01-Basics Coding Question (JS)

Practice Loop, functions & Math Problems:- 🔗 Loops in JavaScript | Math Interview Problems Solved

DSA definition - interview question

Question 1: Sum of all natural numbers from 1 to n

function sumOfNaturalNumber(num){
    let sum = 0;
    for(let i=1; i<=num; i++){
        sum = sum + i;
    }
    return sum;
}

console.log(sumOfNaturalNumber(5)); // 15
console.log(sumOfNaturalNumber(10)); // 55
console.log(sumOfNaturalNumber(8)); // 36

Question 2: Sum of digits of a number

function sumOfDigits(num){
    let sum = 0;
    while(num > 0){
        sum += num%10;
        num = Math.floor(num / 10);
    }
    return sum;
}

console.log(sumOfDigits(1287)); // 18

Question 3: Count the number of digits of a number

function countDigits(num){
    num = Math.abs(num);
    let count = 0;
    do {
        count++;
        num = Math.floor(num / 10);
    } while (num > 0);
    return count;
}

console.log(countDigits(121)); // 3
console.log(countDigits(-1211413131)); // 10

Question 4: Check if a number is palindrome

let isPalindrome = function(x) {
    let copyNum = x, reverseNum = 0;

    while(copyNum > 0){
        const lastDigit = copyNum % 10;
        reverseNum = reverseNum * 10 + lastDigit;
        copyNum = Math.floor(copyNum / 10);
    }

    return x === reverseNum;
};

console.log(isPalindrome(121)); // true
console.log(isPalindrome(1234)); // false

Question 5: Find nth Fibonacci number

The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0 and 1.

let fib = function(n) {
    if(n < 2){
        return n;
    }

    let prev = 0, curr = 1, next;
    for(let i=2; i<= n; i++){
        next = prev + curr;
        prev = curr;
        curr = next;
    }
    return next;
};

// Fibonacci Sequence: 0 1 1 2 3 5 8...
console.log(fib(5)); // 5
console.log(fib(10)); // 55

function fibRecursive(n) {
    if (n < 2) {
        return n;
    }

    return fibRecursive(n - 1) + fibRecursive(n - 2);
}
console.log(fibRecursive(5));

// To print 'n' fibonaci number
for (let i = 0; i <= 10; i++) {
    console.log(fib(i));
    console.log(fibRecursive(i));
}

Question 5.2 Tribonacci Number

class Solution {
    public int tribonacci(int n) {
        int t0 = 0, t1 = 1, t2=1, next=-1;

        // Brute force approach
        if(n==0 || n==1)
            return n;
        else if(n==2)
            return n-1;

        for(int i=3; i<=n; i++){
            next = t0 + t1 + t2;
            t0 = t1;
            t1 = t2;            
            t2 = next;
        }

        return next;
    }
}

Question 6: Missing Number in an Array

Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.

/* arr = [0,1,3];
sum = 4;
n = arr.length = 3
sigma = n * (n + 1) / 2 = 3 * 4 / 2 = 6

missingNumber(?) = sigma - sum;
? = 6 - 4 = 2 Ans */

let missingNum = function (arr) {
    let sum = 0;
    for (let i = 0; i < arr.length; i++) {
        sum += arr[i];
    }

    return (arr.length * (arr.length + 1) / 2) - sum;
}

// One Line Solution: 
let missingNum2 = (arr) => arr.length * (arr.length + 1) / 2 - arr.reduce((acc, val) => val + acc);

console.log(missingNum([3,0,1])); // 2
console.log(missingNum2([9,6,4,2,3,5,7,0,1])); // 8

Question 7: Leap Year

function isLeapYr(y) {
    return (y % 400 === 0) || (y % 4 === 0 && y % 100 !== 0);  
}
console.log(isLeapYr(2022));

Condition 1: y % 400 === 0

• Purpose: Checks if the year y is divisible by 400.

• Explanation: If y is divisible by 400, it is a leap year. For example, 2000 is a leap year.

Condition 2: y % 4 === 0 && y % 100 !== 0

• Purpose: Checks if the year y is divisible by 4 but not by 100.

• Explanation: If y is divisible by 4 but not by 100, it is a leap year, unless it’s also divisible by 400. For example, 1996 is a leap year, while 1900 is not.

