Prefix Sums – GenomicRangeQuery解題紀錄

題目內容

題目頁面: https://app.codility.com/programmers/lessons/5-prefix_sums/genomic_range_query/

題目理解

天啊~我最討厭有故事性的題目了如汽車或DNA，好長阿~~~讓我認真看他在說啥!

DNA和factor可表示為: {‘A’:1,’C’:2,’G’:3,’T’:4}，S = CAGCCTA，P=[2,5,0]，Q=[4,5,6]

然後要找第 K 個查詢（0 ≤ K < M），找到 P[K] 和 Q[K]（含）之間的 DNA 序列中包含的核苷酸的最小影響因子。也就是要計算出一個長度為M(M為P、Q的長度)的陣列，然後返回最小影響因子

例如M[0] = Math.min(S[2],S[3]) = Math.min(‘G’,’C’) = Math.min(2,3) =2

陷阱在這: each element of arrays P and Q is an integer within the range [0..N – 1]; => 所以不一定一樣長

解題思路

先用暴力法寫出一個可行解

// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');

function solution(S, P, Q) {
    // Implement your solution here
    factor = {'A':1,'C':2,'G':3,'T':4}
    var length_M = Math.min(P.length, Q.length)
    var M = []
    for( var i = 0 ; i < length_M; i++){
        var inclusive_S = S.slice(P[i], Q[i]+1).split("")
        var minimum = Infinity
        for(var j =0;j<inclusive_S.length;j++){
            //console.log(minimum,  factor[inclusive_S[j]])
            if(minimum > factor[inclusive_S[j]]){
                minimum = factor[inclusive_S[j]]
            }
        }
        M.push(minimum)
    }
    return M
}

因為邏輯有一點複雜，先把暴力解送出驗證一下正確性，之後再來優化

好的!!62%!!至少邏輯正確，那我們在想想要如何優化。讓我先觀察一下是死在那些效能測試上。分別是GGGGGG..??..GGGGGG..??..GGGGGG、large random string, length、all max ranges，首先先看到almost_all_same_letters，所以先把CAGC對應數字的部分做掉，如: S = CAGCCTA = [2,1,3,2,2,4,1]

下面為改善後的程式碼，可以減少重複將字串轉換為數字的部分。雖然沒有很樂觀，因為還是有先暫存中間值的可能性，先送看看有沒有改善，並評估一下效能瓶頸是否有改變。

function solution(S, P, Q) {
    // Implement your solution here
    factor = {'A':1,'C':2,'G':3,'T':4}
    var length_M = Math.min(P.length, Q.length)
    var S = S.split("").map(x => factor[x])
    var M = []
    for( var i = 0 ; i < length_M; i++){
        M.push(Math.min(...S.slice(P[i], Q[i]+1)))
    }
    return M
}

結果並沒有改善

— M = 0
minimum: Infinity
minimum: 3
minimum: 2
— M = 1
minimum: Infinity
— M = 2
minimum: Infinity
minimum: 2
minimum: 1
minimum: 1
minimum: 1
minimum: 1
minimum: 1

觀察一下現在的迴圈狀況，來找尋可改善的重複步驟，首先最明顯的，就是若遇到1應該可直接break，但我猜這影響不大。另外我發現，因為factor = {‘A’:1,’C’:2,’G’:3,’T’:4}是固定的，且最多就是4種，所以與其在那邊一個個轉換並且比較大小，還不如用4個if…else來從小尋找到最大的值。

function solution(S, P, Q) {
    // Implement your solution here
    //factor = {'A':1,'C':2,'G':3,'T':4}
    var length_M = Math.min(P.length, Q.length)
    var M = []
    for( var i = 0 ; i < length_M; i++){
        var inclusive_S = S.slice(P[i], Q[i]+1).split("")
        //console.log(inclusive_S)
        if(inclusive_S.indexOf('A') >= 0) M.push(1)
        else if(inclusive_S.indexOf('C') >= 0) M.push(2)
        else if(inclusive_S.indexOf('G') >= 0) M.push(3)
        else M.push(4)
    }
    return M
}

仍然沒有好轉

事已至此，感覺卡在一個切入觀點了，所以只好上網搜尋一下: http://myjavawar.blogspot.com/2018/10/genomicrangequery.html

