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The idea is to use a running or cumulative sum and check the remainder when this sum is divided by k. If any two prefix sums have the same remainder, the subarray sum between these indices is divisible by k.
We maintain a hashmap to store each remainder and its index. As we iterate over the array, we update the running sum and check its remainder. If the remainder has been seen before and the subarray length is at least 2, we return true. If not, we continue. If we finish the loop without finding such a subarray, we return false.
Time Complexity: O(n), where n is the length of nums, since we iterate over the array only once.
Space Complexity: O(k), as we use a hashmap (or array) to store remainders which can be 0 through k-1 in the worst case.
1function checkSubarraySum(nums, k) {
2 const remainderMap = new Map();
3 remainderMap.set(0, -1);
4 let sum = 0;
5 for (let i = 0; i < nums.length; i++) {
6 sum += nums[i];
7 let mod = sum % k;
8 if (mod < 0) mod += k;
9 if (remainderMap.has(mod)) {
10 if (i - remainderMap.get(mod) > 1) return true;
11 } else {
12 remainderMap.set(mod, i);
13 }
14 }
15 return false;
16}
17
18console.log(checkSubarraySum([23, 2, 4, 6, 7], 6));
19This JavaScript implementation makes effective use of a Map to store the remainders of cumulative sums modulo k alongside their indices. It handles negative mod results with logical increment by k, to maintain mod values non-negative.
The function seeks to exploit identical remainders indicating subarrays summing to multiples of k, confirming their existence via adequate inter-remainder gap and returning the result appropriately.
This approach attempts every possible subarray of length at least 2, calculating the sum for each and checking if it is a multiple of k. Although not efficient, this serves as a basic solution for smaller inputs.
The process involves iterating over each possible starting point of a subarray, then for each starting point, iterating over each possible ending point to compute the sum, subsequently verifying its divisibility by k.
Time Complexity: O(n²) because of the nested loop structure iterating through potential subarray limits.
Space Complexity: O(1) as no extra space apart from a few variables is used.
Using a brute force approach in Java, this code evaluates every suitable subarray by calculating their sums inside nested loops and assesses divisibility by k.
This method checks all cases thoroughly but suffers from inefficiency for large arrays.