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How long does it take to enter a number for a given keypad
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package main | |
import "fmt" | |
type Keypad struct { | |
matrix [3][3]string | |
} | |
func NewKeypad(keypad string) Keypad { | |
var result Keypad | |
x, y := 0, 0 | |
for _, c := range keypad { | |
result.matrix[y][x] = string(c) | |
x++ | |
if (x > 2) { | |
y++ | |
x = 0 | |
} | |
} | |
return result | |
} | |
func (k *Keypad) Print() { | |
fmt.Print("-------\n") | |
for _, row := range k.matrix { | |
for _, letter := range row { | |
fmt.Print("|" + string(letter)) | |
} | |
fmt.Print("|\n-------\n") | |
} | |
} | |
func (k *Keypad) Coord(key string) (int, int) { | |
for x, row := range k.matrix { | |
for y, letter := range row { | |
if string(letter) == key { | |
return x, y | |
} | |
} | |
} | |
return -1, -1 | |
} | |
func (k *Keypad) Distance(s1, s2 string) int { | |
if s1 == s2 { | |
return 0 | |
} | |
x1, y1 := k.Coord(s1) | |
x2, y2 := k.Coord(s2) | |
// Manhattan distance | |
dist := abs(x2 - x1) + abs(y2 - y1) | |
// Diagonal distance is 2, should be 1 for our case | |
if abs(x2 - x1) == 1 && abs(y2 - y1) == 1 { | |
dist = 1 | |
} | |
return dist | |
} | |
// Absolute value for integers | |
func abs(x int) int { | |
if x < 0 { | |
return -x | |
} | |
return x | |
} | |
func entryTime(s string, keypad string) int32 { | |
var result int32 = 0 | |
k := NewKeypad(keypad) | |
for i := 0; i < len(s)-1; i++ { | |
key := string(s[i]) | |
nextKey := string(s[i+1]) | |
dist := k.Distance(key, nextKey) | |
if dist == 1 { | |
result += 1 | |
} | |
if dist > 1 { | |
result += 2 | |
} | |
} | |
return result | |
} | |
func main() { | |
fmt.Println("= test", entryTime("423692", "923857614"), "expected", 8) | |
fmt.Println("= test", entryTime("5111", "752961348"), "expected", 1) | |
fmt.Println("= test", entryTime("91566165", "639485712"), "expected", 11) | |
} |
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