原來要改變一下思考方向，要把原本的思維從”在一個已被切好的陣列中，搜尋某字元有沒有出現過”的思維，改成”計算從0到M以來某字元出現的次數”。接著若是陣列要從N開始，就直接把M-N，就會是從N~M之間某字元出現的次數”了

function solution(S, P, Q) {
    // Implement your solution here
    //factor = {'A':1,'C':2,'G':3,'T':4}
    var length_M = Math.min(P.length, Q.length)
    var M = []
    var showTimes = []
    var currentShowTime =  {'A':0,'C':0,'G':0,'T':0}
    for( var i = 0 ; i < S.length; i++){
        currentShowTime[S[i]]++
        showTimes.push({'A':currentShowTime['A'],'C':currentShowTime['C'],'G':currentShowTime['G'],'T':currentShowTime['T']})
    }
    for( var i = 0 ; i < length_M; i++){
        //var inclusive_S = S.slice(P[i], Q[i]+1).split("")
        //console.log(inclusive_S)
        if(showTimes[Q[i]]['A'] - showTimes[P[i]]['A'] > 0) M.push(1)
        else if(showTimes[Q[i]]['C'] - showTimes[P[i]]['C'] > 0) M.push(2)
        else if(showTimes[Q[i]]['G'] - showTimes[P[i]]['G'] > 0) M.push(3)
        else M.push(4)
    }
    return M
}

結果雖然效率過了，但是會有錯誤的結果，檢查了一下，這是因為當P[i]與Q[i]一樣大小時，就會出現錯誤，所以要加一個if用別的方法找答案

function solution(S, P, Q) {
    // Implement your solution here
    //factor = {'A':1,'C':2,'G':3,'T':4}
    var length_M = Math.min(P.length, Q.length)
    var M = []
    var showTimes = []
    var currentShowTime =  {'A':0,'C':0,'G':0,'T':0}
    for( var i = 0 ; i < S.length; i++){
        currentShowTime[S[i]]++
        showTimes.push({'A':currentShowTime['A'],'C':currentShowTime['C'],'G':currentShowTime['G'],'T':currentShowTime['T']})
    }
    //console.log(showTimes)
    for( var i = 0 ; i < length_M; i++){
        //var inclusive_S = S.slice(P[i], Q[i]+1).split("")
        //console.log(inclusive_S)
        if(P[i] == Q[i]){
            var factor = {'A':1,'C':2,'G':3,'T':4}
            M.push(factor[S[P[i]]])
        }else{
            if(showTimes[Q[i]]['A'] - showTimes[P[i]]['A'] > 0) M.push(1)
            else if(showTimes[Q[i]]['C'] - showTimes[P[i]]['C'] > 0) M.push(2)
            else if(showTimes[Q[i]]['G'] - showTimes[P[i]]['G'] > 0) M.push(3)
            else if(showTimes[Q[i]]['T'] - showTimes[P[i]]['T'] > 0) M.push(4)
        }
        
    }
    return M
}

結果還是有錯

仔細檢查了一下，錯誤的情況是solution(‘AC’, [0, 0, 1], [0, 1, 1])，發現P[i] 應該要減1，但這樣子P[i]==0時會出現問題，所以要給個預設值

function solution(S, P, Q) {
    // Implement your solution here
    //factor = {'A':1,'C':2,'G':3,'T':4}
    var length_M = Math.min(P.length, Q.length)
    var M = []
    var showTimes = []
    var currentShowTime =  {'A':0,'C':0,'G':0,'T':0}
    for( var i = 0 ; i < S.length; i++){
        currentShowTime[S[i]]++
        showTimes.push({'A':currentShowTime['A'],'C':currentShowTime['C'],'G':currentShowTime['G'],'T':currentShowTime['T']})
    }
    //console.log(showTimes)
    for( var i = 0 ; i < length_M; i++){
        var inclusive_S = S.slice(P[i], Q[i]+1).split("")
        //console.log(inclusive_S)
        if(P[i] == Q[i]){
            var factor = {'A':1,'C':2,'G':3,'T':4}
            M.push(factor[S[P[i]]])
        }else{
            var showTimesMax = showTimes[Q[i]]
            var showTimesMin = P[i] == 0 ? {'A':0,'C':0,'G':0,'T':0}:showTimes[P[i]-1]
            //console.log(showTimesMax, showTimesMin)
            if(showTimesMax['A'] - showTimesMin['A'] > 0) M.push(1)
            else if(showTimesMax['C'] - showTimesMin['C'] > 0) M.push(2)
            else if(showTimesMax['G'] - showTimesMin['G'] > 0) M.push(3)
            else M.push(4)
        }
    }
    return M
}

好了~終於100%了，真是個身心俱疲的一題阿! (https://app.codility.com/demo/results/trainingYXRWFN-H4J/)

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