This is what I have for the min part of the question. Can someone please help me understand the concept?
func MinMaxArray(arr1 [] int, arr2[]int) int {
minA := arr1[0]
minb := arr2[0]
n := len(arr1[] && arr2[])
for i:=0; i<=n ; i++{
if arr1[i] > arr2[i] {
max := arr1[i]
min := arr2[i]
} else {
max := arr2[i]
min := arr1[i]
}
}
return min
}
You should be able to keep a running tally throughout.
func MinMaxArray(arr1 []int, arr2 []int) int {
runningTally := 0
n := len(arr1)
if len(arr2) < n {
n = len(arr2)
}
for i := 0; i < n; i++ {
if arr1[i] > arr2[i] {
runningTally += arr1[i]
} else {
runningTally += arr2[i]
}
}
return runningTally
}