regex - how replace string repeated pattern at his start ,linux terminal? -

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i want edit string using regular expression in linux shell failed.

the thing want @ start of string change repeatedlly occurrence of "00" "1"

let : 00000120001 110100001 failed that.

i tried:

echo 00000120001 | sed 's/^\`\<00/1/g'

but got 1000120001

not sure can single regex; have solution juggling hold , pattern spaces of sed.

in 1 line:

sed 'h;s/[^0].*//;s/00/1/g;x;s/^0*//;x;g;s/\n//' <<< '00000120001'

with details:

sed 'h          # copy input line hold space      s/[^0].*// # keep 00..0 preffix in pattern space      s/00/1/g   # replace double 0s 1s in pattern space      x          # swap hold , pattern spaces (hold 110, pattern 00000120001      s/^0*//    # remove 00..0 prefix pattern space      x          # swap hold , pattern spaces (hold 120001, pattern 110      g          # append hold space pattern space (pattern 110\n120001)      s/\n//'    # remove \n pattern <<< '00000120001'

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