Here is the list of top 10 most hated programming languages in 2022 and what are they used for!
1. Two Sum
Problem
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 have exactly 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 = 9 Output: [0,1] Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].
Example 2:
Input: nums = [3,2,4], target = 6 Output: [1,2]
Example 3:
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.
Solution
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class Solution {
public int[] twoSum(int[] a, int target) {
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for(int i = 0; i< a.length; i++) {
Integer data = map.get(target - a[i]);
if(data == null) {
map.put(a[i], i);
} else {
return new int[]{data, i};
}
}
return null;
}
}
2. Best Time to Buy and Sell Stock
Problem (Leetcode 121)
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 <= 1050 <= prices[i] <= 104
Solution
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class Solution {
public int maxProfit(int[] prices) {
int max = 0;
if(prices.length == 0) {
return max;
}
int min = prices[0];
for(int i = 0; i< prices.length; i++) {
int val = prices[i] - min;
if(val > max) {
max = val;
}
if(min > prices[i]) {
min = prices[i];
}
}
return max;
}
}
