Maximum path sum I
By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
3
7 4
2 4 6
8 5 9 3
That is, 3 + 7 + 4 + 9 = 23.
Find the maximum total from top to bottom of the triangle below:
75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)
Solution
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| #!/usr/bin/python
# -*- coding: utf-8 -*-
def main():
matrix = []
file = open('data_p18.txt')
for line in file.readlines():
matrix.append(map(int, line.replace('\n', '').split(' ')))
(x, y) = (0, 0)
for y in range(len(matrix)):
for x in range(len(matrix[y])):
if y > 0:
greaterParentPathValue = 0
if x > 0:
greaterParentPathValue = matrix[y - 1][x - 1]
if x < len(matrix[y - 1]) and matrix[y - 1][x] \
> greaterParentPathValue:
greaterParentPathValue = matrix[y - 1][x]
matrix[y][x] += greaterParentPathValue
print max(matrix[-1])
if __name__ == '__main__':
main()
|
I’m the 70471st person to have solved this problem.
Maximum path sum II
By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
3
7 4
2 4 6
8 5 9 3
That is, 3 + 7 + 4 + 9 = 23.
Find the maximum total from top to bottom in triangle.txt (right click and ‘Save Link/Target As…’), a 15K text file containing a triangle with one-hundred rows.
NOTE: This is a much more difficult version of Problem 18. It is not possible to try every route to solve this problem, as there are 2^99 altogether! If you could check one trillion (10^12) routes every second it would take over twenty billion years to check them all. There is an efficient algorithm to solve it. ;o)
Solution
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| package main
import (
"bufio"
"fmt"
"io"
"os"
"sort"
"strconv"
"strings"
)
func main() {
f, err := os.Open("data_p67.txt")
if nil != err {
fmt.Println(err)
os.Exit(1)
}
defer f.Close()
matrix := make([][]int, 0)
reader := bufio.NewReader(f)
for {
line, err := reader.ReadString('\n')
if nil != err || io.EOF == err {
break
}
row := make([]int, 0)
numbers := strings.Split(strings.Replace(line, "\n", "", -1), " ")
for i := range numbers {
number, _ := strconv.Atoi(numbers[i])
row = append(row, number)
}
matrix = append(matrix, row)
}
for y := 0; y < len(matrix); y++ {
for x := 0; x < len(matrix[y]); x++ {
if y > 0 {
greaterParentPathValue := 0
if x > 0 {
greaterParentPathValue = matrix[y-1][x-1]
}
if x < len(matrix[y-1]) && matrix[y-1][x] > greaterParentPathValue {
greaterParentPathValue = matrix[y-1][x]
}
matrix[y][x] += greaterParentPathValue
}
}
}
sort.Sort(sort.Reverse(sort.IntSlice(matrix[len(matrix)-1])))
fmt.Println(matrix[len(matrix)-1][0])
}
|
I’m the 50650th person to have solved this problem.