https://algodaily.com/challenges/contiguoussubarraysum
There’s a fairly obvious bruteforce O(n^2) way to do this by looping through
every possible subarray:
function sum(numbers) {
return numbers.reduce((sum, x) => sum + x, 0);
}
function* allSubArrays(numbers) {
for (let i = 0; i <= numbers.length; i++) {
for (let j = 0; j <= numbers.length; j++) {
const slice = numbers.slice(i, j);
if (slice.length > 0) {
yield slice;
}
}
}
}
function subarraySum(numbers, targetSum) {
for (let sub of allSubArrays(numbers)) {
if (sum(sub) === targetSum) {
return true;
}
}
return false;
}
This is O(n^2) , so not great, but I can’t think of a different way to do it.
We could make some small optimisations by skipping a subarray as soon as
possible, but the worst case would still be O(n^2) .
Algo Daily has a more efficient solution using a hashmap to cache previously
calculated subarray sums. If we reach a point where the sum minus the target sum
is in the sums cache, we know that it’s possible to make the target sum using a
subarray of the numbers.
function subarraySum(numbers, targetSum) {
const seenSums = { 0: true };
let cumulativeSum = 0;
for (let num of numbers) {
cumulativeSum += num;
if (seenSums[cumulativeSum  targetSum]) {
return true;
}
seenSums[cumulativeSum] = true;
}
return false;
}
This might be easier to understand with more extreme numbers as the example
(often a good trick):
subarraySum([100, 3, 4], 7);
When we reach the final iteration, the seen sums are
{"0":true,"100":true,"103":true} and the cumulative sum is 103 + 4 = 107 .
We check if 107  7 = 100 appears in the seen sums, because we can exclude any
irrelevant earlier sums by subtracting the target sum from the whole running
total. As 107  7 = 100 does indeed appear in the seen sums, we know that we
can achieve the target sum with the other numbers.
View post:
AlgoDaily 21: Contiguous Subarray Sum
