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LESSON에 해당되는 글 5건
- 2022.06.29 codility Lesson 풀어보기 3
- 2022.06.26 codility Lesson 풀어보기 2
글
codility Lesson 풀어보기 3
Codiliy Lesson 풀어보기 Day3.
- 아직은 Easy 위주로 진행.
Codility Lesson 3. Time Complexity - TapeEquilibrium
public int solution(int[] A) {
// write your code in Java SE 8
int minNum = Integer.MAX_VALUE;
int totalSum = 0, leftSum=0;
for(int i : A)
totalSum += i;
for(int i=0; i<A.length-1; i++){
leftSum += A[i];
minNum = Math.min(minNum, Math.abs(leftSum - (totalSum-leftSum)));
}
return minNum;
}
- for문 내에서 합계를 매번 구하도록 하면 타임아웃이 난다.
https://app.codility.com/demo/results/trainingA9H573-HVU/
Test results - Codility
A non-empty array A consisting of N integers is given. Array A represents numbers on a tape. Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1]. The difference betw
app.codility.com
Codility Lesson 4. Counting Elements - FrogRiverOne
public int solution(int X, int[] A) {
// write your code in Java SE 8
int[] chkList = new int[X+1];
for(int i=0, cnt=0; i<A.length; i++) {
if(chkList[A[i]] != 1 && A[i]<= X){
chkList[A[i]] = 1;
cnt++;
}
if(cnt == X) return i;
}
return -1;
}
- 모든 배열을 비교할 필요는 없고, X 까지 값 채워지는 순간까지만 비교하면 OK
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글
codility Lesson 풀어보기 2
Codiliy Lesson 풀어보기 Day2.
- Day1(06.19)에 Lesson 2까지 전부 풀었으나, 작성한 글이 날아가서 Lesson2 Arrays 먼저 작성 후 Lesson3 Time Complexity 문제 풀이 진행
- 아직은 Easy 위주로 진행.
Codility Lesson 2. Arrays - CyclicRotation
public int[] solution(int[] A, int K) {
LinkedList<Integer> list = new LinkedList<>();
boolean checkValue;
//validation value
checkValue = checkValue(A, K);
if(!checkValue) {
return A;
}
//int[] to LinkedList
for(int a : A){
list.add(a);
}
for(int i=0; i<K; i++) {
list.addFirst(list.pollLast());
}
return list.stream().mapToInt(Integer::intValue).toArray();
}
private boolean checkValue(int[] a, int k) {
if(a.length < 1) {
return false;
}
if(k < 1 ) {
return false;
}
return true;
}
Codility Lesson 2. Arrays - OddOccurencesInArray
public int solution(int[] A) {
// write your code in Java SE 8
Map<Integer, Integer> map = new HashMap<>();
//make map
for(int a : A) {
map.put(a, map.getOrDefault(a, 0) + 1);
}
//find odd num
for(int key : map.keySet()) {
if(map.get(key)%2 == 1) return key;
}
return -1;
}
Codility Lesson 3. Time Complexity - CyclicRotation
public int solution(int X, int Y, int D) {
// write your code in Java SE 8
return (int)Math.ceil((double)(Y-X)/D);
}
https://app.codility.com/demo/results/trainingA5PMWS-AFH/
Test results - Codility
A small frog wants to get to the other side of the road. The frog is currently located at position X and wants to get to a position greater than or equal to Y. The small frog always jumps a fixed distance, D. Count the minimal number of jumps that the smal
app.codility.com
Codility Lesson 3. Time Complexity - PermMissingElem
public int solution(int[] A) {
// write your code in Java SE 8
int[] matchNum = new int[A.length+2];
for(int i=0; i<A.length; i++){
matchNum[A[i]] = 1;
}
for(int i=1; i<matchNum.length; i++){
if(matchNum[i] == 0) return i;
}
return -1;
}
https://app.codility.com/demo/results/training626AX7-JTG/
Test results - Codility
An array A consisting of N different integers is given. The array contains integers in the range [1..(N + 1)], which means that exactly one element is missing. Your goal is to find that missing element. Write a function: class Solution { public int solutio
app.codility.com
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