diff --git a/1937.Maximum-Number-of-Points-with-Cost/memo.md b/1937.Maximum-Number-of-Points-with-Cost/memo.md
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+# 1937. Maximum Number of Points with Cost
+
+## step1
+
+dp を使えそうだと思った。計算量 O(mn^2) の解法しか思いつかず、TLE した解法: step1_TLE.py
+
+28mほど経過して諦めて答えを見る
+
+# step2
+
+https://leetcode.com/problems/maximum-number-of-points-with-cost/solutions/1344888/c-dp-from-om-n-n-to-om-n-by-npes87184-waex/?envType=problem-list-v2&envId=7p55wqm
+
+各行の更新にもDPをつかう。これで正しいのか分からなかったので考えてみる
+
+ left[c]
+= max_{i<=c}(dp_prev[i]  - (c-i))
+= max_{i<=c}(dp_prev[i]  + i) - c
+= max(max_{i<=c-1}(dp_prev[i]  + i), dp_prev[c] + c) - c
+= max(max_{i<=c-1}(dp_prev[i]  + i) - (c - 1) - 1, dp_prev[c])
+= max(left[c-1] - 1, dp_prev[c])
+
+1マスずれるたびに一律に1ペナルティが発生するから前の列の値を使える
+
+rightも同様
+
+---
+
+running_max という変数を使って配列の生成コストをなくす
+
+これ以外の解法は思いつかない
+
+## step3
+TODO
+
+
diff --git a/1937.Maximum-Number-of-Points-with-Cost/step1_TLE.py b/1937.Maximum-Number-of-Points-with-Cost/step1_TLE.py
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+++ b/1937.Maximum-Number-of-Points-with-Cost/step1_TLE.py
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+class Solution:
+    def maxPoints(self, points: List[List[int]]) -> int:
+        if not points or not points[0]:
+            return 0
+
+        num_rows = len(points)
+        num_cols = len(points[0])
+
+        dp_prev = points[0][:]
+        dp = [0] * num_cols
+
+        for r in range(1, num_rows):
+            for c in range(num_cols):
+                dp[c] = max([dp_prev[c_prev] - abs(c - c_prev) for c_prev in range(num_cols)]) + points[r][c]
+            dp, dp_prev = dp_prev, dp
+
+        return max(dp_prev)
+
+
diff --git a/1937.Maximum-Number-of-Points-with-Cost/step2.py b/1937.Maximum-Number-of-Points-with-Cost/step2.py
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+++ b/1937.Maximum-Number-of-Points-with-Cost/step2.py
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+class Solution:
+    def maxPoints(self, points: List[List[int]]) -> int:
+        if not points or not points[0]:
+            return 0
+
+        num_rows = len(points)
+        num_cols = len(points[0])
+
+        dp_prev = points[0][:]
+        dp = [0] * num_cols
+
+        for r in range(1, num_rows):
+            left = [0] * num_cols
+            left[0] = dp_prev[0]
+            for c in range(1, num_cols):
+                left[c] = max(left[c - 1] - 1, dp_prev[c])
+
+            right = [0] * num_cols
+            right[-1] = dp_prev[-1]
+            for c in range(num_cols - 2, -1, -1):
+                right[c] = max(right[c + 1] - 1, dp_prev[c])
+
+            for c in range(num_cols):
+                dp[c] = points[r][c] + max(left[c], right[c])
+
+            dp_prev, dp = dp, dp_prev
+
+        return max(dp_prev)
diff --git a/1937.Maximum-Number-of-Points-with-Cost/step2_running_max.py b/1937.Maximum-Number-of-Points-with-Cost/step2_running_max.py
new file mode 100644
index 0000000..348295a
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+++ b/1937.Maximum-Number-of-Points-with-Cost/step2_running_max.py
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+class Solution:
+    def maxPoints(self, points: List[List[int]]) -> int:
+        if not points or not points[0]:
+            return 0
+
+        num_rows = len(points)
+        num_cols = len(points[0])
+
+        dp_prev = points[0][:]
+        dp = [0] * num_cols
+
+        for r in range(1, num_rows):
+            running_max = 0
+            for c in range(num_cols):
+                running_max = max(running_max - 1, dp_prev[c])
+                dp[c] = running_max
+
+            running_max = 0
+            for c in range(num_cols - 1, -1, -1):
+                running_max = max(running_max - 1, dp_prev[c])
+                dp[c] = max(dp[c], running_max) + points[r][c]
+
+            dp_prev, dp = dp, dp_prev
+
+        return max(dp_prev)