Added english tutorial for LOJ-1068 Investigation

1068/en.md

@@ -0,0 +1,157 @@
+# LOJ-1068 – Investigation  
+---  
+**Tags:** `digit dp`, `dynamic programming`, `modulo math`
+
+---
+
+### Helpful Resources
+
+- [Digit DP Tutorial – Codeforces Blog](https://codeforces.com/blog/entry/53960 "Digit DP – Codeforces")
+
+---
+
+## Problem Summary
+
+You're given two integers `A` and `B`, and an integer `K`. Your task is to count how many numbers between `A` and `B` (inclusive) are divisible by `K` **and** whose sum of digits is also divisible by `K`.
+
+This is a classic case for applying **Digit Dynamic Programming (Digit DP)**, since a brute-force approach will time out due to the constraint on the range (`A`, `B` ≤ `2^31`).
+
+---
+
+## Key Observations
+
+1. We need to **track two properties** for a number:
+   - Is the number divisible by `K`?
+   - Is the **sum of its digits** divisible by `K`?
+
+2. We can use **Digit DP** to solve this efficiently by building the number digit by digit, while tracking:
+   - Current position in the digit array
+   - Remainder of the number modulo `K`
+   - Remainder of the digit sum modulo `K`
+   - A `tight` flag to restrict digits to the current bound if we’re matching the prefix
+
+3. We cannot take `K` directly in the state array due to memory limits.  
+   However, we can observe that:
+   - The maximum digit sum for any number ≤ `2^31` is `9 × 10 = 90`
+   - So, if `K > 90`, **no valid number exists** in the range.
+
+---
+
+## Approach
+
+- Extract digits of a number to process digit-by-digit with tight bound control
+- Use memoization to cache overlapping subproblems (`pos`, `modNum`, `modSum`, `tight`)
+- Count valid numbers between `1` to `B`, then subtract valid numbers between `1` to `A-1` to get the count in range `[A, B]`
+
+---
+
+## Time Complexity
+
+- The number of states is bounded by: `pos (≤ 10) × modNum (≤ K) × modSum (≤ K) × tight (2)`
+- Overall time complexity is **O(10 × K × K × 2)** = `O(K^2)` (acceptable since `K ≤ 90`)
+
+---
+
+## Solution (Java)
+
+```java
+import java.util.*;
+
+public class Solution {
+    static int[][][][] dp;
+    static int[] digits;
+    static int K;
+
+    public static void main(String[] args) {
+        Scanner sc = new Scanner(System.in);
+        int T = sc.nextInt();
+        for (int tc = 1; tc <= T; tc++) {
+            int A = sc.nextInt();
+            int B = sc.nextInt();
+            K = sc.nextInt();
+
+            if (K > 90) {
+                System.out.println("Case " + tc + ": 0"); 
+                continue;
+            }
+
+            int res = countValid(B) - countValid(A - 1);
+            System.out.println("Case " + tc + ": " + res);
+        }
+    }
+
+    static int countValid(int x) {
+        digits = extractDigits(x);
+        int n = digits.length;
+        dp = new int[n][K][K][2]; // pos, mod_num, mod_sum, tight
+        for (int[][][] d1 : dp)
+            for (int[][] d2 : d1)
+                for (int[] d3 : d2)
+                    Arrays.fill(d3, -1);
+
+        return dfs(0, 0, 0, 1);
+    }
+
+    static int dfs(int pos, int modNum, int modSum, int tight) {
+        if (pos == digits.length)
+            return (modNum == 0 && modSum == 0) ? 1 : 0;
+
+        if (dp[pos][modNum][modSum][tight] != -1)
+            return dp[pos][modNum][modSum][tight];
+
+        int limit = (tight == 1) ? digits[pos] : 9;
+        int res = 0;
+
+        for (int d = 0; d <= limit; d++) {
+            int newTight = (tight == 1 && d == limit) ? 1 : 0;
+            int newModNum = (modNum * 10 + d) % K;
+            int newModSum = (modSum + d) % K;
+
+            res += dfs(pos + 1, newModNum, newModSum, newTight);
+        }
+
+        return dp[pos][modNum][modSum][tight] = res;
+    }
+
+    static int[] extractDigits(int x) {
+        List<Integer> list = new ArrayList<>();
+        if (x == 0) 
+            list.add(0); 
+        while (x > 0) {
+            list.add(x % 10);
+            x /= 10;
+        }
+        Collections.reverse(list);
+        return list.stream()
+                    .mapToInt(i -> i)
+                    .toArray();
+    }
+}
+```
+## Additional Notes
+
+### Memory Limit Issues (MLE)
+
+If you're facing **Memory Limit Exceeded (MLE)** on platforms like **LightOJ**, consider the following optimizations:
+
+- **Use `Integer` instead of `int` in DP array** only if absolutely needed (not usually required here).
+- Switch from a 4D array to a **HashMap with composite key encoding** like `pos * K * K * 2 + modNum * K * 2 + modSum * 2 + tight`. This reduces memory overhead at the cost of some performance.
+- Use **`Integer[][][][]` DP** and `null` instead of `-1` if required to avoid initialization overhead.
+- Avoid printing inside the DP recursion — this can blow up memory if prints are numerous.
+- Avoid constructing large intermediate strings (like via `StringBuilder`) in recursive calls.
+
+### Time Limit Exceeded (TLE)
+
+If you’re running into **TLE**:
+
+- Precompute and reuse DP table per test case wherever possible.
+- Ensure `extractDigits()` and `countValid()` are efficient.
+- Avoid unnecessary object creation inside recursive calls.
+- Use `FastIO` for input/output on platforms with tight I/O constraints.
+
+---
+
+## Author
+
+- Sandesh Atulya
+- Find me at [LinkedIn](https://www.linkedin.com/in/sandesh-atulya/)