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Commit09b5413

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Added Practise Problems for Loop & Math topics
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‎Basics/README.md‎

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#Practise Loop, functions & Math Problems
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###Question 1: sum of all natural numbers from 1 to n
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```javascript
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functionsumOfNaturalNumber(num){
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let sum=0;
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for(let i=1; i<=num; i++){
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sum= sum+ i;
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}
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return sum;
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}
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console.log(sumOfNaturalNumber(5))//15
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console.log(sumOfNaturalNumber(10))//55
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console.log(sumOfNaturalNumber(8))//36
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```
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###Question 2: Sum of digits of a number
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```javascript
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functionsumOfDigits(num){
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let sum=0;
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while(num>0){
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sum+= num%10;
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num=Math.floor(num/10);
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}
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return sum;
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}
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console.log(sumOfDigits(1287))//18
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```
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###Question 3: count the number of digits of a number
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```javascript
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functioncountDigits(num){
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num=Math.abs(num);
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let count=0;
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do {
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count++;
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num=Math.floor(num/10);
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}while (num>0);
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return count;
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}
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console.log(countDigits(121))//3
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console.log(countDigits(-1211413131))//10
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```
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###Question 4: Given an integer x, return true if x is a palindrome, and false otherwise.
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A palindrome number is a number that remains the same when digits are reversed
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```javascript
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letisPalindrome=function(x) {
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let copyNum= x, reverseNum=0;
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while(copyNum>0){
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constlastDigit= copyNum%10;
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reverseNum= reverseNum*10+ lastDigit;
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copyNum=Math.floor(copyNum/10)
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}
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return x=== reverseNum
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};
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console.log(isPalindrome(121))//true
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console.log(isPalindrome(1234))//false
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```
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###Question 5: Find nth fibonacci number
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The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence,
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such that each number is the sum of the two preceding ones, starting from 0 and 1.
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```javascript
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letfib=function(n) {
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if(n<2){
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return n;
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}
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let prev=0, curr=1, next;
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for(let i=2; i<= n; i++){
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next= prev+ curr;
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prev= curr;
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curr= next;
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}
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return next;
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};
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// Fibonacci Sequence: 0 1 1 2 3 5 8...
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console.log(fib(5))// 5
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console.log(fib(10))//55
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```
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###Question 6: Missing Number
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Given an array nums containing n distinct numbers in the range[0, n],
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return the only number in the range that is missing from the array.
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```javascript
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letmissingNumber=function(nums) {
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let sum=0;
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for(let i=0; i<nums.length; i++){
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sum+= nums[i];
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}
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returnnums.length*(nums.length+1)/2- sum;
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};
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// One Line Solution:
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letmissingNumber= (nums)=>nums.length*(nums.length+1)/2-nums.reduce((acc,num)=> num+ acc);
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console.log(missingNumber([3,0,1]))//2
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console.log(missingNumber([9,6,4,2,3,5,7,0,1]))//8
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```
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#Practice Questions
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-[Count odd Numbers in an Interval Range](https://leetcode.com/problems/count-odd-numbers-in-an-interval-range/)
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-[Fizz Buzz](https://leetcode.com/problems/fizz-buzz/)
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-[Power of Two](https://leetcode.com/problems/power-of-two/)
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-[Find Square root of a Number](https://leetcode.com/problems/sqrtx/)

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