E. Sponsor of Your Problems
dp; greedy; implementation; strings
For two integers \(a\) and \(b\), we define \(f(a, b)\) as the number of positions in the decimal representation of the numbers \(a\) and \(b\) where their digits are the same. For example, \(f(12, 21) = 0\), \(f(31, 37) = 1\), \(f(19891, 18981) = 2\), \(f(54321, 24361) = 3\).You are given two integers \(l\) and \(r\) of the same length in decimal representation. Consider all integers \(l \leq x \leq r\). Your task is to find the minimum value of \(f(l, x) + f(x, r)\).
Each test consists of multiple test cases. The first line contains a single integer \(t\) (\(1 \leq t \leq 10^4\)) — the number of test cases. The description of the test cases follows.Each test case consists of a single line containing two integers \(l\) and \(r\) (\(1 \leq l \leq r < 10^9\)).It is guaranteed that the numbers \(l\) and \(r\) have the same length in decimal representation and do not have leading zeros.
For each test case, output the minimum value of \(f(l, x) + f(x, r)\) among all integer values \(l \leq x \leq r\).
In the first test case, you can choose \(x = 1\). Then \(f(1, 1) + f(1, 1) = 1 + 1 = 2\).In the second test case, you can choose \(x = 2\). Then \(f(2, 2) + f(2, 3) = 1 + 0 = 1\).In the third test case, you can choose \(x = 5\). Then \(f(4, 5) + f(5, 6) = 0 + 0 = 0\).In the fourth test case, you can choose \(x = 15\). Then \(f(15, 15) + f(15, 16) = 2 + 1 = 3\).In the fifth test case, you can choose \(x = 18\). Then \(f(17, 18) + f(18, 19) = 1 + 1 = 2\).In the sixth test case, you can choose \(x = 200\). Then \(f(199, 200) + f(200, 201) = 0 + 2 = 2\).In the seventh test case, you can choose \(x = 900\). Then \(f(899, 900) + f(900, 999) = 0 + 1 = 1\).In the eighth test case, you can choose \(x = 1992\). Then \(f(1990, 1992) + f(1992, 2001) = 3 + 0 = 3\).