This includes the possibility of a sub-array having a length of 0, which would correspond to a sum of 0.

Intuition

To tackle this problem, we need to understand a few important concepts:

Kadane's Algorithm: This algorithm is used for finding the maximum sum of a contiguous sub-array within a single unmodified array. It does this by looking for all positive contiguous segments of the array (max_ending_here) and keeping track of the maximum sum contiguous segment among all positive segments (max_so_far). The algorithm increments the sum when the running sum is positive, and resets it to zero when the running sum becomes negative.
Prefix Sum and Suffix Sum: The prefix sum is the sum of elements from the start of the array up to a certain index, and the suffix sum is the sum of elements from a certain index to the end of the array. These can be useful when handling repeated arrays, as the maximum sum can span across the boundary between two repeated segments.

Considering these concepts, the intuition for the solution approach can be broken down into steps:

The Kadane's Algorithm is used to find the maximum sum of a contiguous sub-array from the original arr.
The sum of the entire array (s) is calculated to be used in checking if we can increase the maximum sum by including sums from multiple repetitions of arr.
We find the maximum prefix sum (mx_pre), i.e., the largest sum obtained from the start of an array up to any index.
We find the maximum suffix sum (mx_suf), i.e., the largest sum that starts at any index and goes to the end of an array.
Based on the value of k, we decide how to combine these sums to compute the final answer:
- If k is equal to 1, the maximum sub-array sum is only within the single original array and it's the result obtained from Kadane's Algorithm.
- If k is greater than 1 and the total sum of the array (s) is positive, the maximum sum could potentially span across the middle arrays completely, so we consider the array sum multiplied by (k-2) and add both the maximum prefix and suffix sums.
- If k is greater than 1 and the total sum of the array is non-positive, we just need to consider the maximum prefix and suffix sums, as spanning across copies would not increase the overall sub-array sum.

Finally, since the answer can be very large, we return the result modulo 10^9 + 7 to ensure it stays within the integer limits for the problem.

Learn more about Dynamic Programming patterns.

Solution Approach

In the provided solution, the implementation walks through the array and applies the concepts mentioned in the Intuition section.

Here's a breakdown of the logic used:

Initialization: We start by initializing variables to store the sum of the array (s), the maximum prefix sum (mx_pre), the minimum prefix sum (mi_pre), and the maximum sub-array sum (mx_sub).
Single Pass for Kadane's Algorithm and Prefix Sums:
- We loop through each element in arr:
  - Update the sum of the array (s) by adding the current element x.
  - Update the maximum prefix sum (mx_pre) to the maximum of mx_pre and the current sum s.
  - Update the minimum prefix sum (mi_pre) to the minimum of mi_pre and the current sum s.
  - Update the maximum sub-array sum (mx_sub) to the maximum of the current mx_sub and the difference between the current sum s and the minimum prefix sum (mi_pre), which represents Kadane's algorithm execution for the sub-array ending at the current element.
Handling Multiple Concatenations:
- After the loop, we calculate the suffix maximum sum (mx_suf) by subtracting the minimum prefix sum (mi_pre) from the total sum of the array (s).
- The base answer variable ans is initialized with the value obtained from Kadane's Algorithm (mx_sub).
- If k equals 1, which means the array is not concatenated, we use the answer gotten from the Kadane's pass earlier and return it modulo 10^9 + 7.
- For the case where k is greater than 1, we explore the options by combining different parts of the array:
  - First, we use the maximum prefix sum plus the maximum suffix sum (mx_pre + mx_suf) and update ans if this is greater than the current ans.
  - Next, if the sum of the array (s) is positive, we explore the possibility that the maximum sum spans across the entire middle part of the concatenated array. This leads us to consider (k - 2) * s + mx_pre + mx_suf. We then again update ans if this is greater than the current ans.
Returning the Result:
- In the final step, we return the answer ans modulo 10^9 + 7.

This approach effectively handles the possibility of maximizing the sub-array sum by including the sum of the entire array when it is beneficial (i.e., when the total sum is positive) due to the concatenation specified by k. It ensures that the maximum sub-array sum is found even if it spans across multiple copies of the array.

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Example Walkthrough

Let's walk through an example to illustrate the solution approach:

Suppose our given array is arr = [3, -1, 2] and k = 2. That means we need to find the maximum sub-array sum in an array that would look like [3, -1, 2, 3, -1, 2] after concatenating arr to itself once (as k = 2).

