## Sunday, May 15, 2022

### Easy_Question1 : Best Time to Buy and Sell Stock

Given an integer array, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

subarray is a contiguous part of an array.

Example 1:

Input: nums = [-2,1,-3,4,-1,2,1,-5,4]
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.


Example 2:

Input: nums = [1]
Output: 1


Example 3:

Input: nums = [5,4,-1,7,8]
Output: 23


Constraints:

• 1 <= nums.length <= 105
• -104 <= nums[i] <= 104

## Solution Article

We need to find out the maximum difference (which will be the maximum profit) between two numbers in the given array. Also, the second number (selling price) must be larger than the first one (buying price).

In formal terms, we need to find $\max(\text{prices[j]} - \text{prices[i]})$, for every $i$ and $j$ such that $j > i$.

#### Approach 1: Brute Force:

Public class Solution {
public int maxProfit(int prices[]) {
int maxprofit = 0;
for (int i = 0; i < prices.length - 1; i++) {
for (int j = i + 1; j < prices.length; j++) {
int profit = prices[j] - prices[i];
if (profit > maxprofit)
maxprofit = profit;
}
}
return maxprofit;
}}


Complexity Analysis

• Time complexity: $O(n^2)$. Loop runs $\dfrac{n (n-1)}{2}$ times.

• Space complexity: $O(1)$. Only two variables - $\text{maxprofit}$ and $\text{profit}$ are used.

#### Approach 2: One Pass

Algorithm

Say the given array is:

[7, 1, 5, 3, 6, 4]

If we plot the numbers of the given array on a graph, we get:

The points of interest are the peaks and valleys in the given graph. We need to find the largest peak following the smallest valley.

We can maintain two variables - minprice and maxprofit corresponding to the smallest valley and maximum profit (maximum difference between selling price and minprice) obtained so far respectively.

Complexity Analysis

• Time complexity: $O(n)$. Only a single pass is needed.

• Space complexity: $O(1)$. Only two variables are used.

public class Solution {
public int maxProfit(int prices[]) {
int minprice = Integer.MAX_VALUE;
int maxprofit = 0;
for (int i = 0; i < prices.length; i++) {
if (prices[i] < minprice)
minprice = prices[i];
else if (prices[i] - minprice > maxprofit)
maxprofit = prices[i] - minprice;
}
return maxprofit;
}
}

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