oyekanmiayo

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

jirevwe

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()

joshbuchea

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