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[GangBean] Week 6 #884

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Jan 19, 2025
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45 changes: 45 additions & 0 deletions container-with-most-water/GangBean.java
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투 포인터가 줄어들 때 마다, 전체 면적이 줄어 들 때 내부 while 문을 사용하면 시간을 좀 더 효율적으로 가져갈 수도 있을것 같아요 :)

Original file line number Diff line number Diff line change
@@ -0,0 +1,45 @@
class Solution {
public int maxArea(int[] height) {
/**
1. understanding
- with two pair of line, each can contain "min(line1,line2) * (distance)" mount of water
- find maximum amount of water can contain
2. strategy
- brute force
- for each pair of lines, calculate amount and update maximum amount.
- it can takes O(N), where N is the count of lines.
- N is 10^5 at most, so it can takes 10^10, which can takes about 10 seconds
- you need to swap out unnecessary calculation
- so, let's find a unnecessary calculation in this problem.
3. complexity
- time: O(N)
- space: O(1)
*/
int l = 0;
int r = height.length - 1;
int maxAmount = amountOf(height, l, r); // Math.min(height[l], height[r], r-l);
while (l < r) { // O(N)
maxAmount = Math.max(maxAmount, amountOf(height, l, r));
if (height[l] < height[r]) {
l++;
} else if (height[l] > height[r]) {
r--;
} else {
int nextLeftAmount = amountOf(height, l+1, r);
int nextRightAmount = amountOf(height, l, r-1);
if (nextLeftAmount < nextRightAmount) {
r--;
} else {
l++;
}
}
}

return maxAmount;
}

private int amountOf(int[] height, int left, int right) {
return (Math.min(height[left], height[right]) * (right - left));
}
}

38 changes: 38 additions & 0 deletions valid-parentheses/GangBean.java
Original file line number Diff line number Diff line change
@@ -0,0 +1,38 @@
class Solution {
public boolean isValid(String s) {
/**
1. understanding
- given string contains character which has paired characters.
- validate all given characters in input string has each pair, and valid in order.
2. strategy
- use stack, to save left brackets, and if you encounter right bracket, then pop from stack, and compare two characters are paired.
- if not paired on any right character, or stack is empty then return false.
- all characters are iterated and stack is not empty, then return false.
3. complexity
- time: O(N), N is the length of input string s.
- space: O(N), for stack variable memory.
*/
Stack<Character> leftBracket = new Stack<>();
for (char c : s.toCharArray()) {
if (c == '(' || c == '{' || c == '[') {
// when character is left bracket, then push to stack.
leftBracket.push(c);
} else if (c == ')' || c == '}' || c == ']') {
if (leftBracket.isEmpty()) return false;
char left = leftBracket.pop();
if (isPair(left, c)) continue;
return false;
} else {
throw new RuntimeException(String.format("Not valid input character: %c in input %s", c, s));
}
Comment on lines +25 to +27
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예외처리 보다는 false를 바로 던지시면 되지 않을까요?
해당 문제에서 return 은 true, false 만 사용하면 되니까요 :)

}
return leftBracket.isEmpty();
}

private boolean isPair(char left, char right) {
return (left == '(' && right == ')')
|| (left == '{' && right == '}')
|| (left == '[' && right == ']');
}
}

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