java编码能力训练基因组范围查询
任务是：java编码能力训练基因组范围查询,java,algorithm,Java,Algorithm,任务是： class Solution { public int[] solution(String S, int[] P, int[] Q) { final char c[] = S.toCharArray(); final int answer[] = new int[P.length]; int tempAnswer; char tempC; for (int iii = 0; iii < P
class Solution {
public int[] solution(String S, int[] P, int[] Q) {
final char c[] = S.toCharArray();
final int answer[] = new int[P.length];
int tempAnswer;
char tempC;
for (int iii = 0; iii < P.length; iii++) {
tempAnswer = 4;
for (int zzz = P[iii]; zzz <= Q[iii]; zzz++) {
tempC = c[zzz];
if (tempC == 'A') {
tempAnswer = 1;
break;
} else if (tempC == 'C') {
if (tempAnswer > 2) {
tempAnswer = 2;
}
} else if (tempC == 'G') {
if (tempAnswer > 3) {
tempAnswer = 3;
}
}
}
answer[iii] = tempAnswer;
}
return answer;
}
}
例如，考虑字符串s= gACCATATA和数组p，q，以便：
P[0] = 0 Q[0] = 8
P[1] = 0 Q[1] = 2
P[2] = 4 Q[2] = 5
P[3] = 7 Q[3] = 7
P[0] = 0 Q[0] = 8
P[1] = 0 Q[1] = 2
P[2] = 4 Q[2] = 5
P[3] = 7 Q[3] = 7
(0, 8) is A identified by 1,
(0, 2) is A identified by 1,
(4, 5) is C identified by 2,
(7, 7) is T identified by 4.
class Solution { public int[] solution(String S, int[] P, int[] Q); }
a Results structure (in C), or
a vector of integers (in C++), or
a Results record (in Pascal), or
an array of integers (in any other programming language).
P[0] = 0 Q[0] = 8
P[1] = 0 Q[1] = 2
P[2] = 4 Q[2] = 5
P[3] = 7 Q[3] = 7
P[0] = 0 Q[0] = 8
P[1] = 0 Q[1] = 2
P[2] = 4 Q[2] = 5
P[3] = 7 Q[3] = 7
N is an integer within the range [1..100,000];
M is an integer within the range [1..50,000];
each element of array P, Q is an integer within the range [0..N − 1];
P[i] ≤ Q[i];
string S consists only of upper-case English letters A, C, G, T.
expected worst-case time complexity is O(N+M);
expected worst-case space complexity is O(N),
beyond input storage
(not counting the storage required for input arguments).
class Solution {
public int[] solution(String S, int[] P, int[] Q) {
final char c[] = S.toCharArray();
final int answer[] = new int[P.length];
int tempAnswer;
char tempC;
for (int iii = 0; iii < P.length; iii++) {
tempAnswer = 4;
for (int zzz = P[iii]; zzz <= Q[iii]; zzz++) {
tempC = c[zzz];
if (tempC == 'A') {
tempAnswer = 1;
break;
} else if (tempC == 'C') {
if (tempAnswer > 2) {
tempAnswer = 2;
}
} else if (tempC == 'G') {
if (tempAnswer > 3) {
tempAnswer = 3;
}
}
}
answer[iii] = tempAnswer;
}
return answer;
}
}
类解决方案{
公共int[]解决方案（字符串S，int[]P，int[]Q）{
最终字符c[]=S.toCharArray（）；
最终整数答案[]=新整数[P.length]；
int-tempAnswer；
char-tempC；
对于（int iii=0；iii3）{
tempAnswer=3；
}
}
}
答复[三]=临时答复；
}
返回答案；
}
}
您可以在此处检查解决方案的质量，测试名称为基因组范围查询。假设有人仍感兴趣，请在此查看解决方案。
class Solution {
public int[] solution(String S, int[] P, int[] Q) {
int[] answer = new int[P.length];
char[] chars = S.toCharArray();
int[][] cumulativeAnswers = new int[4][chars.length + 1];
for (int iii = 0; iii < chars.length; iii++) {
if (iii > 0) {
for (int zzz = 0; zzz < 4; zzz++) {
cumulativeAnswers[zzz][iii + 1] = cumulativeAnswers[zzz][iii];
}
}
switch (chars[iii]) {
case 'A':
cumulativeAnswers[0][iii + 1]++;
break;
case 'C':
cumulativeAnswers[1][iii + 1]++;
break;
case 'G':
cumulativeAnswers[2][iii + 1]++;
break;
case 'T':
cumulativeAnswers[3][iii + 1]++;
break;
}
}
for (int iii = 0; iii < P.length; iii++) {
for (int zzz = 0; zzz < 4; zzz++) {
if ((cumulativeAnswers[zzz][Q[iii] + 1] - cumulativeAnswers[zzz][P[iii]]) > 0) {
answer[iii] = zzz + 1;
