Find the index S such that the leaders of the sequences A[0], A[1], ..., A[S] and A[S + 1], A[S + 2], ..., A[N - 1] are the same.
A non-empty array A consisting of N integers is given.
The leader of this array is the value that occurs in more than half of the elements of A.
An equi leader is an index S such that 0 ≤ S < N − 1 and two sequences A[0], A[1], ..., A[S] and A[S + 1], A[S + 2], ..., A[N − 1] have leaders of the same value.
For example, given array A such that:
A[0] = 4
A[1] = 3
A[2] = 4
A[3] = 4
A[4] = 4
A[5] = 2
we can find two equi leaders:
- 0, because sequences: (4) and (3, 4, 4, 4, 2) have the same leader, whose value is 4.
- 2, because sequences: (4, 3, 4) and (4, 4, 2) have the same leader, whose value is 4. The goal is to count the number of equi leaders.
Write a function:
func Solution(A []int) int
that, given a non-empty array A consisting of N integers, returns the number of equi leaders.
For example, given:
A[0] = 4
A[1] = 3
A[2] = 4
A[3] = 4
A[4] = 4
A[5] = 2
the function should return 2, as explained above.
Write an efficient algorithm for the following assumptions:
N is an integer within the range [1..100,000]; each element of array A is an integer within the range [−1,000,000,000..1,000,000,000]. Copyright 2009–2021 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
選定一個下標, 將一個數組分爲左右兩個子數組, 使得兩個子數組都有相同的leader, 則稱此下標爲EquiLeader, 要求返回給定數組中EquiLeader的個數n. 事實上, 若一個數同時是左子數組的leader, 它必然也是整個數組的leader.
需要先找出序列中的leader, 記錄其出現的次數. 然後再遍歷整個數組，枚舉分割點，記錄下左子數組leader出現的次數s, 看s與n-s是否能使得leader在左右子數組中仍爲leader.
- https://app.codility.com/programmers/lessons/8-leader/equi_leader/
- https://github.com/Anfany/Codility-Lessons-By-Python3/blob/master/L8_Leader/8.1%20EquiLeader.md
package EquiLeader
func Solution(A []int) int {
leaderDict := make(map[int]int)
for i := 0; i < len(A); i++ {
if _, ok := leaderDict[A[i]]; ok {
leaderDict[A[i]]++
} else {
leaderDict[A[i]] = 1
}
}
leader := 0
times := 0
for k, v := range leaderDict {
if v > times {
times = v
leader = k
}
}
equiCount := 0
count := 0 // 超頻數已出現的次數
for index, v := range A {
if v == leader {
count++
}
if count > (index+1)/2 && (times-count) > (len(A)-(index+1))/2 {
equiCount++
}
}
return equiCount
}