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[YoungSeok-Choi] Week 1 Solutions #1163

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19 changes: 19 additions & 0 deletions contains-duplicate/YoungSeok-Choi.java
Original file line number Diff line number Diff line change
@@ -0,0 +1,19 @@
import java.util.HashMap;
import java.util.Map;

class Solution {
// 시간복잡도 O(n)
public boolean containsDuplicate(int[] nums) {
Map<Integer, Boolean> dupMap = new HashMap<>();

for(int i = 0; i < nums.length; i++) {
if(dupMap.containsKey(nums[i])) {
return true;
}

dupMap.put(nums[i], true);
}

return false;
}
}
84 changes: 84 additions & 0 deletions house-robber/YoungSeok-Choi.java
Original file line number Diff line number Diff line change
@@ -0,0 +1,84 @@
import java.util.HashMap;
import java.util.Map;

// 시간복잡도: O(n)
// TODO: DP 방식으로 풀어보기
class Solution {
public Map<Integer, Integer> robMap = new HashMap<>();
public int rob(int[] nums) {
return dfs(nums, 0);
}

public int dfs(int[] nums, int index) {
if(nums.length == 0) {
return 0;
}

if(index >= nums.length) {
return 0;
}

// 이미 털었던 집이라면, 해
if(robMap.containsKey(index)) {
return robMap.get(index);
}

// 이번 집을 털게되는 경우
int robThis = nums[index] + dfs(nums, index + 2);

// 이번 집을 털지않고 건너뛰는 경우,.
int skipThis = dfs(nums, index + 1);

robMap.put(index, Math.max(robThis, skipThis));

return robMap.get(index);
}
}

// TODO: 비효율적으로 작성한 알고리즘의 동작 방식을 도식화 해서 그려보기.
// NOTE: dfs를 사용한 완전탐색
// 탐색 방식이 매우 비효율적이라, 정답은 맞추지만 N이 커지면 시간초과
// 시간복잡도: O(2^n) + alpha(중복탐색)
class WrongSolution {
public boolean[] visit;
public int mx = -987654321;
public int curSum = 0;

public int rob(int[] nums) {
if(nums.length == 1) {
return nums[0];
}

visit = new boolean[nums.length];
dfs(nums, 0);
dfs(nums, 1);

return mx;
}

public void dfs(int[] arr, int idx) {
int len = arr.length;
int prevIdx = idx - 1;
int nextIdx = idx + 1;


if(idx == 0) {
if(visit[idx]) return;
} else {
if(idx >= len || visit[idx] || visit[prevIdx]) {
return;
}
}

visit[idx] = true;
curSum += arr[idx];
mx = Math.max(mx, curSum);

for(int i = idx; i < len; i++) {
dfs(arr, i);
}

visit[idx] = false;
curSum -= arr[idx];
}
}
34 changes: 34 additions & 0 deletions longest-consecutive-sequence/YoungSeok-Choi.java
Original file line number Diff line number Diff line change
@@ -0,0 +1,34 @@
import java.util.Arrays;

class Solution {
public int longestConsecutive(int[] nums) {
int curSeq = 1;
int maxSeq = -987654321;

if(nums.length == 0) {
return 0;
}

Arrays.sort(nums);

int cur = nums[0];
for(int i = 1; i < nums.length; i++) {
if(cur == nums[i]) {
continue;
}

if(cur < nums[i] && Math.abs(nums[i] - cur) == 1) {
curSeq++;
cur = nums[i];
continue;
}

// NOTE: 수열의 조건이 깨졌을 때
maxSeq = Math.max(maxSeq, curSeq);
curSeq = 1;
cur = nums[i];
}

return Math.max(maxSeq, curSeq);
}
}
47 changes: 47 additions & 0 deletions product-of-array-except-self/YoungSeok-Choi.java
Original file line number Diff line number Diff line change
@@ -0,0 +1,47 @@
class Solution {
// 시간복잡도: O(3n) -> O(n)
public int[] productExceptSelf(int[] nums) {

int zeroCount = 0;
int[] result = new int[nums.length];
int productExceptZero = 1;
int zeroIdx = 0;

for(int i = 0; i < nums.length; i++) {
if(nums[i] == 0) {
zeroCount++;
}
}

// NOTE: 0이 두개 이상일 때, 모든 배열의 원소가 0;
if(zeroCount >= 2) {
return result;
}

// NOTE: 0이 1개일 때, 0인 index만을 제외하고 모두 곱
if(zeroCount == 1) {
for(int i = 0; i < nums.length; i++) {
if(nums[i] == 0) {
zeroIdx = i;
continue;
}
productExceptZero *= nums[i];
}

result[zeroIdx] = productExceptZero;
return result;
}

// NOTE: 0이 없을 때 모든수를 곱한 뒤 idx를 나누기.
for(int i = 0; i < nums.length; i++) {
productExceptZero *= nums[i];
}

for(int i = 0; i < nums.length; i++) {
int copy = productExceptZero;
result[i] = copy / nums[i];
}

return result;
}
}
23 changes: 23 additions & 0 deletions two-sum/YoungSeok-Choi.java
Original file line number Diff line number Diff line change
@@ -0,0 +1,23 @@
// time complexity: O(n);
// 특정 nums[i] 가 target이 되기위한 보수 (target - nums[i])를 candidate Map에서 찾으면 종료

import java.util.HashMap;
import java.util.Map;

class Solution {
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> candidate = new HashMap<>();

for(int i = 0; i < nums.length; i++) {

int diff = target - nums[i];

if(candidate.containsKey(diff)) {
return new int[] { candidate.get(diff), i };
}

candidate.put(nums[i], i);
}
return new int[0];
}
}