C ] leetCode 2125 - Number of Laser Beams in a Bank

 

 

 

리트코드 2125 문제

 

Anti-theft security devices are activated inside a bank. You are given a 0-indexed binary string array bank representing the floor plan of the bank, which is an m x n 2D matrix. bank[i] represents the ith row, consisting of '0's and '1's. '0' means the cell is empty, while'1' means the cell has a security device.

 

There is one laser beam between any two security devices if both conditions are met:

• The two devices are located on two different rows: r1 and r2, where r1 < r2.
• For each row i where r1 < i < r2, there are no security devices in the ith row.

 

Laser beams are independent, i.e., one beam does not interfere nor join with another.

Return the total number of laser beams in the bank.

 

 

Example 1:

 

 

• Input: bank = ["011001","000000","010100","001000"]
• Output: 8
• Explanation: Between each of the following device pairs, there is one beam.
• In total, there are 8 beams:

* bank[0][1] -- bank[2][1]

* bank[0][1] -- bank[2][3]

* bank[0][2] -- bank[2][1]

* bank[0][2] -- bank[2][3]

* bank[0][5] -- bank[2][1]

* bank[0][5] -- bank[2][3]

* bank[2][1] -- bank[3][2]

* bank[2][3] -- bank[3][2]

• Note that there is no beam between any device on the 0th row with any on the 3rd row. This is because the 2nd row contains security devices, which breaks the second condition.

 

 

Example 2:

 

• Input: bank = ["000","111","000"]
• Output: 0
• Explanation: There does not exist two devices located on two different rows.

 

 

Constraints:

• m == bank.length
• n == bank[i].length
• 1 <= m, n <= 500
• bank[i][j] is either '0' or '1'.

 

 

 

레이저 빔의 총합을 구하면 되는 문제인데 일단 bank[0]에는 빔이 없고 bank[1]부터 bank[bankSize-1]까지 등장하는 빔은 (현재 행의 디바이스 개수 * 존재하는 디바이스 장치가 0이 아니면서 가장 인접한 직전 행의 디바이스 개수)로 구해진다. 이를 누적합해서 리턴하면 되는 간단한 문제이다.

 

 

 

 

#include <stdio.h>

int numberOfBeams(char** bank, int bankSize) {
    int before = 0;
    int ret = 0;

    for(int i = 0; i < bankSize; i++) {
        int temp = 0;
        char *p = bank[i];
        while(*p != '\0') {
            temp += *p - '0';
            p++;
        }
        ret += before * temp;
        before = temp == 0 ? before : temp;
    }
    
    return ret;
}

int main()
{
    char *bank[] = {"011001","000000","010100","001000"};
    
    printf("Result = %d\n", numberOfBeams(bank, sizeof(bank) / sizeof(bank[0])));

    return 0;
}