The problem
Full the strategy which returns the quantity which is most frequent within the given enter array. If there’s a tie for probably the most frequent quantity, return the most important quantity amongst them.
Word: no empty arrays will likely be given.
Examples
[12, 10, 8, 12, 7, 6, 4, 10, 12] --> 12
[12, 10, 8, 12, 7, 6, 4, 10, 12, 10] --> 12
[12, 10, 8, 8, 3, 3, 3, 3, 2, 4, 10, 12, 10] --> 3
The answer in Golang
Possibility 1:
bundle resolution
func HighestRank(nums []int) int {
mii, maxK, maxV := map[int]int{}, 0, 0
for _, v := vary nums {
mii[v]++
if mii[v] > maxV || (mii[v] == maxV && v > maxK) {
maxK = v
maxV = mii[v]
}
}
return maxK
}
Possibility 2:
bundle resolution
func HighestRank(nums []int) int {
fs := make(map[int]int)
for _, n := vary nums {
fs[n]++
}
maxf, maxfn := 0, 0
for n, f := vary fs {
if f >= maxf && (f > maxf || n > maxfn) {
maxf, maxfn = f, n
}
}
return maxfn
}
Possibility 3:
bundle resolution
func HighestRank(nums []int) int {
occurrences := make(map[int]int)
var max, val int
for _, worth := vary nums {
occurrences[value]++
}
max, val = occurrences[nums[0]], nums[0]
for worth, occasions := vary occurrences {
if occasions > max {
val = worth
max = occasions
} else if occasions == max {
if worth > val {
val = worth
}
}
}
return val
}
Take a look at circumstances to validate our resolution
bundle solution_test
import (
. "github.com/onsi/ginkgo"
. "github.com/onsi/gomega"
)
var _ = Describe("Exams", func() {
Describe("Pattern assessments", func() {
It("Pattern check 1: 12, 10, 8, 12, 7, 6, 4, 10, 12", func() {
Anticipate(HighestRank([]int{12, 10, 8, 12, 7, 6, 4, 10, 12 })).To(Equal(12))
})
})
})
