package com.fishercoder.solutions;

import java.util.HashMap;
import java.util.List;
import java.util.Map;

/**
 * There is a brick wall in front of you. The wall is rectangular and has several rows of bricks. The bricks have the same height but different width. You want to draw a vertical line from the top to the bottom and cross the least bricks.

 The brick wall is represented by a list of rows. Each row is a list of integers representing the width of each brick in this row from left to right.

 If your line go through the edge of a brick, then the brick is not considered as crossed. You need to find out how to draw the line to cross the least bricks and return the number of crossed bricks.

 You cannot draw a line just along one of the two vertical edges of the wall, in which case the line will obviously cross no bricks.

 Example:
 Input:
 [[1,2,2,1],
 [3,1,2],
 [1,3,2],
 [2,4],
 [3,1,2],
 [1,3,1,1]]
 Output: 2
 Explanation:

 Note:
 The width sum of bricks in different rows are the same and won't exceed INT_MAX.
 The number of bricks in each row is in range [1,10,000]. The height of wall is in range [1,10,000]. Total number of bricks of the wall won't exceed 20,000.
 */
public class _554 {
    //credit to: https://leetcode.com/articles/brick-wall/

    /**we make use of a HashMap
     map which is used to store entries in the form:
     (sum,count). Here,
     sum refers to the cumulative sum of the bricks' widths encountered in the current row, and
     count refers to the number of times the corresponding sum is obtained. Thus,
     sum in a way, represents the positions of the bricks's boundaries relative to the leftmost boundary.

     This is done based on the following observation:
     We will never obtain the same value of sum twice while traversing over a particular row.
     Thus, if the sum value is repeated while traversing over the rows, it means some row's brick boundary coincides with some previous row's brick boundary.
     This fact is accounted for by incrementing the corresponding count value.

     But, for every row, we consider the sum only upto the second last brick, since the last boundary isn't a valid boundary for the solution.*/

    public int leastBricks(List<List<Integer>> wall) {
        Map<Integer, Integer> map = new HashMap();
        for (List<Integer> row : wall) {
            int sum = 0;
            for (int i = 0; i < row.size() - 1; i++) {
                //NOTE: i < row.size()-1
                sum += row.get(i);
                if (map.containsKey(sum)) {
                    map.put(sum, map.get(sum) + 1);
                } else {
                    map.put(sum, 1);
                }
            }
        }
        int result = wall.size();
        for (int key : map.keySet()) {
            result = Math.min(result, wall.size() - map.get(key));
        }
        return result;
    }

}

package com.fishercoder.solutions;

import java.util.HashMap;
import java.util.List;
import java.util.Map;

/**
 * 554. Brick Wall
 *
 * There is a brick wall in front of you. The wall is rectangular and has several rows of bricks.
 * The bricks have the same height but different width.
 * You want to draw a vertical line from the top to the bottom and cross the least bricks.
 * The brick wall is represented by a list of rows.
 * Each row is a list of integers representing the width of each brick in this row from left to right.
 * If your line go through the edge of a brick,
 * then the brick is not considered as crossed.
 * You need to find out how to draw the line to cross the least bricks and return the number of crossed bricks.
 * You cannot draw a line just along one of the two vertical edges of the wall, in which case the line will obviously cross no bricks.

 Example:
 Input:
 [[1,2,2,1],
 [3,1,2],
 [1,3,2],
 [2,4],
 [3,1,2],
 [1,3,1,1]]

 Output: 2
 Explanation:

 Note:
 The width sum of bricks in different rows are the same and won't exceed INT_MAX.
 The number of bricks in each row is in range [1,10,000]. The height of wall is in range [1,10,000]. Total number of bricks of the wall won't exceed 20,000.
 */
public class _554 {
    //credit to: https://leetcode.com/articles/brick-wall/

    /**we make use of a HashMap
     map which is used to store entries in the form:
     (sum,count). Here,
     sum refers to the cumulative sum of the bricks' widths encountered in the current row, and
     count refers to the number of times the corresponding sum is obtained. Thus,
     sum in a way, represents the positions of the bricks's boundaries relative to the leftmost boundary.

     This is done based on the following observation:
     We will never obtain the same value of sum twice while traversing over a particular row.
     Thus, if the sum value is repeated while traversing over the rows, it means some row's brick boundary coincides with some previous row's brick boundary.
     This fact is accounted for by incrementing the corresponding count value.

     But, for every row, we consider the sum only upto the second last brick, since the last boundary isn't a valid boundary for the solution.*/

    public int leastBricks(List<List<Integer>> wall) {
        Map<Integer, Integer> map = new HashMap();
        for (List<Integer> row : wall) {
            int sum = 0;
            for (int i = 0; i < row.size() - 1; i++) {
                //NOTE: i < row.size()-1
                sum += row.get(i);
                if (map.containsKey(sum)) {
                    map.put(sum, map.get(sum) + 1);
                } else {
                    map.put(sum, 1);
                }
            }
        }
        int result = wall.size();
        for (int key : map.keySet()) {
            result = Math.min(result, wall.size() - map.get(key));
        }
        return result;
    }

}

refactor 554