Fish
N voracious fish are moving along a river. Calculate how many fish are alive.
You are given two non-empty arrays A and B consisting of N integers. Arrays A and B represent N voracious fish in a river, ordered downstream along the flow of the river.
The fish are numbered from 0 to N − 1. If P and Q are two fish and P < Q, then fish P is initially upstream of fish Q. Initially, each fish has a unique position.
Fish number P is represented by A[P] and B[P]. Array A contains the sizes of the fish. All its elements are unique. Array B contains the directions of the fish. It contains only 0s and/or 1s, where:
0 represents a fish flowing upstream, 1 represents a fish flowing downstream. If two fish move in opposite directions and there are no other (living) fish between them, they will eventually meet each other. Then only one fish can stay alive − the larger fish eats the smaller one. More precisely, we say that two fish P and Q meet each other when P < Q, B[P] = 1 and B[Q] = 0, and there are no living fish between them. After they meet:
If A[P] > A[Q] then P eats Q, and P will still be flowing downstream, If A[Q] > A[P] then Q eats P, and Q will still be flowing upstream. We assume that all the fish are flowing at the same speed. That is, fish moving in the same direction never meet. The goal is to calculate the number of fish that will stay alive.
For example, consider arrays A and B such that:
A[0] = 4 B[0] = 0 A[1] = 3 B[1] = 1 A[2] = 2 B[2] = 0 A[3] = 1 B[3] = 0 A[4] = 5 B[4] = 0 Initially all the fish are alive and all except fish number 1 are moving upstream. Fish number 1 meets fish number 2 and eats it, then it meets fish number 3 and eats it too. Finally, it meets fish number 4 and is eaten by it. The remaining two fish, number 0 and 4, never meet and therefore stay alive.
Write a function:
func Solution(A []int, B []int) int
that, given two non-empty arrays A and B consisting of N integers, returns the number of fish that will stay alive.
For example, given the arrays shown above, 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 [0..1,000,000,000]; each element of array B is an integer that can have one of the following values: 0, 1; the elements of A are all distinct.
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題目大意
最開始每個魚都有特定的起始位置 A: 活魚的大小, B: 魚游的方向. 如果於相遇的話大魚會吃掉小魚. 返回剩下魚的數量
解題思路
從 B 開始找, 當值為1 存入stack. 代表向下游的魚. 來進行把活魚吃掉. 如果把列表的活魚都吃掉. 則結果+1 如果值為0且stack為空, 代表沒遇到下游的魚所以活魚++
來源
https://app.codility.com/programmers/lessons/7-stacks_and_queues/fish/
解答
package Fish
import "LeetcodeGolang/Utility/structures"
func Solution(A []int, B []int) int {
stack := structures.NewArrayStack()
aliveFish := 0
for idx, val := range B {
if val == 1 {
stack.Push(A[idx])
} else {
// 繼續往下游
for !stack.IsEmpty() {
if stack.Top().(int) < A[idx] {
// stack的魚比遇到的魚還小, stack被吃掉
stack.Pop()
} else {
break
}
}
if stack.IsEmpty() {
aliveFish++
}
}
}
return aliveFish + stack.Size()
}