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Count of different numbers divisible by 3 that can be obtained by changing at most one digit
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# '0081' -> 11 | |
# '23' -> 7 | |
# '022' -> 9 | |
def is_div(s: str) -> bool: | |
#print(int(s)) | |
if (int(s) % 3 == 0): | |
return True | |
return False | |
# O(n^2) | |
def solve1(nstr: str) -> int: | |
res = 0 | |
for i in range(len(nstr)): | |
#print(int(nstr[i])) | |
temp = list(nstr) | |
for x in range(10): | |
if x == int(nstr[i]): | |
continue | |
#print(str(x)) | |
temp[i] = str(x) | |
if is_div("".join(temp)): | |
res += 1 | |
if is_div(nstr): | |
res += 1 | |
print("Result: ", res) | |
return res | |
# O(n) | |
def solve2(nstr: str) -> int: | |
result = 0 | |
sum = 0 | |
for c in nstr: | |
sum += int(c) | |
rest = sum % 3 | |
# Check if initial string is divisible by 3 | |
if rest == 0: | |
result += 1 | |
# Changing a digit means that sum of all digits | |
# may change by minimum of -9 and maximum of 9. | |
# We want to add (or subtract) only those numbers, | |
# that keep the sum divisible by 3, | |
# hence we use (inclusive) range | |
# [-9-rest, 9-rest], where step is 3. | |
for inc in range(-9-rest, 9-rest+1, 3): | |
if inc == 0: | |
# We have already taken into account initial string. | |
continue | |
for i in range(len(nstr)): | |
x = int(nstr[i]) + inc | |
if x >= 0 and x <= 9: | |
result += 1 | |
print("Result: ", result) | |
return result | |
def main(): | |
assert(solve2('23') == 7) | |
assert(solve2('0081') == 11) | |
assert(solve2('022') == 9) | |
assert(solve2('235') == 9) | |
if __name__ == '__main__': | |
main() |
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