84. Largest Rectangle in Histogram

84. Largest Rectangle in Histogram.md

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+## step1 とりあえず解く
+- 計算量
+  - 時間：O(N^2)
+  - 空間：O(N)
+- Acceptされなかったコード
+- 考えたこととしては、どこまでそのバーの高さを維持できるか調べる
+  - 高さが下がればStackから取り出して適宜最大面積を計算していく
+  - 最後にStackに残った要素を計算する
+- このコードしか思いつかなかったが、ACできなかった理由としては
+  - addFirstしているため単調増加Stackを保てなかった
+    - 落ちるケース[3,6,5,7,4,8,1,0]
+    - 単調増加できないとpeekLastメソッドの判定がうまくいかない
+
+```java
+class Solution {
+    private final record ContinuousBar(int index, int height) {};
+    public int largestRectangleArea(int[] heights) {
+        ArrayDeque<ContinuousBar> continuousBars = new ArrayDeque<>();
+        int maxSize = 0;
+        int heightsLength = heights.length;
+        for (int i = 0; i < heightsLength; i++) {
+            while (!continuousBars.isEmpty() && continuousBars.peekLast().height() > heights[i]) {
+                ContinuousBar bar = continuousBars.pollLast();
+                maxSize = Math.max(maxSize, (i - bar.index()) * bar.height());
+                continuousBars.addFirst(new ContinuousBar(bar.index(), heights[i]));
+                System.out.println("maxSize : " + maxSize + " " + bar.index() + " " + bar.height());
+            }
+            continuousBars.add(new ContinuousBar(i, heights[i]));
+        }
+        while (!continuousBars.isEmpty()) {
+            ContinuousBar bar = continuousBars.pollLast();
+            maxSize = Math.max(maxSize, (heightsLength - bar.index()) * bar.height());
+            System.out.println("maxSize2 : " + maxSize + " " + bar.index() + " : " + bar.height());
+        }
+        return maxSize;
+    }
+}
+
+```
+
+## step2 他の人の回答を見る
+- 計算量
+  - 時間：O(N)
+  - 空間：O(N)
+- step1で自分が間違えたコードの修正版のようなコード
+- 単調増加Stackを作ることで時間計算量が抑えられる
+  - 各要素はadd, pollが一度しかされずO(N)
+
+```java
+class Solution {
+
+    private final record RectangleCandidate(int index, int height) {};
+
+    public int largestRectangleArea(int[] heights) {
+        ArrayDeque<RectangleCandidate> rectangleCandidate = new ArrayDeque<>(); 
+        int maxArea = 0;
+
+        for (int i = 0; i < heights.length; i++) {
+            int start = i;
+            while (!rectangleCandidate.isEmpty() && rectangleCandidate.peekLast().height() > heights[i]) {
+                RectangleCandidate candidate = rectangleCandidate.pollLast();
+                maxArea = Math.max(maxArea, candidate.height() * (i - candidate.index()));
+                start = candidate.index();
+            }
+            rectangleCandidate.addLast(new RectangleCandidate(start, heights[i]));
+        }
+
+        while (!rectangleCandidate.isEmpty()) {
+            RectangleCandidate candidate = rectangleCandidate.pollLast();
+            maxArea = Math.max(maxArea, candidate.height() * (heights.length - candidate.index()));
+        }
+
+        return maxArea;
+    }
+}
+
+```
+
+- 全探索
+- 計算量
+  - 時間：O(N^2)
+  - 空間：O(1)
+- i番目のバーから見て右、左にどれだけ伸ばせるかを考えるコード
+- 時間計算量が大きいので今回の最適解ではないと感じた
+
+```java
+public class Solution {
+    public int largestRectangleArea(int[] heights) {
+        int n = heights.length;
+        int maxArea = 0;
+
+        for (int i = 0; i < n; i++) {
+            int height = heights[i];
+
+            int rightMost = i + 1;
+            while (rightMost < n && heights[rightMost] >= height) {
+                rightMost++;
+            }
+
+            int leftMost = i;
+            while (leftMost >= 0 && heights[leftMost] >= height) {
+                leftMost--;
+            }
+
+            rightMost--;
+            leftMost++;
+            maxArea = Math.max(maxArea, height * (rightMost - leftMost + 1));
+        }
+        return maxArea;
+    }
+}
+```
+
+## step3 3回とく
+- Step2で見たStackを使うやり方
+
+```java
+class Solution {
+
+    private final record RectangleCandidate(int index, int height) {};
+
+    public int largestRectangleArea(int[] heights) {
+        ArrayDeque<RectangleCandidate> rectangleCandidate = new ArrayDeque<>();
+        int maxArea = 0;
+        for (int i = 0; i < heights.length; i++) {
+            int startIndex = i;
+            while (!rectangleCandidate.isEmpty() && rectangleCandidate.peekLast().height() > heights[i]) {
+                RectangleCandidate candidate = rectangleCandidate.pollLast();
+                maxArea = Math.max(maxArea, candidate.height() * (i - candidate.index()));
+                startIndex = candidate.index();
+            }
+            rectangleCandidate.add(new RectangleCandidate(startIndex, heights[i]));
+        }
+        while (!rectangleCandidate.isEmpty()) {
+            RectangleCandidate candidate = rectangleCandidate.pollLast();
+            maxArea = Math.max(maxArea, candidate.height() * (heights.length - candidate.index()));
+        }
+
+        return maxArea;
+    }
+}
+
+```
+
+
+## step4 いただいた指摘を元に修正
+>heights の後ろに0を追加すると、後ろのコード不要になりそうですね。(番兵。)
+
+- 上記コメントを元に修正
+- `System.arraycopy` を用いて配列コピーして最後に0を追加してもよかったが、配列コピーのためにメモリがO(N)追加で必要になるため採用しなかった
+
+```java
+class Solution {
+
+    private final record RectangleCandidate(int index, int height) {};
+
+    public int largestRectangleArea(int[] heights) {
+        ArrayDeque<RectangleCandidate> rectangleCandidate = new ArrayDeque<>();
+        int maxArea = 0;
+        for (int i = 0; i <= heights.length; i++) {
+            int startIndex = i;
+            int currentHeight;
+            if (i == heights.length) {
+                currentHeight = 0;
+            } else {
+                currentHeight = heights[i];
+            }
+            while (!rectangleCandidate.isEmpty() && rectangleCandidate.peekLast().height() > currentHeight) {
+                RectangleCandidate candidate = rectangleCandidate.pollLast();
+                maxArea = Math.max(maxArea, candidate.height() * (i - candidate.index()));
+                startIndex = candidate.index();
+            }
+            rectangleCandidate.add(new RectangleCandidate(startIndex, currentHeight));
+        }
+
+        return maxArea;
+    }
+}
+
+```