Create a new array C of size n, and initialize C[i] = A[i] — B[i], where A and B are the two given arrays. Then the problem boils down to finding a segment in C whose sum is equal to 0. This is a standard problem which can be solved efficiently in O(n) or O(nlogn) using map.

Try representing Sum(L, R) in terms of Prefix Sums

We want to make Sum(L, R) equal for both arrays, Can we rearrange the equations to solve this problem?

Using, Sum(L, R) = Prefix[R] - Prefix[L] + Arr[L]
We want to check, there is any[L, R] such that: Sum1(L, R) = Sum2(L, R)

Sum1(L, R) = Sum2(L, R)

Prefix1[R] - Prefix1[L] + Arr1[L] = Prefix2[R] - Prefix2[L] + Arr2[L]

(Prefix1[R] - Prefix2[R]) = (Prefix1[L] - Arr1[L]) - (Prefix2[L] - Arr2[L])

Let f(L) = (Prefix1[L] - Arr1[L]) - (Prefix2[L] - Arr2[L])

So, the above eqn becomes: (Prefix1[R] - Prefix2[R]) = f(L)

=> For every R, we can check, is there any f(L) = (Prefix1[R] - Prefix2[R])

int prefix_sum1 = 0;
int prefix_sum2 = 0;

auto f = [ & ](int L) {
    return (prefix_sum1 - arr1[L]) - (prefix_sum2 - arr2[L]);
};

unordered_set < int > seen;

for (int R = 0; R < n; ++R) {
    prefix_sum1 += arr1[R];
    prefix_sum2 += arr2[R];

    seen.insert(f(R));

    if (seen.count(prefix_sum1 - prefix_sum2)) {
        return true;
    }
}