Day 29 Of Dsa Problem Solving

✅ Question 1: Sum of Digits

🧾 Problem Statement:

Given an array of integers called prices, return the sum of all elements in the array.

📥 Input: An array of integers.
📤 Output: A single integer — the sum of all elements in the array.

Example:

Input: [1, 2, 3]
Output: 6

🧩 Approach

First, initialize a variable sum with a value of 0.
Loop through each element in the array using a for loop.
In each iteration, add the current element to the sum.
After the loop ends, return the final value of sum.

💻 Code:

var SumOfDigits = function(prices) {
    let sum = 0;
    for (let i = 0; i < prices.length; i++) {
        sum += prices[i];
    }
    return sum;
};

console.log(SumOfDigits([1, 2, 3])); // Output: 6

🧪 Edge Cases:

✅ Empty array [] → Output should be 0 since there are no elements to sum.
✅ Array with one element [10] → Output should be 10.
✅ Array with negative numbers [-1, -2, 3] → Output should be 0.

Complexity

⏱️ Time Complexity:

O(n) where n is the number of elements in the array.

📦 Space Complexity:

O(1) – No extra space used apart from a constant sum variable.

✅ Question 2: Best Time to Buy Stock

The goal is to find the maximum profit from a single buy and sell operation on a stock, where the buy must happen before the sell.

To solve this, we:
- Keep track of the minimum price seen so far.
- At each step, calculate the profit if we sell at the current price.
- If this profit is greater than the current maximum profit, we update it.

Example:


Input: [1, 2, 3]
Output: 6

🧩 Approach

Initialize two variables:
- min_Price to store the lowest price so far (starting with prices[0])
- max_Profit to store the highest profit so far (starting at 0)
Iterate through the prices array** starting from index 1:
- At each step:
  - If prices[i] - min_Price (profit) is greater than max_Profit, update max_Profit
  - If prices[i] is less than min_Price, update min_Price
Return max_Profit after the loop ends.

This approach ensures that we always buy at the lowest price seen so far and sell at the highest price that comes after, achieving the maximum profit from one transaction.

💻 Code:


var maxProfit = function(prices) {
    let min_Price = prices[0]
    let max_Price = 0;


  for(let x = 1 ; x < prices.length ; x++){

       if(prices[x] - min_Price >  max_Price){
           max_Price =  prices[x] - min_Price

      }

    if(prices[x] < min_Price){
     min_Price  = prices[x]
    }

  }  

  return max_Price;


};