taoc6 - aoc22 - advent of code 2022 solutions
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       ---
       taoc6 (259B)
       ---
            1 #!/bin/awk -f
            2 {
            3         for (i=4; i<=length($0); i++) {
            4                 m1 = substr($0, i-3, 1)
            5                 m2 = substr($0, i-2, 1)
            6                 m3 = substr($0, i-1, 1)
            7                 m4 = substr($0, i, 1)
            8                 if (m1 != m2 && m1 != m3 && m1 != m4 && m2 != m3 && m2 != m4 && m3 != m4) {
            9                         print(i)
           10                         break
           11                 }
           12         }
           13 }