702765. Mr. Modulo and Arrays
Mr. Modulo and Arrays
Summary
Mr. Modulo is a clever fellow who loves arrays. He has an array of N integers, where each element is between 1 and K (inclusive). His task is to find the maximum sum of any subarray within the given range [L, R] such that all elements in the subarray are congruent modulo M.
This problem requires a combination of array manipulation, binary search, and mathematical concepts. It's an interesting challenge that tests your understanding of arrays, modular arithmetic, and algorithmic thinking.
Detailed Explanation
To solve this problem, we'll need to first understand the concept of congruence modulo M. For two integers x and y, x is congruent to y modulo M if their remainder when divided by M is the same. In other words, x % M == y % M.
We can use binary search to find the maximum sum of any subarray within the range [L, R] that satisfies the given condition. Here's a step-by-step breakdown of the solution:
- Initialize a variable
max_sumto store the maximum sum found so far. - Perform a binary search on the array to find the smallest element in the range [L, R] that is congruent modulo M with the first element of the array.
- Calculate the sum of all elements in the subarray from the found element to the end of the range [R].
- Update
max_sumif the calculated sum is greater than the current maximum sum. - Repeat steps 2-4 until the binary search space is exhausted.
- Return the maximum sum found (
max_sum) as the result.
Here's a diagram illustrating the problem:
+---------------+
| L (inclusive) |
+---------------+
| ... |
+---------------+
| R (exclusive) |
+---------------+
Find the maximum sum of any subarray within [L, R]
such that all elements are congruent modulo M.
Time complexity: O(N log K) Space complexity: O(1)
Optimized Solutions
Java
public int maxSum(int[] arr, int L, int R, int M) {
int max_sum = 0;
for (int i = L; i <= R; i++) {
int remain = i % M;
int sum = 0;
for (int j = i; j < R && arr[j] % M == remain; j++) {
sum += arr[j];
}
max_sum = Math.max(max_sum, sum);
}
return max_sum;
}