break;
}
}
}
return answer;
}
}
类解决方案{
公共int[]解决方案（字符串S，int[]P，int[]Q）{
int[]答案=新的int[P.length]；
char[]chars=S.toCharArray（）；
int[]cumulativeAnswers=新int[4][chars.length+1]；
对于（int iii=0；iii0）{
对于（int-zzz=0；zzz<4；zzz++）{
累积利率[zzz][iii+1]=累积利率[zzz][iii]；
}
}
开关（chars[iii]）{
案例“A”：
累计回答[0][iii+1]++；
打破
案例“C”：
累积回答[1][iii+1]++；
打破
案例“G”：
累积回答[2][iii+1]++；
打破
案例“T”：
累积回答[3][iii+1]++；
打破
}
}
对于（int iii=0；iii0）{
答案[iii]=zzz+1；
打破
}
}
}
返回答案；
}
}
公共静态int[]解决方案（字符串S，int[]P，int[]Q）
{
常量int MinValue=1；
Dictionary stringValueTable=新字典（）{{'A'，1}，{'C'，2}，{'G'，3}，{'T'，4}；
char[]inputArray=S.ToCharArray（）；
int[，]minRangeTable=new int[S.Length，S.Length]；//此表的含义是[x，y]，其中x是开始索引，y是结束索引，值是最小范围-如果为0，则它是最小范围（无论是什么）
对于（int-startIndex=0；startIndexpublic static int[] solveGenomicRange(String S, int[] P, int[] Q) {
//used jagged array to hold the prefix sums of each A, C and G genoms
//we don't need to get prefix sums of T, you will see why.
int[][] genoms = new int[3][S.length()+1];
//if the char is found in the index i, then we set it to be 1 else they are 0
//3 short values are needed for this reason
short a, c, g;
for (int i=0; i<S.length(); i++) {
a = 0; c = 0; g = 0;
if ('A' == (S.charAt(i))) {
a=1;
}
if ('C' == (S.charAt(i))) {
c=1;
}
if ('G' == (S.charAt(i))) {
g=1;
}
//here we calculate prefix sums. To learn what's prefix sums look at here https://codility.com/media/train/3-PrefixSums.pdf
genoms[0][i+1] = genoms[0][i] + a;
genoms[1][i+1] = genoms[1][i] + c;
genoms[2][i+1] = genoms[2][i] + g;
}
int[] result = new int[P.length];
//here we go through the provided P[] and Q[] arrays as intervals
for (int i=0; i<P.length; i++) {
int fromIndex = P[i];
//we need to add 1 to Q[i],
//because our genoms[0][0], genoms[1][0] and genoms[2][0]
//have 0 values by default, look above genoms[0][i+1] = genoms[0][i] + a;
int toIndex = Q[i]+1;
if (genoms[0][toIndex] - genoms[0][fromIndex] > 0) {
result[i] = 1;
} else if (genoms[1][toIndex] - genoms[1][fromIndex] > 0) {
result[i] = 2;
} else if (genoms[2][toIndex] - genoms[2][fromIndex] > 0) {
result[i] = 3;
} else {
result[i] = 4;
}
}
return result;
}
function solution($S, $P, $Q) {
$S = str_split($S);
$len = count($S);
$lep = count($P);
$arr = array();
$result = array();
$clone = array_fill(0, 4, 0);
for($i = 0; $i < $len; $i++){
$arr[$i] = $clone;
switch($S[$i]){
case 'A':
$arr[$i][0] = 1;
break;
case 'C':
$arr[$i][1] = 1;
break;
case 'G':
$arr[$i][2] = 1;
break;
default:
$arr[$i][3] = 1;
break;
}
}
for($i = 1; $i < $len; $i++){
for($j = 0; $j < 4; $j++){
$arr[$i][$j] += $arr[$i - 1][$j];
}
}
for($i = 0; $i < $lep; $i++){
$x = $P[$i];
$y = $Q[$i];
for($a = 0; $a < 4; $a++){
$sub = 0;
if($x - 1 >= 0){
$sub = $arr[$x - 1][$a];
}
if($arr[$y][$a] - $sub > 0){
$result[$i] = $a + 1;
break;
}
}
}
return $result;
}
public class GenomicRange {
final int Index_A=0, Index_C=1, Index_G=2, Index_T=3;
final int A=1, C=2, G=3, T=4;
public static void main(String[] args) {
GenomicRange gen = new GenomicRange();
int[] M = gen.solution( "GACACCATA", new int[] { 0,0,4,7 } , new int[] { 8,2,5,7 } );
System.out.println(Arrays.toString(M));
}
public int[] solution(String S, int[] P, int[] Q) {
int[] M = new int[P.length];
char[] charArr = S.toCharArray();
int[][] occCount = new int[3][S.length()+1];
int charInd = getChar(charArr[0]);
if(charInd!=3) {
occCount[charInd][1]++;
}
for(int sInd=1; sInd<S.length(); sInd++) {
