Press "Enter" to skip to content

Posts tagged as “Easy”

LeetCode 121. Best Time to Buy and Sell Stock (javascript)

By stone on March 27, 2021

You are given an array prices where prices[i] is the price of a given stock on the ith day.

You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.

Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.

Example 1:

Input: prices = [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.

Example 2:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transactions are done and the max profit = 0.

Constraints:

  • 1 <= prices.length <= 105
  • 0 <= prices[i] <= 104

Idea:

Tracking two variables – minStock(min price of the stock) and maxGain(max profit)

Solution:

/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function(prices) {
    let maxGain = 0;
    let minStock = prices[0];
    
    for (let i = 0; i < prices.length; i++) {
        minStock = Math.min(minStock, prices[i]);
        maxGain = Math.max(maxGain, prices[i] - minStock);
    }
    return maxGain;
};

LeetCode 26. Remove Duplicates from Sorted Array (javascript)

By stone on March 26, 2021

Given a sorted array nums, remove the duplicates in-place such that each element appears only once and returns the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means a modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

Example 1:

Input: nums = [1,1,2]
Output: 2, nums = [1,2]
Explanation: Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the returned length.

Example 2:

Input: nums = [0,0,1,1,1,2,2,3,3,4]
Output: 5, nums = [0,1,2,3,4]
Explanation: Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively. It doesn't matter what values are set beyond the returned length.

Constraints:

  • 0 <= nums.length <= 3 * 104
  • -104 <= nums[i] <= 104
  • nums is sorted in ascending order.

Solution:

/**
 * @param {number[]} nums
 * @return {number}
 */
var removeDuplicates = function(nums) {
    if(nums === null || nums.length === 0) return 0;
    if(nums.length == 1) return 1;
    var count = 0;
    for(var i = 0; i < nums.length; i++) {
        if(nums[i] !== nums[i+1]){
            count++;
            nums[count] = nums[i+1];
        }
    }
    return count;
};

LeetCode 387. First Unique Character in a String

By stone on March 18, 2021

Given a string, find the first non-repeating character in it and return its index. If it doesn’t exist, return -1.

Examples:

s = "leetcode"
return 0.

s = "loveleetcode"
return 2.

Note: You may assume the string contains only lowercase English letters.

Idea: Hashtable

Solution:

/**
 * @param {string} s
 * @return {number}
 */
var firstUniqChar = function(s) {
    let map = new Map;
    
    for (let i = 0; i < s.length; i++) {
        if (map.has(s[i])) {
            map.set(s[i], map.get(s[i]) + 1);
        } else {
            map.set(s[i], 1);
        }
    }
    
    for (let i = 0; i < s.length; i++) {
        if (map.get(s[i]) === 1) return i;
    }
    return -1;
};

LeetCode 1. Two Sum

By stone on March 18, 2021

Problem:

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Input: nums = [2,7,11,15], target = 9

Output: [0,1]

Output: Because nums[0] + nums[1] == 9, we return [0, 1].

Idea:

Hashtable

Solution:

Hashtable / Javascript

Time complexity: O(n)

Space complexity: O(n)

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(nums, target) {
    let map = new Map();
    
    for (let i = 0; i < nums.length; i++) {
        let diff = target - nums[i];
        if (map.has(diff)) {
            return [i, map.get(diff)];
        } else {
            map.set(nums[i],i);
        }
    }
};