Question 8: Find Largest of 3 Numbers

let a = 110, b = 105, c = 17;

if (a > b && a > c) {
    console.log(a, ' A is greater');
}
else if (b > a && b > c) {
    console.log(b, " B is greater");
}
else {
    console.log(c, " C is greater");
}

Question 9: Swapping (using 3rd Variable & without using)

let n1 = 16, n2 = 64, temp;

console.log(`a: ${n1} and b: ${n2}`);
temp = n1;
n1 = n2;
n2 = temp;
console.log(`a: ${n1} and b: ${n2}`);

// WITHOUT USING THIRD VARIABLE
console.log(`a: ${n1} and b: ${n2}`);
n1 = n1 + n2;
n2 = n1 - n2;
n1 = n1 - n2;
console.log(`a: ${n1} and b: ${n2}`);

Concept: Comparison between iterative and recursive approaches

Question 10: Factorial - (iterative & recursive method)

// Iterative method
let factorial = function (n) {
    let result = 1;
    for (let i = 2; i <= 5; i++) {
        result *= i;
    }
    return result;
}
console.log(factorial(5));  // 120

// Recursive Method
let fact = function (n) {
    // Base case: if n is 0 or 1, factorial is 1
    if (n === 0 || n === 1) {
        return 1;
    }
    // Recursive case: calculate factorial by multiplying 'n' with factorial of (n-1)
    else {
        return n * fact(n - 1);
    }
};
console.log(fact(4));   // 24

Question 11: Print all the factors of a number & SUM OF ALL FACTOR

let factors = function (n) {
    let result = [];
    for (let i = 1; i < n; i++) {
        if (n % i === 0) {
            // console.log(i);
            result.push(i);
        }
    }
    return result;
}

console.log(factors(15));    // [1, 3, 5]

// One Line Solution: SUM OF ALL FACTORS
console.log(  factors(15).reduce((acc, item) => acc + item)  );    // 9

Question 12: Perfect Number // 6, 28, 496, 8128, ...

A positive integer that is equal to the sum of its proper divisors, excluding itself.

let perfectNum = function (n) {
    let result = 0;
    for (let i = 1; i < n; i++) {
        if (n % i === 0) 
            result += i;        
    }
    return result === n;
}

console.log(perfectNum(28));  // true

Question 13: Prime Number // 2, 3, 5, 7, 11, 13, 17, 19, ...

A prime number is a number divisible only by 1 and itself.

let isPrime = function (n) {
    for (let i = 2; i <= Math.sqrt(n); i++){
        if (n % i === 0)
        return false;
    }
    return true;
}
console.log(isPrime(3));

// List all Prime Number till 100 
for (let i = 2; i <= 50; i++){
    (isPrime(i) ? console.log(i) : null);
}

Question 14: Armstrong Number

• abcd… = aⁿ + bⁿ + cⁿ + dⁿ + …

• Where, n = order(length/digits in number)

• Eg. 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634, 8208, 9474, ...

let isArmstrong = function (n) {
    let sum = 0, originalNum = n, count = String(n).length;

    for (let i = 0; i < count; i++){
        let remainder = n % 10;
        sum += Math.pow(remainder, count)
        n = Math.floor(n/10);
    }
    return sum === originalNum;
}
console.log(isArmstrong(407));    // true

// List some 'n' armstrong numbers
for (let i = 1; i < 1700; i++){
    if (isArmstrong(i))
        console.log(i);
}

Question 15: Check for Perfect Square:

A perfect square number is a non-negative integer that is the square of an integer.

// To check for perfect square
let isPerfectSquare = function (n) {
    let sr = Math.floor(Math.sqrt(n));
    return sr * sr === n;
}

console.log(isPerfectSquare(25));

// To print some 'n' perfect square numbers
for (let i = 1; i <= 100; i++){
    isPerfectSquare(i) ? console.log(i) : null;
}

Question 16: Binary to Decimal to conversion & vice versa

let decimalToBinnary = function (decimal) {
    let binary = [], remainder;
    while (decimal>0) {
        remainder = Math.floor(decimal % 2);        
        // console.log(remainder);
        binary.unshift(remainder);

        decimal = Math.floor(decimal / 2);
        // console.log(decimal);
    }
    return binary.join('');;
}

console.log(decimalToBinnary(16));    // 10000

let BinToDec = function (bin) {
    let decimal = 0;
    bin = String(bin).split('').reverse();
    console.log(bin);

    bin.forEach((elem, i) => {
        if (elem === '1')
            decimal += Math.pow(2, i);
    })
    return decimal;
}

console.log(BinToDec(1100));    // 12

Question 18: Reverse Number

let reverseNum = function (n) {
    let sum = 0, digit;
    while (n > 0) {
        digit = n % 10;
        sum = sum * 10 + digit;
        n = Math.floor(n / 10);
    }
    return sum;
}
console.log(reverseNum(123));

// ONE LINE SOLUTION:
let reverseNum2 = function(n) {
    return +String(n).split('').reverse().join('');
}
console.log(reverseNum2(123));

Practice Questions

• Count Odd Numbers in an Interval Range

• Fizz Buzz

• Power of Two

• Find Square root of a Number