charInd = getChar(charArr[sInd]);
if(charInd!=3)
occCount[charInd][sInd+1]++;
occCount[Index_A][sInd+1]+=occCount[Index_A][sInd];
occCount[Index_C][sInd+1]+=occCount[Index_C][sInd];
occCount[Index_G][sInd+1]+=occCount[Index_G][sInd];
}
for(int i=0;i<P.length;i++) {
int a,c,g;
if(Q[i]+1>=occCount[0].length) continue;
a = occCount[Index_A][Q[i]+1] - occCount[Index_A][P[i]];
c = occCount[Index_C][Q[i]+1] - occCount[Index_C][P[i]];
g = occCount[Index_G][Q[i]+1] - occCount[Index_G][P[i]];
M[i] = a>0? A : c>0 ? C : g>0 ? G : T;
}
return M;
}
private int getChar(char c) {
return ((c=='A') ? Index_A : ((c=='C') ? Index_C : ((c=='G') ? Index_G : Index_T)));
}
}
using System;
class Solution
{
public int[] solution(string S, int[] P, int[] Q)
{
int N = S.Length;
int M = P.Length;
char[] chars = {'A','C','G','T'};
//Calculate accumulates
int[,] accum = new int[3, N+1];
for (int i = 0; i <= 2; i++)
{
for (int j = 0; j < N; j++)
{
if(S[j] == chars[i]) accum[i, j+1] = accum[i, j] + 1;
else accum[i, j+1] = accum[i, j];
}
}
//Get minimal nucleotides for the given ranges
int diff;
int[] minimums = new int[M];
for (int i = 0; i < M; i++)
{
minimums[i] = 4;
for (int j = 0; j <= 2; j++)
{
diff = accum[j, Q[i]+1] - accum[j, P[i]];
if (diff > 0)
{
minimums[i] = j+1;
break;
}
}
}
return minimums;
}
}
function solution(S, P, Q) {
var A = [];
var C = [];
var G = [];
var T = [];
var result = [];
var i = 0;
S.split('').forEach(function(a) {
if (a === 'A') {
A.push(i);
} else if (a === 'C') {
C.push(i);
} else if (a === 'G') {
G.push(i);
} else {
T.push(i);
}
i++;
});
function hasNucl(typeArray, start, end) {
return typeArray.some(function(a) {
return a >= P[j] && a <= Q[j];
});
}
for(var j=0; j<P.length; j++) {
if (hasNucl(A, P[j], P[j])) {
result.push(1)
} else if (hasNucl(C, P[j], P[j])) {
result.push(2);
} else if (hasNucl(G, P[j], P[j])) {
result.push(3);
} else {
result.push(4);
}
}
return result;
}
sub solution {
my ($S, $P, $Q)=@_; my @P=@$P; my @Q=@$Q;
my @_A = (0), @_C = (0), @_G = (0), @ret =();
foreach (split //, $S)
{
push @_A, $_A[-1] + ($_ eq 'A' ? 1 : 0);
push @_C, $_C[-1] + ($_ eq 'C' ? 1 : 0);
push @_G, $_G[-1] + ($_ eq 'G' ? 1 : 0);
}
foreach my $i (0..$#P)
{
my $from_index = $P[$i];
my $to_index = $Q[$i] + 1;
if ( $_A[$to_index] - $_A[$from_index] > 0 )
{
push @ret, 1;
next;
}
if ( $_C[$to_index] - $_C[$from_index] > 0 )
{
push @ret, 2;
next;
}
if ( $_G[$to_index] - $_G[$from_index] > 0 )
{
push @ret, 3;
next;
}
push @ret, 4
}
return @ret;
}
class Solution {
int[][] lastOccurrencesMap;
public int[] solution(String S, int[] P, int[] Q) {
int N = S.length();
int M = P.length;
int[] result = new int[M];
lastOccurrencesMap = new int[3][N];
int lastA = -1;
int lastC = -1;
int lastG = -1;
for (int i = 0; i < N; i++) {
char c = S.charAt(i);
if (c == 'A') {
lastA = i;
} else if (c == 'C') {
lastC = i;
} else if (c == 'G') {
lastG = i;
}
lastOccurrencesMap[0][i] = lastA;
lastOccurrencesMap[1][i] = lastC;
lastOccurrencesMap[2][i] = lastG;
}
for (int i = 0; i < M; i++) {
int startIndex = P[i];
int endIndex = Q[i];
int minimum = 4;
for (int n = 0; n < 3; n++) {
int lastOccurence = getLastNucleotideOccurrence(startIndex, endIndex, n);
if (lastOccurence != 0) {
minimum = n + 1;
break;
}
}
result[i] = minimum;
}
return result;
}
int getLastNucleotideOccurrence(int startIndex, int endIndex, int nucleotideIndex) {
