For two integers a and b , we define f(a,b) as t...

java
import java.io.*;
import java.util.*;

public class Main {
    static String L, R;
    static int N;
    // memo[pos][tightLow?1:0][tightHigh?1:0]
    static int[][][] memo;
    static int[] lDigits, rDigits;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        int t = Integer.parseInt(br.readLine().trim());
        while (t-- > 0) {
            StringTokenizer st = new StringTokenizer(br.readLine());
            L = st.nextToken();
            R = st.nextToken();
            N = L.length();
            lDigits = new int[N];
            rDigits = new int[N];
            for (int i = 0; i < N; i++) {
                lDigits[i] = L.charAt(i) - '0';
                rDigits[i] = R.charAt(i) - '0';
            }
            memo = new int[N][2][2];
            for (int i = 0; i < N; i++) {
                for (int a = 0; a < 2; a++) {
                    for (int b = 0; b < 2; b++) {
                        memo[i][a][b] = -1;
                    }
                }
            }
            int answer = dp(0, 1, 1);
            out.println(answer);
        }
        out.flush();
    }

    // pos: current digit position [0..N)
    // tightLow: if prefix so far equals L's prefix (1=true, 0=false)
    // tightHigh: if prefix so far equals R's prefix (1=true, 0=false)
    static int dp(int pos, int tightLow, int tightHigh) {
        if (pos == N) {
            return 0;
        }
        if (memo[pos][tightLow][tightHigh] != -1) {
            return memo[pos][tightLow][tightHigh];
        }
        int res = Integer.MAX_VALUE;
        int low = tightLow == 1 ? lDigits[pos] : 0;
        int high = tightHigh == 1 ? rDigits[pos] : 9;
        for (int d = low; d <= high; d++) {
            int nextTightLow = (tightLow == 1 && d == lDigits[pos]) ? 1 : 0;
            int nextTightHigh = (tightHigh == 1 && d == rDigits[pos]) ? 1 : 0;
            int cost = 0;
            if (d == lDigits[pos]) cost++;
            if (d == rDigits[pos]) cost++;
            int sub = dp(pos + 1, nextTightLow, nextTightHigh);
            res = Math.min(res, cost + sub);
        }
        memo[pos][tightLow][tightHigh] = res;
        return res;
    }
}

Explanation of the approach:

1. Digit DP setup. We process the numbers as digit arrays of equal length $N$.
2. State definition. Let dp(pos, tightLow, tightHigh) be the minimum total cost from digit position pos to the end, given whether we’re still “tight” to the lower bound $L$ (tightLow) and to the upper bound $R$ (tightHigh).
3. Transitions. At each position, choose a digit $d$ between the allowed range:
   • If we’ve been tight to $L$, $d$ must be $\ge L[pos]$; otherwise $d$ can be 0.
   • If we’ve been tight to $R$, $d$ must be $\le R[pos]$; otherwise $d$ can be 9.
4. Cost per digit. Choosing digit $d$ incurs: $$(\,d == L[pos]\,) + (\,d == R[pos]\,)$$ which exactly matches the definition $f(L, x)$ and $f(x, R)$ at that position.
5. Tightness updates. After choosing $d$, we remain tight to $L$ if we were tight and $d = L[pos]$; similarly for $R$.
6. Memoization. We cache results in a 3D array of size $\le 10 \times 2 \times 2$, yielding $O(N \times 2 \times 2 \times 10)$ per test case, which is efficient even for $10^4$ test cases.

This solves for the minimum value of $f(L, x) + f(x, R)$ over all $L \le x \le R$.