String Question | JPMorgan

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String question | JPMorgan

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Can you post the question link??

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I was asked this in the interview, is anything unclear?

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I didn’t get the string conversion.

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You need to make the minimum number of changes in string S such that it has x number of 1′s in the beginning, followed by the remaining (length−x) number of 0′s.

You can change any 1 to 0 and vice versa

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for every index, store the number of 0’s to its left and number of 1’s to its right. we just have to minimize the sum.

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It’s literally just this

s=input()
curr=s.count('1')
ans=curr
for c in s:
    if(c=='0'):
        curr+=1
    else:
        curr-=1
        ans=min(ans, curr)
print(ans)

x=0 case, we convert all to 0s. then we go forward. If it’s a 0, then for this x, we’ll need one more change, if it’s a 1, we’ll need one less change, and take minimum.

2 Likes
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Can you pls elaborate

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if x=0, then we convert every ‘1’ to a ‘0’. So base answer = s.count(‘1’)
Then if we increase x, we have to look at the next value. If it’s a ‘0’, We’ll have to convert this also to a ‘1’, so the current answer will increase. If it’s a ‘1’, Then we won’t have to convert it to a ‘0’, so the current answer will decrease. Just take minimum of that.

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@vai53
Dude like literally it’s exactly the same question from hackerearth circuits 2 problem.
There it was a string of A’s and B’s and here u changed it as 1’s and 0’s :unamused: :unamused: