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Someth else solution in Clear category for Greedy Number by U.V
def greedy_number(line:str, leng: int) -> str:
ans=list(line)
while len(ans) > leng:
for i in range(len(ans) - 1):
if ans[i] < ans[i + 1]:
ans.pop(i)
break
else:
ans.pop()
return ''.join(ans)
Sept. 30, 2023