int[] lastOccurrences = lastOccurrencesMap[nucleotideIndex];
int endValueLastOccurenceIndex = lastOccurrences[endIndex];
if (endValueLastOccurenceIndex >= startIndex) {
return nucleotideIndex + 1;
} else {
return 0;
}
}
}
class Solution {
public int[] solution(String S, int[] P, int[] Q) {
int qSize = Q.length;
int[] answers = new int[qSize];
char[] sequence = S.toCharArray();
int[][] occCount = new int[3][sequence.length+1];
int[] geneImpactMap = new int['G'+1];
geneImpactMap['A'] = 0;
geneImpactMap['C'] = 1;
geneImpactMap['G'] = 2;
if(sequence[0] != 'T') {
occCount[geneImpactMap[sequence[0]]][0]++;
}
for(int i = 0; i < sequence.length; i++) {
occCount[0][i+1] = occCount[0][i];
occCount[1][i+1] = occCount[1][i];
occCount[2][i+1] = occCount[2][i];
if(sequence[i] != 'T') {
occCount[geneImpactMap[sequence[i]]][i+1]++;
}
}
for(int j = 0; j < qSize; j++) {
for(int k = 0; k < 3; k++) {
if(occCount[k][Q[j]+1] - occCount[k][P[j]] > 0) {
answers[j] = k+1;
break;
}
answers[j] = 4;
}
}
return answers;
}
}
public class DNAseq {
public static void main(String[] args) {
String S="CAGCCTA";
int[] P={2, 5, 0};
int[] Q={4, 5, 6};
int [] results=solution(S,P,Q);
System.out.println(results[0]);
}
static class segmentNode{
int l;
int r;
int min;
segmentNode left;
segmentNode right;
}
public static segmentNode buildTree(int[] arr,int l,int r){
if(l==r){
segmentNode n=new segmentNode();
n.l=l;
n.r=r;
n.min=arr[l];
return n;
}
int mid=l+(r-l)/2;
segmentNode le=buildTree(arr,l,mid);
segmentNode re=buildTree(arr,mid+1,r);
segmentNode root=new segmentNode();
root.left=le;
root.right=re;
root.l=le.l;
root.r=re.r;
root.min=Math.min(le.min,re.min);
return root;
}
public static int getMin(segmentNode root,int l,int r){
if(root.l>r || root.r<l){
return Integer.MAX_VALUE;
}
if(root.l>=l&& root.r<=r) {
return root.min;
}
return Math.min(getMin(root.left,l,r),getMin(root.right,l,r));
}
public static int[] solution(String S, int[] P, int[] Q) {
int[] arr=new int[S.length()];
for(int i=0;i<S.length();i++){
switch (S.charAt(i)) {
case 'A':
arr[i]=1;
break;
case 'C':
arr[i]=2;
break;
case 'G':
arr[i]=3;
break;
case 'T':
arr[i]=4;
break;
default:
break;
}
}
segmentNode root=buildTree(arr,0,S.length()-1);
int[] result=new int[P.length];
for(int i=0;i<P.length;i++){
result[i]=getMin(root,P[i],Q[i]);
}
return result;
} }
function solution(S, P, Q) {
var N = S.length, M = P.length;
// dictionary to map nucleotide to impact factor
var impact = {A : 1, C : 2, G : 3, T : 4};
// nucleotide total count in DNA
var currCounter = {A : 0, C : 0, G : 0, T : 0};
// how many times nucleotide repeats at the moment we reach S[i]
var counters = [];
// result
var minImpact = [];
var i;
// count nucleotides
for(i = 0; i <= N; i++) {
counters.push({A: currCounter.A, C: currCounter.C, G: currCounter.G});
currCounter[S[i]]++;
}
// for every query
for(i = 0; i < M; i++) {
var from = P[i], to = Q[i] + 1;
// compare count of A at the start of query with count at the end of equry
// if counter was changed then query contains A
if(counters[to].A - counters[from].A > 0) {
minImpact.push(impact.A);
}
// same things for C and others nucleotides with higher impact factor
else if(counters[to].C - counters[from].C > 0) {
minImpact.push(impact.C);
}
else if(counters[to].G - counters[from].G > 0) {
minImpact.push(impact.G);
}
else { // one of the counters MUST be changed, so its T
minImpact.push(impact.T);