Initialization:
- s = 0 (sum of the array)
- mx_pre = 0 (maximum prefix sum)
- mi_pre = 0 (minimum prefix sum)
- mx_sub = 0 (maximum sub-array sum)
Single Pass for Kadane's Algorithm and Prefix Sums:
- For the first element 3:
  - s = 3
  - mx_pre = 3
  - mi_pre = 0
  - mx_sub = 3
- For the second element -1:
  - s = 3 - 1 = 2
  - mx_pre remains 3
  - mi_pre remains 0
  - mx_sub remains 3
- For the third element 2:
  - s = 2 + 2 = 4
  - mx_pre = 4
  - mi_pre remains 0
  - mx_sub = 4 (since 4 - 0 > 3)
Handling Multiple Concatenations:
- After the loop, we calculate mx_suf which is s - mi_pre = 4 - 0 = 4.
- The base answer ans is mx_sub which is currently 4.
- Since k > 1, we examine the maximum sum using concatenation:
  - We calculate mx_pre + mx_suf = 4 + 4 = 8 and compare it with ans, thus ans becomes 8.
  - Since the sum of the array s is positive, we explore the maximum sum crossing the entire middle part which would be (k - 2) * s + mx_pre + mx_suf. Since k = 2, multiplying by k - 2 equals 0, so this step does not change ans.
Returning the Result:
- The function would return ans modulo 10^9 + 7, which is 8 in this case, as the maximum sum sub-array is [3, -1, 2, 3, -1, 2] itself when k = 2, with the sum being 8.

This example demonstrates that even when our given array has negative numbers, by strategically utilizing the concatenation of the array k times, we can maximize the sub-array sum without including the negative sub-arrays. The technique of combining prefix and suffix sums with Kadane's algorithm and handling different cases based on the total sum s and the count k leads to a comprehensive solution.

Solution Implementation

1from typing import List
2
3class Solution:
4    def kConcatenationMaxSum(self, arr: List[int], k: int) -> int:
5        # Initialize variables
6        total_sum = max_prefix = min_prefix = max_subarray_sum = 0
7      
8        # Calculate max subarray sum for a single array iteration
9        for num in arr:
10            total_sum += num
11            max_prefix = max(max_prefix, total_sum)
12            min_prefix = min(min_prefix, total_sum)
13            max_subarray_sum = max(max_subarray_sum, total_sum - min_prefix)
14      
15        # The result after a single iteration
16        result = max_subarray_sum
17        mod = 10**9 + 7
18      
19        # If k is 1, return the result of a single array's max subarray sum
20        if k == 1:
21            return result % mod
22      
23        # Calculate the maximum suffix sum for potential use in concatenated arrays
24        max_suffix = total_sum - min_prefix
25      
26        # Update result for potential double array combination
27        result = max(result, max_prefix + max_suffix)
28      
29        # If the array sum is positive, calculate the max sum when array is concatenated k times
30        if total_sum > 0:
31            result = max(result, ((k - 2) * total_sum) + max_prefix + max_suffix)
32      
33        return result % mod  # Return the result modulo the provided modulus
34

1class Solution {
2
3    // Computes the maximum sum of a subsequence in an array that can be achieved by
4    // concatenating the array k times.
5    public int kConcatenationMaxSum(int[] arr, int k) {
6        long sum = 0; // Total sum of the array elements.
7        long maxPrefixSum = 0; // Maximum prefix sum found so far.
8        long minPrefixSum = 0; // Minimum prefix sum found so far.
9        long maxSubarraySum = 0; // Maximum subarray sum found so far.
10
11        // Iterate over the array to find the maximum subarray sum, maximum prefix,
12        // and minimum prefix sums.
13        for (int value : arr) {
14            sum += value;
15            maxPrefixSum = Math.max(maxPrefixSum, sum);
16            minPrefixSum = Math.min(minPrefixSum, sum);
17            maxSubarraySum = Math.max(maxSubarraySum, sum - minPrefixSum);
18        }
19
20        long answer = maxSubarraySum; // This holds the result, which is initialized to maxSubarraySum.
21        final int mod = (int) 1e9 + 7; // Module to perform the answer under modulo operation.
22
23        // If there's only one concatenation, simply return the max subarray sum modulo mod.
24        if (k == 1) {
25            return (int) (answer % mod);
26        }
27
28        long maxSuffixSum = sum - minPrefixSum; // Maximum suffix sum after one traversal.
29      
30        // Check if adding the entire array sum (suffix and prefix) is better.
31        answer = Math.max(answer, maxPrefixSum + maxSuffixSum);
32
33        // If the sum of the array is positive, the best option might be to take the sum k-2 times,
34        // then add the maxPrefix and maxSuffix sums.
35        if (sum > 0) {
36            answer = Math.max(answer, (k - 2) * sum + maxPrefixSum + maxSuffixSum);
37        }
38
39        // Return the maximal sum found under modulo mod.
40        return (int) (answer % mod);
41    }
42}
43

