[codilty] MaxProfit

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문제

An array A consisting of N integers is given. It contains daily prices of a stock share for a period of N consecutive days. If a single share was bought on day P and sold on day Q, where 0 ≤ P ≤ Q < N, then the profit of such transaction is equal to A[Q] − A[P], provided that A[Q] ≥ A[P]. Otherwise, the transaction brings loss of A[P] − A[Q].

For example, consider the following array A consisting of six elements such that:

A[0] = 23171 A[1] = 21011 A[2] = 21123 A[3] = 21366 A[4] = 21013 A[5] = 21367

If a share was bought on day 0 and sold on day 2, a loss of 2048 would occur because A[2] − A[0] = 21123 − 23171 = −2048. If a share was bought on day 4 and sold on day 5, a profit of 354 would occur because A[5] − A[4] = 21367 − 21013 = 354. Maximum possible profit was 356. It would occur if a share was bought on day 1 and sold on day 5.

Write a function,

 

class Solution {

   public int solution(int[] A);

}

 

that, given an array A consisting of N integers containing daily prices of a stock share for a period of N consecutive days, returns the maximum possible profit from one transaction during this period. The function should return 0 if it was impossible to gain any profit.

For example, given array A consisting of six elements such that:

A[0] = 23171 A[1] = 21011 A[2] = 21123 A[3] = 21366 A[4] = 21013 A[5] = 21367

the function should return 356, as explained above.

Write an efficient algorithm for the following assumptions:

• N is an integer within the range [0..400,000];
• each element of array A is an integer within the range [0..200,000].

 

나의 첫번째 풀이 

쉽다고 생각하고 이렇게 하고 제출.. 했으나..

public static int solution(int[] A) {
    // Implement your solution here
    int ret = 0;
    int min = A[0];
    int max = 0;
    for (int i = 0; i < A.length; i++){
        if (min > A[i]){
            min = A[i];
            max = A[i];
        }
        if (max < A[i]){
            max = A[i];
        }
    }
    return max - min < 0 ? ret : max - min;
}

그래 이렇게 쉬울리가 읍지

나의 풀이 

public static int solution(int[] A) {
    int ret = 0;
    if (A.length < 2) {
        return ret;
    }
    // Implement your solution here
    int min = A[0];
    int max = 0;
    for (int i = 0; i < A.length; i++) {
        if (min > A[i]) {
            min = A[i];
            max = A[i];
        }
        if (max < A[i]) {
            max = A[i];
        }
        if(max - min > ret) {
            ret = max - min;
        }
    }
    return ret;
}