Maximum subarray sum modulo M

We can do this as follow:

Maintaining an array sum which at index ith, it contains the modulus sum from 0 to ith.

For each index ith, we need to find the maximum sub sum that end at this index:

For each subarray (start + 1 , i ), we know that the mod sum of this sub array is

int a = (sum[i] - sum[start] + M) % M

So, we can only achieve a sub-sum larger than sum[i] if sum[start] is larger than sum[i] and as close to sum[i] as possible.

This can be done easily if you using a binary search tree.

Pseudo code:

int[] sum;
sum[0] = A[0];
Tree tree;
tree.add(sum[0]);
int result = sum[0];
for(int i = 1; i < n; i++){
    sum[i] = sum[i - 1] + A[i];
    sum[i] %= M;
    int a = tree.getMinimumValueLargerThan(sum[i]);
    result = max((sum[i] - a + M) % M, result);
    tree.add(sum[i]);
}
print result;

Time complexity :O(n log n)

Let A be our input array with zero-based indexing. We can reduce A modulo M without changing the result.

First of all, let's reduce the problem to a slightly easier one by computing an array P representing the prefix sums of A, modulo M:

A = 6 6 11 2 12 1
P = 6 12 10 12 11 12

Now let's process the possible left borders of our solution subarrays in decreasing order. This means that we will first determine the optimal solution that starts at index n - 1, then the one that starts at index n - 2 etc.

In our example, if we chose i = 3 as our left border, the possible subarray sums are represented by the suffix P[3..n-1] plus a constant a = A[i] - P[i]:

a = A[3] - P[3] = 2 - 12 = 3 (mod 13)
P + a = * * * 2 1 2

The global maximum will occur at one point too. Since we can insert the suffix values from right to left, we have now reduced the problem to the following:

Given a set of values S and integers x and M, find the maximum of S + x modulo M

This one is easy: Just use a balanced binary search tree to manage the elements of S. Given a query x, we want to find the largest value in S that is smaller than M - x (that is the case where no overflow occurs when adding x). If there is no such value, just use the largest value of S. Both can be done in O(log |S|) time.

Total runtime of this solution: O(n log n)

Here's some C++ code to compute the maximum sum. It would need some minor adaptions to also return the borders of the optimal subarray:

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

int max_mod_sum(const vector<int>& A, int M) {
    vector<int> P(A.size());
    for (int i = 0; i < A.size(); ++i)
        P[i] = (A[i] + (i > 0 ? P[i-1] : 0)) % M;
    set<int> S;
    int res = 0;
    for (int i = A.size() - 1; i >= 0; --i) {
        S.insert(P[i]);
        int a = (A[i] - P[i] + M) % M;
        auto it = S.lower_bound(M - a);
        if (it != begin(S))
            res = max(res, *prev(it) + a);
        res = max(res, (*prev(end(S)) + a) % M);
    }
    return res;
}

int main() {
    // random testing to the rescue
    for (int i = 0; i < 1000; ++i) {
        int M = rand() % 1000 + 1, n = rand() % 1000 + 1;
        vector<int> A(n);
        for (int i = 0; i< n; ++i)
            A[i] = rand() % M;
        int should_be = 0;
        for (int i = 0; i < n; ++i) {
            int sum = 0;
            for (int j = i; j < n; ++j) {
                sum = (sum + A[j]) % M;
                should_be = max(should_be, sum);
            }
        }
        assert(should_be == max_mod_sum(A, M));
    }
}

For me, all explanations here were awful, since I didn't get the searching/sorting part. How do we search/sort, was unclear.

We all know that we need to build prefixSum, meaning sum of all elems from 0 to i with modulo m

I guess, what we are looking for is clear. Knowing that subarray[i][j] = (prefix[i] - prefix[j] + m) % m (indicating the modulo sum from index i to j), our maxima when given prefix[i] is always that prefix[j] which is as close as possible to prefix[i], but slightly bigger.

E.g. for m = 8, prefix[i] being 5, we are looking for the next value after 5, which is in our prefixArray.

For efficient search (binary search) we sort the prefixes.

What we can not do is, build the prefixSum first, then iterate again from 0 to n and look for index in the sorted prefix array, because we can find and endIndex which is smaller than our startIndex, which is no good.

Therefore, what we do is we iterate from 0 to n indicating the endIndex of our potential max subarray sum and then look in our sorted prefix array, (which is empty at the beginning) which contains sorted prefixes between 0 and endIndex.

def maximumSum(coll, m):
    n = len(coll)
    maxSum, prefixSum = 0, 0
    sortedPrefixes = []

    for endIndex in range(n):
        prefixSum = (prefixSum + coll[endIndex]) % m
        maxSum = max(maxSum, prefixSum)

        startIndex = bisect.bisect_right(sortedPrefixes, prefixSum)
        if startIndex < len(sortedPrefixes): 
            maxSum = max(maxSum, prefixSum - sortedPrefixes[startIndex] + m)

        bisect.insort(sortedPrefixes, prefixSum)

    return maxSum

From your question, it seems that you have created an array to store the cumulative sums (Prefix Sum Array), and are calculating the sum of the sub-array arr[i:j] as (sum[j] - sum[i] + M) % M. (arr and sum denote the given array and the prefix sum array respectively)

Calculating the sum of every sub-array results in a O(n*n) algorithm.

The question that arises is -

Do we really need to consider the sum of every sub-array to reach the desired maximum?

No!

For a value of j the value (sum[j] - sum[i] + M) % M will be maximum when sum[i] is just greater than sum[j] or the difference is M - 1.

This would reduce the algorithm to O(nlogn).

You can take a look at this explanation! https://www.youtube.com/watch?v=u_ft5jCDZXk