- Notifications
You must be signed in to change notification settings - Fork1
Adrinlol/Leetcode-JavaScript-Algorithm-Solutions
Folders and files
Name | Name | Last commit message | Last commit date | |
---|---|---|---|---|
Repository files navigation
An algorithm is a series of step-by-step instructions that describe how to do something.
To write an effective algorithm, it helps to break a problem down into smaller parts and think carefully about how to solve each part with code.
This repository contains JavaScript algorithm examples and solutions from Leetcode.
All of these examples are from FreeCodeCamp'sproblemset.
Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.
You may assume that each input would haveexactly one solution, and you may not use the same element twice.
You can return the answer in any order.
Example 1:
Input:nums=[2,7,11,15],target=9Output:[0,1]Explanation:Becausenums[0]+nums[1]==9,wereturn[0,1].Example2:
Input:nums=[3,2,4],target=6Output:[1,2]Example3:
Input:(nums=[3,3]),(target=6);Output:[0,1];
Constraints:
- 2 <= nums.length <= 104
- -109 <= nums[i] <= 109
- -109 <= target <= 109
- Only one valid answer exists.
The space complexity for this solution is a constant O(1).
The time complexity for this solution is O(N)^2, as we are looping over the array twice.
vartwoSum=function(nums,target){for(leti=0;i<nums.length;i++){for(letj=i+1;j<nums.length;j++){if(nums[i]+nums[j]===target){return[i,j];}}}};
To improve the first solution we can improve time complexity by sacrificing space and creating an object and storing the difference between the index of nums array and the target value.
The space complexity for this solution is O(N).
The time complexity for this solution is O(N).
vartwoSum=function(nums,target){letnumsStorage={};for(leti=0;i<nums.length;i++){if(numsStorage[nums[i]]===undefined){numsStorage[target-nums[i]]=i;}else{return[numsStorage[nums[i]],i];}}};
Given an integer x, return true if x is palindrome integer.
An integer is a palindrome when it reads the same backward as forward.
For example, 121 is a palindrome while 123 is not.
Example 1:
Input:x=121Output:trueExplanation:121readsas121fromlefttorightandfromrighttoleft.
Example 2:
Input:x=-121Output:falseExplanation:Fromlefttoright,itreads-121.Fromrighttoleft,itbecomes121-.Thereforeitisnotapalindrome.
Example 3:
Input:x=10Output:falseExplanation:Reads01fromrighttoleft.Thereforeitisnotapalindrome.
Constraints:
- -231 <= x <= 231 - 1
The space complexity for this solution is O(N).
The time complexity for this solution is O(N).
varisPalindrome=function(x){letsplitNumber=x.toString().split("");letreverseNumberArr=[];for(leti=splitNumber.length;i>-1;i--){reverseNumberArr.push(splitNumber[i]);}returnNumber(reverseNumberArr.join(""))===x;};
The space complexity for this solution is a constant O(1).
The time complexity for this solution is also O(1).
varisPalindrome=function(x){returnx.toString()===x.toString().split("").reverse().join("");};
Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.
SymbolValueI1V5X10L50C100D500M1000
For example, 2 is written as II in Roman numeral, just two ones added together. 12 is written as XII, which is simply X + II. The number 27 is written as XXVII, which is XX + V + II.
Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:
I can be placed before V (5) and X (10) to make 4 and 9.X can be placed before L (50) and C (100) to make 40 and 90.C can be placed before D (500) and M (1000) to make 400 and 900.Given a roman numeral, convert it to an integer.
Example 1:
Input:s="III";Output:3;Explanation:III=3;
Example 2:
Input:s="LVIII";Output:58;Explanation:(L=50),(V=5),(III=3);
Example3:Input:s="MCMXCIV"Output:1994Explanation:M=1000,CM=900,XC=90andIV=4.
Constraints:
- 1 <= s.length <= 15
- s contains only the characters ('I', 'V', 'X', 'L', 'C', 'D', 'M').
- It is guaranteed that s is a valid roman numeral in the range [1, 3999].
varromanToInt=(s)=>{letresult=0;constromanVal={I:1,V:5,X:10,L:50,C:100,D:500,M:1000};for(leti=0;i<s.length;i++){if(romanVal[s[i]]<romanVal[s[i+1]]){result-=romanVal[s[i]];}else{result+=romanVal[s[i]];}}returnresult;};
Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string "".
Example 1:
Input:strs=["flower","flow","flight"];Output:"fl";
Example 2:
Input:strs=["dog","racecar","car"]Output:""Explanation:Thereisnocommonprefixamongtheinputstrings.
Constraints:
- 1 <= strs.length <= 200
- 0 <= strs[i].length <= 200
- strs[i] consists of only lowercase English letters.
varlongestCommonPrefix=function(strs){if(!strs.length)return"";for(leti=0;i<strs[0].length;i++){for(letj=1;j<strs.length;j++){if(strs[0][i]!==strs[j][i]){returnstrs[0].slice(0,i);}}}returnstrs[0];};
About
This repository contains JavaScript algorithm examples and solutions from Leetcode.
Topics
Resources
Uh oh!
There was an error while loading.Please reload this page.