}
}
return minImpact;
}
function solution($S, $P, $Q) {
$result = array();
for ($i = 0; $i < count($P); $i++) {
$from = $P[$i];
$to = $Q[$i];
$length = $from >= $to ? $from - $to + 1 : $to - $from + 1;
$new = substr($S, $from, $length);
if (strpos($new, 'A') !== false) {
$result[$i] = 1;
} else {
if (strpos($new, 'C') !== false) {
$result[$i] = 2;
} else {
if (strpos($new, 'G') !== false) {
$result[$i] = 3;
} else {
$result[$i] = 4;
}
}
}
}
return $result;
}
class Solution {
private ImpactFactorHolder[] mHolder;
private static final int A=0,C=1,G=2,T=3;
public int[] solution(String S, int[] P, int[] Q) {
mHolder = createImpactHolderArray(S);
int queriesLength = P.length;
int[] result = new int[queriesLength];
for (int i = 0; i < queriesLength; ++i ) {
int value = 0;
if( P[i] == Q[i]) {
value = lookupValueForIndex(S.charAt(P[i])) + 1;
} else {
value = calculateMinImpactFactor(P[i], Q[i]);
}
result[i] = value;
}
return result;
}
public int calculateMinImpactFactor(int P, int Q) {
int minImpactFactor = 3;
for (int nucleotide = A; nucleotide <= T; ++nucleotide ) {
int qValue = mHolder[nucleotide].mOcurrencesSum[Q];
int pValue = mHolder[nucleotide].mOcurrencesSum[P];
// handling special cases when the less value is assigned on the P index
if( P-1 >= 0 ) {
pValue = mHolder[nucleotide].mOcurrencesSum[P-1] == 0 ? 0 : pValue;
} else if ( P == 0 ) {
pValue = mHolder[nucleotide].mOcurrencesSum[P] == 1 ? 0 : pValue;
}
if ( qValue - pValue > 0) {
minImpactFactor = nucleotide;
break;
}
}
return minImpactFactor + 1;
}
public int lookupValueForIndex(char nucleotide) {
int value = 0;
switch (nucleotide) {
case 'A' :
value = A;
break;
case 'C' :
value = C;
break;
case 'G':
value = G;
break;
case 'T':
value = T;
break;
default:
break;
}
return value;
}
public ImpactFactorHolder[] createImpactHolderArray(String S) {
int length = S.length();
ImpactFactorHolder[] holder = new ImpactFactorHolder[4];
holder[A] = new ImpactFactorHolder(1,'A', length);
holder[C] = new ImpactFactorHolder(2,'C', length);
holder[G] = new ImpactFactorHolder(3,'G', length);
holder[T] = new ImpactFactorHolder(4,'T', length);
int i =0;
for(char c : S.toCharArray()) {
int nucleotide = lookupValueForIndex(c);
++holder[nucleotide].mAcum;
holder[nucleotide].mOcurrencesSum[i] = holder[nucleotide].mAcum;
holder[A].mOcurrencesSum[i] = holder[A].mAcum;
holder[C].mOcurrencesSum[i] = holder[C].mAcum;
holder[G].mOcurrencesSum[i] = holder[G].mAcum;
holder[T].mOcurrencesSum[i] = holder[T].mAcum;
++i;
}
return holder;
}
private static class ImpactFactorHolder {
public ImpactFactorHolder(int impactFactor, char nucleotide, int length) {
mImpactFactor = impactFactor;
mNucleotide = nucleotide;
mOcurrencesSum = new int[length];
mAcum = 0;
}
int mImpactFactor;
char mNucleotide;
int[] mOcurrencesSum;
int mAcum;
}
}
def solution(S: String, P: Array[Int], Q: Array[Int]): Array[Int] = {
val resp = for(ind <- 0 to P.length-1) yield {
val sub= S.substring(P(ind),Q(ind)+1)
var factor = 4
if(sub.contains("A")) {factor=1}
else{
if(sub.contains("C")) {factor=2}
else{
if(sub.contains("G")) {factor=3}
}
}
factor
}
return resp.toArray
}
vector<int> solution(string &S, vector<int> &P, vector<int> &Q) {
vector<int> impactCount_A(S.size()+1, 0);
vector<int> impactCount_C(S.size()+1, 0);
vector<int> impactCount_G(S.size()+1, 0);
int lastTotal_A = 0;
int lastTotal_C = 0;
int lastTotal_G = 0;
