Problem: You have given a stock prices for next 10 days. Find out: max benefit in best time complexity in buy and sell 1 share ?
Condition: Share must be sold any day after buying date.
For ex:
Share price in thousands
5 1 4 6 7 8 4 3 7 9
Max benefit 9 – 1 = 8 thousand
Solution:
The max benefit is the case where you buy at the lowest price and sell at the highest, with the condition that the buy has to be before sell.
public static int getMaxBenefit(int []S)
{
int maxBenefit = 0;
int minPrice = Integer.MAX_VALUE;
for (int i = 0; i < S.length; i++)
{
maxBenefit = Math.max(maxBenefit, S[i] - minPrice);
minPrice = Math.min(S[i], minPrice);
}
return maxBenefit;
}
Time Complexity: O(N)
Here is answer :
public class Question1 {
/*Problem: You have given a stock prices for next 10 days. Find out: max benefit in best time complexity in buy and sell 1 share ?
Condition: Share must be sold any day after buying date.
For ex:
Share price in thousands
5 1 4 6 7 8 4 3 7 9
Max benefit 9 – 1 = 8 thousand*/
public static void main(String[] args) {
int[] shares = {5,3,4,25,9,7};
int maxbenifit = 0;
int minbuy=0;
int maxSell=0;
int buyPrice = 0;
int sellPrice = 0;
for(int i=0;i<shares.length;i++){
buyPrice = shares[i];
for (int j = i+1; j 0 && sellPrice – buyPrice > maxbenifit){
maxbenifit = sellPrice – buyPrice;
minbuy = buyPrice;
maxSell = sellPrice;
/*System.out.println( ” min minbuy :” + minbuy );
System.out.println( ” max maxSell :” + maxSell );
System.out.println( ” max benifit :” + maxbenifit );
System.out.println( “————————-”);*/
}
}
}
System.out.println( ” min pirce :” + minbuy );
System.out.println( ” max pirce :” + maxSell );
}
}
int a[]={7,6,23,19,10,11,9,3,15};
Arrays.sort(a);
int i =a[a.length-1]-a[0];
System.out.println(a[a.length-1]+” , “+a[0]+” and difference is “+i);