Mohsin Kaleem

I'm a developer

Gold Rush

func goldRush(gMap Map) int {
	max := 0
	for i, col := range gMap {
		for j, _ := range col {
    		max = variadicMax(max, goldRushPath(gMap, i, j, 0))
		}
	}
	return max
}

func goldRushPath(gMap Map, x, y, sum int) int {
	if x < 0 || y < 0 || x >= len(gMap) || y >= len(gMap[0]) || gMap[x][y] == 0 {
		return sum
	}

	sum += gMap[x][y]
	defer (func(val int) { gMap[x][y] = val })(gMap[x][y])
	gMap[x][y] = 0

	return variadicMax(
		goldRushPath(gMap, x+1, y, sum),
		goldRushPath(gMap, x-1, y, sum),
		goldRushPath(gMap, x, y+1, sum),
		goldRushPath(gMap, x, y-1, sum),
	)
}

#Challenge

SourceLanguage
DailyBytego

#Solution

Given a 2D matrix representing a gold mine, where each cells value represents an amount of gold, return the maximum amount of gold you can collect given the following rules:

  1. You may start or stop collecting gold at any poisition.
  2. You can never visit a cell containing 0 gold.
  3. You cannot visit the same cell more than once.
  4. FRom the current cell you may walk one cell left, right, up or down.

For example, given the following gold mine:

[0 2 0]
[8 6 3]
[0 9 0]

We should get 23 (starting from 9, moving to 6 then 8 respectively).

#Utils

type Map = [][]int

func variadicMax(nums ...int) int {
	if len(nums) == 0 {
		panic("must provide at least one num")
	}

	val := 0
	for _, num := range nums {
		if num > val {
			val = num
		}
	}
	return val
}