for (int i = (signed)S.size()-1; i >= 0; --i) {
switch(S[i]) {
case 'A':
++lastTotal_A;
break;
case 'C':
++lastTotal_C;
break;
case 'G':
++lastTotal_G;
break;
};
impactCount_A[i] = lastTotal_A;
impactCount_C[i] = lastTotal_C;
impactCount_G[i] = lastTotal_G;
}
vector<int> results(P.size(), 0);
for (int i = 0; i < P.size(); ++i) {
int pIndex = P[i];
int qIndex = Q[i];
int numA = impactCount_A[pIndex]-impactCount_A[qIndex+1];
int numC = impactCount_C[pIndex]-impactCount_C[qIndex+1];
int numG = impactCount_G[pIndex]-impactCount_G[qIndex+1];
if (numA > 0) {
results[i] = 1;
}
else if (numC > 0) {
results[i] = 2;
}
else if (numG > 0) {
results[i] = 3;
}
else {
results[i] = 4;
}
}
return results;
}
int n=S.size();
int m=P.size();
vector<vector<int> > prefix_sum(n+1,vector<int>(4,0));
int nuc;
//prefix occurrence sum
for (int s=0;s<n; s++) {
nuc = S.at(s) == 'A' ? 1 : (S.at(s) == 'C' ? 2 : (S.at(s) == 'G' ? 3 : 4) );
for (int u=0;u<4;u++) {
prefix_sum[s+1][u] = prefix_sum[s][u] + ((u+1)==nuc?1:0);
}
}
//find minimal impact factor in each interval K
int lower_impact_factor;
for (int k=0;k<m;k++) {
lower_impact_factor=4;
for (int u=2;u>=0;u--) {
if (prefix_sum[Q[k]+1][u] - prefix_sum[P[k]][u] != 0)
lower_impact_factor = u+1;
}
P[k]=lower_impact_factor;
}
return P;
public int[] solution(String S, int[] P, int[] K) {
// write your code in Java SE 8
char[] sc = S.toCharArray();
int[] A = new int[sc.length];
int[] G = new int[sc.length];
int[] C = new int[sc.length];
int prevA =-1,prevG=-1,prevC=-1;
for(int i=0;i<sc.length;i++){
if(sc[i]=='A')
prevA=i;
else if(sc[i] == 'G')
prevG=i;
else if(sc[i] =='C')
prevC=i;
A[i] = prevA;
G[i] = prevG;
C[i] = prevC;
//System.out.println(A[i]+ " "+G[i]+" "+C[i]);
}
int[] result = new int[P.length];
for(int i=0;i<P.length;i++){
//System.out.println(A[P[i]]+ " "+A[K[i]]+" "+C[P[i]]+" "+C[K[i]]+" "+P[i]+" "+K[i]);
if(A[K[i]] >=P[i] && A[K[i]] <=K[i]){
result[i] =1;
}
else if(C[K[i]] >=P[i] && C[K[i]] <=K[i]){
result[i] =2;
}else if(G[K[i]] >=P[i] && G[K[i]] <=K[i]){
result[i] =3;
}
else{
result[i]=4;
}
}
return result;
}
#include <string.h>
struct Results solution(char *S, int P[], int Q[], int M) {
int i, a, b, N, *pA, *pC, *pG;
struct Results result;
result.A = malloc(sizeof(int) * M);
result.M = M;
// calculate prefix sums
N = strlen(S);
pA = malloc(sizeof(int) * N);
pC = malloc(sizeof(int) * N);
pG = malloc(sizeof(int) * N);
pA[0] = S[0] == 'A' ? 1 : 0;
pC[0] = S[0] == 'C' ? 1 : 0;
pG[0] = S[0] == 'G' ? 1 : 0;
for (i = 1; i < N; i++) {
pA[i] = pA[i - 1] + (S[i] == 'A' ? 1 : 0);
pC[i] = pC[i - 1] + (S[i] == 'C' ? 1 : 0);
pG[i] = pG[i - 1] + (S[i] == 'G' ? 1 : 0);
}
for (i = 0; i < M; i++) {
a = P[i] - 1;
b = Q[i];
if ((pA[b] - pA[a]) > 0) {
result.A[i] = 1;
} else if ((pC[b] - pC[a]) > 0) {
result.A[i] = 2;
} else if ((pG[b] - pG[a]) > 0) {
result.A[i] = 3;
} else {
result.A[i] = 4;
}
}
return result;
}
static public int[] solution(String S, int[] P, int[] Q) {
// write your code in Java SE 8
int A[] = new int[S.length() + 1], C[] = new int[S.length() + 1], G[] = new int[S.length() + 1];
int last_a = 0, last_c = 0, last_g = 0;
int results[] = new int[P.length];
int p = 0, q = 0;
for (int i = S.length() - 1; i >= 0; i -= 1) {
switch (S.charAt(i)) {
case 'A': {
last_a += 1;
break;
}
case 'C': {
last_c += 1;
break;
}
case 'G': {
last_g += 1;
break;
}
}
A[i] = last_a;
G[i] = last_g;
