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Format: <type>(<scope>): <subject>
<scope>
is optional
class Solution { | |
// This approach is O(n), where n = length of string s | |
// The sliding window approach uses two pointers to traverse a string in linear time. Each character is visited | |
// at most twice | |
// There's a method `mapHasAllChars` whose time complexity may not be intuitive, so it's worth mentioning. At most | |
// each map will contain all the characters from the ascii set (0 - 255), so we know that at most we will always visit 256 | |
// keys every time we call this method. So note this => | |
// AS LONG AS WE KNOW THE WORST CASE FOR NUMBER OF LOOP ITERATIONS BEFOREHAND, THAT LOOP'S TIME COMPLEXITY IS CONSTANT | |
public String minWindow(String s, String t) { | |
//Edge Cases |
const start_time = performance.now(); | |
const count: number = 1000; | |
const wallets: Wallet[] = []; | |
const walletCount = await WalletModel.find({}).estimatedDocumentCount(); | |
// create a cursor for every `currentIndex` documents in the collection | |
const getNthCursor = async (currentIndex: number) => | |
await WalletModel.find({ deleted_at: undefined }, { timeout: false }) | |
.lean() |