1class Solution {
2public:
3    int kConcatenationMaxSum(vector<int>& arr, int k) {
4        long sumOfArray = 0, maxPrefixSum = 0, minPrefixSum = 0, maxSubarraySum = 0;
5        const int MOD = 1e9 + 7; // Define the modulus for the answer
6
7        // Calculate the maximum subarray sum for one array
8        for (int num : arr) {
9            sumOfArray += num; // Sum of elements so far
10            maxPrefixSum = max(maxPrefixSum, sumOfArray); // Max sum from the start to current
11            minPrefixSum = min(minPrefixSum, sumOfArray); // Min sum from the start to current
12            maxSubarraySum = max(maxSubarraySum, sumOfArray - minPrefixSum); // Kadane's algorithm update
13        }
14
15        long result = maxSubarraySum; // Initialize the result with max subarray sum
16      
17        // Handle the case when the array is concatenated only once
18        if (k == 1) {
19            return result % MOD;
20        }
21
22        long maxSuffixSum = sumOfArray - minPrefixSum; // Sum of max suffix
23        result = max(result, maxPrefixSum + maxSuffixSum); // Max of result and sum of max prefix and suffix
24
25        // If the sum of the array is positive, we can take the whole arrays k-2 times along with maxPrefix and maxSuffix
26        if (sumOfArray > 0) {
27            result = max(result, maxPrefixSum + (k - 2) * sumOfArray + maxSuffixSum);
28        }
29
30        // Return the result modulo 10^9 + 7
31        return result % MOD;
32    }
33};
34

1const MOD: number = 1e9 + 7; // Define the modulus for the answer
2
3function kConcatenationMaxSum(arr: number[], k: number): number {
4    let sumOfArray: number = 0, maxPrefixSum: number = 0, minPrefixSum: number = 0, maxSubarraySum: number = 0;
5
6    // Calculate the maximum subarray sum for one array
7    for (const num of arr) {
8        sumOfArray = (sumOfArray + num) % MOD; // Keep updating the sum of elements
9        maxPrefixSum = Math.max(maxPrefixSum, sumOfArray); // Max sum from start to current position
10        minPrefixSum = Math.min(minPrefixSum, sumOfArray); // Min sum from start to current position
11        maxSubarraySum = Math.max(maxSubarraySum, (sumOfArray - minPrefixSum + MOD) % MOD); // Kadane's algorithm update
12    }
13
14    let result: number = maxSubarraySum; // Initialize the result with max subarray sum
15
16    // Handle the case when the array is concatenated only once
17    if (k === 1) {
18        return result;
19    }
20
21    let maxSuffixSum: number = (sumOfArray - minPrefixSum + MOD) % MOD; // Sum of the max suffix
22    result = Math.max(result, (maxPrefixSum + maxSuffixSum) % MOD); // Max of result and the sum of max prefix and max suffix
23
24    // If the sum of the array is positive, we include the whole arrays (k - 2) times along with maxPrefix and maxSuffix
25    if (sumOfArray > 0) {
26        result = Math.max(result, (maxPrefixSum + ((k - 2) * sumOfArray + maxSuffixSum) % MOD) % MOD);
27    }
28
29    // Return the result modulo 10^9 + 7
30    return result;
31}
32

Time and Space Complexity

The given Python code defines a method kConcatenationMaxSum which finds the maximum sum of a subarray in the K-concatenated array formed by repeating the given array k times.

Time Complexity

The function iterates once through the array arr, performing a constant amount of work in each iteration, including finding maximum and minimum prefixes, and computing the maximum subarray sum ending at any element. Therefore, the time complexity of iterating the array is O(n) where n is the length of arr.

After this iteration, there is a constant amount of work done to compute mx_suf and the maximum of ans, mx_pre + mx_suf, and (k - 2) * s + mx_pre + mx_suf if s > 0. These operations take O(1) time.

Hence, the overall time complexity of the function is O(n + 1), which simplifies to O(n).

Space Complexity

In terms of space, the function allocates a few variables (s, mx_pre, mi_pre, mx_sub, and mod), which use O(1) space as their number does not scale with the input size.

The inputs arr and k are used without any additional space being allocated that depends on their size (no extra arrays or data structures are created). Thus, the space used is constant.

As such, the space complexity is O(1).

Combining the analysis above, the code has a linear time complexity and constant space complexity.

Learn more about how to find time and space complexity quickly using problem constraints.

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