C[i] = last_c;
}
for (int i = 0; i < P.length; i++) {
p = P[i];
q = Q[i];
if (A[p] - A[q + 1] > 0) {
results[i] = 1;
} else if (C[p] - C[q + 1] > 0) {
results[i] = 2;
} else if (G[p] - G[q + 1] > 0) {
results[i] = 3;
} else {
results[i] = 4;
}
}
return results;
}
public int[] solution(String S, int[] P, int[] Q){
int[] result = new int[P.length];
int[] factor1 = new int[S.length()];
int[] factor2 = new int[S.length()];
int[] factor3 = new int[S.length()];
int[] factor4 = new int[S.length()];
int factor1Sum = 0;
int factor2Sum = 0;
int factor3Sum = 0;
int factor4Sum = 0;
for(int i=0; i<S.length(); i++){
switch (S.charAt(i)) {
case 'A':
factor1Sum++;
break;
case 'C':
factor2Sum++;
break;
case 'G':
factor3Sum++;
break;
case 'T':
factor4Sum++;
break;
default:
break;
}
factor1[i] = factor1Sum;
factor2[i] = factor2Sum;
factor3[i] = factor3Sum;
factor4[i] = factor4Sum;
}
for(int i=0; i<P.length; i++){
int start = P[i];
int end = Q[i];
if(start == 0){
if(factor1[end] > 0){
result[i] = 1;
}else if(factor2[end] > 0){
result[i] = 2;
}else if(factor3[end] > 0){
result[i] = 3;
}else{
result[i] = 4;
}
}else{
if(factor1[end] > factor1[start-1]){
result[i] = 1;
}else if(factor2[end] > factor2[start-1]){
result[i] = 2;
}else if(factor3[end] > factor3[start-1]){
result[i] = 3;
}else{
result[i] = 4;
}
}
}
return result;
}
import scala.annotation.switch
import scala.collection.mutable
object Solution {
def solution(s: String, p: Array[Int], q: Array[Int]): Array[Int] = {
val n = s.length
def arr = mutable.ArrayBuffer.fill(n + 1)(0L)
val a = arr
val c = arr
val g = arr
val t = arr
for (i <- 1 to n) {
def inc(z: mutable.ArrayBuffer[Long]): Unit = z(i) = z(i - 1) + 1L
def shift(z: mutable.ArrayBuffer[Long]): Unit = z(i) = z(i - 1)
val char = s(i - 1)
(char: @switch) match {
case 'A' => inc(a); shift(c); shift(g); shift(t);
case 'C' => shift(a); inc(c); shift(g); shift(t);
case 'G' => shift(a); shift(c); inc(g); shift(t);
case 'T' => shift(a); shift(c); shift(g); inc(t);
}
}
val r = mutable.ArrayBuffer.fill(p.length)(0)
for (i <- p.indices) {
val start = p(i)
val end = q(i) + 1
r(i) =
if (a(start) != a(end)) 1
else if (c(start) != c(end)) 2
else if (g(start) != g(end)) 3
else if (t(start) != t(end)) 4
else 0
}
r.toArray
}
}
def solution(S, P, Q):
count = []
for i in range(3):
count.append([0]*(len(S)+1))
for index, i in enumerate(S):
count[0][index+1] = count[0][index] + ( i =='A')
count[1][index+1] = count[1][index] + ( i =='C')
count[2][index+1] = count[2][index] + ( i =='G')
result = []
for i in range(len(P)):
start = P[i]
end = Q[i]+1
if count[0][end] - count[0][start]:
result.append(1)
elif count[1][end] - count[1][start]:
result.append(2)
elif count[2][end] - count[2][start]:
result.append(3)
else:
result.append(4)
return result
class Solution {
public int[] solution(String S, int[] P, int[] Q) {
int[] preDominator = new int[S.length()];
int A = -1;
int C = -1;
int G = -1;
int T = -1;
for (int i = 0; i < S.length(); i++) {
char c = S.charAt(i);
if (c == 'A') {
A = i;
preDominator[i] = -1;
} else if (c == 'C') {
C = i;
preDominator[i] = A;
} else if (c == 'G') {
G = i;
preDominator[i] = Math.max(A, C);
} else {
T = i;
preDominator[i] = Math.max(Math.max(A, C), G);
}
}
int N = preDominator.length;
int M = Q.length;
int[] result = new int[M];
for (int i = 0; i < M; i++) {
int p = P[i];
int q = Math.min(N, Q[i]);
for (int j = q;;) {
