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Blog History / / 2020. 4. 20. 23:01

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불러오는 중입니다...

신청 완료. 참, 고맙네. 경제가 잘 돌도록 좋은 소상공인 사업자에 잘 써야 겠다.

 

A non-empty zero-indexed array A consisting of N integers is given. Array A represents numbers on a tape.

Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].

The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])|

In other words, it is the absolute difference between the sum of the first part and the sum of the second part.

For example, consider array A such that:

A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3

We can split this tape in four places:

  • P = 1, difference = |3 − 10| = 7 
  • P = 2, difference = |4 − 9| = 5 
  • P = 3, difference = |6 − 7| = 1 
  • P = 4, difference = |10 − 3| = 7 

Write a function:

class Solution { public int solution(int[] A); }

that, given a non-empty zero-indexed array A of N integers, returns the minimal difference that can be achieved.

For example, given:

A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3

the function should return 1, as explained above.

Assume that:

  • N is an integer within the range [2..100,000];
  • each element of array A is an integer within the range [−1,000..1,000].

Complexity:

  • expected worst-case time complexity is O(N);
  • expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

// you can also use imports, for example:

// import java.util.*;

 

// you can write to stdout for debugging purposes, e.g.

// System.out.println("this is a debug message");

 

 

class Solution {

    public int solution(int[] A) {

        // write your code in Java SE 8

        int partAlange = 0;

        int minValue = Integer.MAX_VALUE;

       

        int partAsum = 0;

        int partBsum = 0;

       

        for(int i=0; i < A.length ; i++) {

            if(i<=partAlange) partAsum += A[i];

            else partBsum += A[i];

            minValue = Math.min(Math.abs(partAsum - partBsum), minValue);

            partAsum = 0;

            partBsum = 0;

            partAlange++;

        }

       

        return minValue;

    }

}

 

class Solution {

    public int solution(int[] A) {

        // write your code in Java SE 8

        int partAlange = 0;

        int minValue = Integer.MAX_VALUE;

       

        int partAsum = 0;   

       

        for(int j=0; j < A.length ; j++) {

        for(int i=0; i < A.length ; i++) {

            if(i<=partAlange) partAsum += A[i];

            else partAsum -= A[i];           

            }

       

        minValue = Math.min(Math.abs(partAsum), minValue);

            partAsum = 0;           

            partAlange++;

        }

       

        return minValue;

    }

}

 

class Solution { public int solution(int[] A) { // write your code in Java SE 8 int minValue = Integer.MAX_VALUE; int head = A[0]; int tail = 0; for(int i=1; i < A.length ; i++) tail += A[i]; for(int j=1; j < A.length ; j++) { minValue = Math.min(Math.abs(head-tail), minValue); head += A[j]; tail -= A[j]; } return minValue; } }

 

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