Do anything with Arrays

This post will be updated regularly as I learn

Loop Through Array

let i = 0; i < nums.length; i++

Problem 1: Maximum Consecutive Ones

Input: nums = [1,1,0,1,1,1]
Output: 3
Explanation: The first two digits or the last three digits are consecutive 1s. The maximum number of consecutive 1s is 3.

Knowledge Check - JavaScript

• Math.max(result, count) => Keep counter updated with max value
• arr.length => Used to loop an array

Solution - JavaScript

var findMaxConsecutiveOnes = function(nums) {
    let count = 0;
    let result = 0;

    for (let i = 0; i < nums.length; i++) {
        // Reset count when 0 is found
        if (nums[i] == 0)
            count = 0;

        // If 1 is found, increment
        else {
            // increase count
            count++;
            result = Math.max(result, count);
        }
    }

    return result;
};

Solution - Java

Math.max(result, count) and arr.length are in java too 🤯

class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int count = 0; //initialize count
        int result = 0; //initialize max

        for (int i = 0; i < nums.length; i++)
        {

            // Reset count when 0 is found
            if (nums[i] == 0)
                count = 0;

            // If 1 is found, increment count
            else
            {
                count++;//increase count
                result = Math.max(result, count);
            }
        }
        return result;
    }
}

Problem 2 : Find Numbers with Even Number of Digits

Input: nums = [12,345,2,6,7896]
Output: 2
Explanation: 
12 contains 2 digits (even number of digits). 
345 contains 3 digits (odd number of digits). 
2 contains 1 digit (odd number of digits). 
6 contains 1 digit (odd number of digits). 
7896 contains 4 digits (even number of digits). 
Therefore only 12 and 7896 contain an even number of digits.

Knowledge Check - JavaScript

• Math.floor(n / 10) - Can be used for count digits until it becomes 0
• Math.floor(numberOfDigits % 2) == 0 - Checking for even

Solution - Javascript

var findNumbers = function(nums) {
    let result = 0
    // loop through numbers
     for (let i = 0; i < nums.length; i++){
         // find number of digits
         const num = nums[i]
         const numberOfDigits = countDigit(num)
         const isEven = Math.floor(numberOfDigits % 2) == 0

         // if even increse counter
         isEven && result ++
     }
    return result
};

function countDigit(n){
    let count = 0;
    while (n != 0) {
        n = Math.floor(n / 10);
        count ++;
    }
    return count;
}

Sorting an array

• arr.sort((a, b)=>{return a - b}) can be used for numeric sorting (By default, the sort() function sorts values as strings.)

Squares of a Sorted Array

Input: nums = [-4,-1,0,3,10]
Output: [0,1,9,16,100]
Explanation: After squaring, the array becomes [16,1,0,9,100].
After sorting, it becomes [0,1,9,16,100].

var sortedSquares = function(nums) {
    let squaredArray = []
    for (let i =0 ; i < nums.length; i++){
        const num = nums[i]
        const square = num * num
        squaredArray.push(square)
    }
    const result = squaredArray.sort((a, b)=>{return a - b})
    return result
};