oyekanmiayo

#include <stdlib.h>

int main(){

    // Read integer from line
    int t; 
    scanf("%d",&t);
    
    // Read integer from line times
    for(int a0 = 0; a0 < t; a0++){

oyekanmiayo

// See https://developers.google.com/apps-script/guides/services/external
// and https://developers.google.com/apps-script/reference/url-fetch/url-fetch-app
function fetchListOfChannels() {
  var options = {
    'method' : 'get',
    'contentType': 'application/json',
    'headers': {
        'Authorization': 'Bearer <access_token>',
    }
  };

oyekanmiayo

// Enter value in a cell
var valueRange = Sheets.newValueRange();
valueRange.values = [["text"]];
const result = Sheets.Spreadsheets.Values.update(valueRange,
  SHEET_ID, A1_NOTATION, {valueInputOption: "RAW"});
  
// Fetches values as a multidimensional array
Sheets.Spreadsheets.Values.get(SHEET_ID, RANGE)

oyekanmiayo

Remove MySQL completely

1. Open the Terminal

2. Use mysqldump to backup your databases

3. Check for MySQL processes with: ps -ax | grep mysql

4. Stop and kill any MySQL processes

5. Analyze MySQL on HomeBrew:

   brew remove mysql
   

oyekanmiayo

// reference: https://stackoverflow.com/questions/46310113/consume-a-delete-endpoint-from-golang
package main

import (
    "fmt"
    "io/ioutil"
    "net/http"
)

func sendRequest() {

oyekanmiayo

import javax.crypto.Cipher;
import javax.crypto.spec.IvParameterSpec;
import javax.crypto.spec.SecretKeySpec;

public class AESCBCDecryption {
    
    static final String cipher_type = "AES/CBC/PKCS5Padding";

    public static void main(String[] args) {
        String key = "57067125438768260656188878670043";

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

oyekanmiayo

class Solution {
    
    /*
    Approach:
    - Find start position
    - Count number of visitable nodes in the grids
    - A path is only valid if we read end node && all visitable nodes have been visited
    - Perform dfs, everytime we visit a cell, we mark as visited. 
    - We cannot visit a cell more than once
    - Handle boundary edge cases too

oyekanmiayo

/**
 * // This is the robot's control interface.
 * // You should not implement it, or speculate about its implementation
 * interface Robot {
 *     // Returns true if the cell in front is open and robot moves into the cell.
 *     // Returns false if the cell in front is blocked and robot stays in the current cell.
 *     public boolean move();
 *
 *     // Robot will stay in the same cell after calling turnLeft/turnRight.
 *     // Each turn will be 90 degrees.

oyekanmiayo

class Solution {
    // Final solution
    public int findMin(int[] nums) {
        
        int start = 0;
        int end = nums.length - 1;

        while(start < end){
            int mid = start + (end - start) / 2;
            if(nums[mid] < nums[end]){