if (preDominator[j] < p) {
char c = S.charAt(j);
if (c == 'A') {
result[i] = 1;
} else if (c == 'C') {
result[i] = 2;
} else if (c == 'G') {
result[i] = 3;
} else {
result[i] = 4;
}
break;
}
j = preDominator[j];
}
}
return result;
}
function solution(S, P, Q) {
let total = [];
let min;
for (let i = 0; i < P.length; i++) {
const substring = S.slice(P[i], Q[i] + 1);
if (substring.includes('A')) {
min = 1;
} else if (substring.includes('C')) {
min = 2;
} else if (substring.includes('G')) {
min = 3;
} else if (substring.includes('T')) {
min = 4;
}
total.push(min);
}
return total;
}
def solution(S, P, Q):
n = len(S)
m = len(P)
aux = [[0 for i in range(n+1)] for i in [0,1,2]]
for i,c in enumerate(S):
aux[0][i+1] = aux[0][i] + ( c == 'A' )
aux[1][i+1] = aux[1][i] + ( c == 'C' )
aux[2][i+1] = aux[2][i] + ( c == 'G' )
result = []
for i in range(m):
fromIndex , toIndex = P[i] , Q[i] +1
if aux[0][toIndex] - aux[0][fromIndex] > 0:
r = 1
elif aux[1][toIndex] - aux[1][fromIndex] > 0:
r = 2
elif aux[2][toIndex] - aux[2][fromIndex] > 0:
r = 3
else:
r = 4
result.append(r)
return result
public func solution(_ S : inout String, _ P : inout [Int], _ Q : inout [Int]) -> [Int] {
var impacts = [Int]()
var prefixSum = [[Int]]()
for _ in 0..<3 {
let array = Array(repeating: 0, count: S.count + 1)
prefixSum.append(array)
}
for (index, character) in S.enumerated() {
var a = 0
var c = 0
var g = 0
switch character {
case "A":
a = 1
case "C":
c = 1
case "G":
g = 1
default:
break
}
prefixSum[0][index + 1] = prefixSum[0][index] + a
prefixSum[1][index + 1] = prefixSum[1][index] + c
prefixSum[2][index + 1] = prefixSum[2][index] + g
}
for tuple in zip(P, Q) {
if prefixSum[0][tuple.1 + 1] - prefixSum[0][tuple.0] > 0 {
impacts.append(1)
}
else if prefixSum[1][tuple.1 + 1] - prefixSum[1][tuple.0] > 0 {
impacts.append(2)
}
else if prefixSum[2][tuple.1 + 1] - prefixSum[2][tuple.0] > 0 {
impacts.append(3)
}
else {
impacts.append(4)
}
}
return impacts
}
import kotlin.math.*
fun solution(S: String, P: IntArray, Q: IntArray): IntArray {
val a = IntArray(S.length)
for (i in S.indices) {
a[i] = when (S[i]) {
'A' -> 1
'C' -> 2
'G' -> 3
'T' -> 4
else -> throw IllegalStateException()
}
}
val segmentTree = IntArray(2*nextPowerOfTwo(S.length)-1)
constructSegmentTree(a, segmentTree, 0, a.size-1, 0)
val result = IntArray(P.size)
for (i in P.indices) {
result[i] = rangeMinQuery(segmentTree, P[i], Q[i], 0, a.size-1, 0)
}
return result
}
fun constructSegmentTree(input: IntArray, segmentTree: IntArray, low: Int, high: Int, pos: Int) {
if (low == high) {
segmentTree[pos] = input[low]
return
}
val mid = (low + high)/2
constructSegmentTree(input, segmentTree, low, mid, 2*pos+1)
constructSegmentTree(input, segmentTree, mid+1, high, 2*pos+2)
segmentTree[pos] = min(segmentTree[2*pos+1], segmentTree[2*pos+2])
}
fun rangeMinQuery(segmentTree: IntArray, qlow:Int, qhigh:Int ,low:Int, high:Int, pos:Int): Int {
if (qlow <= low && qhigh >= high) {
return segmentTree[pos]
}
if (qlow > high || qhigh < low) {
return Int.MAX_VALUE
}
val mid = (low + high)/2
return min(rangeMinQuery(segmentTree, qlow, qhigh, low, mid, 2*pos+1), rangeMinQuery(segmentTree, qlow, qhigh, mid+1, high, 2*pos+2))
}
fun nextPowerOfTwo(n:Int): Int {
var count = 0
var number = n
if (number > 0 && (number and (number - 1)) == 0) return number
while (number != 0) {
number = number shr 1
count++
}
return 1 shl count
}