In C, how do you prove that if no character appears more than floor(n/2) times, we can always achieve n%2 length of the final string? (where 'n' is the length of the given string) Please help.
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In C, how do you prove that if no character appears more than floor(n/2) times, we can always achieve n%2 length of the final string? (where 'n' is the length of the given string) Please help.
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Let
First, notice that we can do some operation exactly when maxf1<n.
Case 1: n\equiv0\pmod{2}
Claim: If \mathrm{maxf}_1 \le n/2 and n > 0, we can do an operation such that after the operation, \mathrm{maxf}_1 \le n/2.
Proof:
This means that if we have \mathrm{maxf}_1 \le n/2 and \mathrm{maxf}_1 < n, we can do an operation and keep the first condition satisfied. Thus, we can do operations while n/2 < n\ \Leftrightarrow\ 0 < n/2\ \Leftrightarrow\ n > 0. Thus we can always reach the state with n = 0. \square
Case 2: n\equiv1\pmod{2}
Claim: If \mathrm{maxf}_1 \le \lceil n/2\rceil and n > 1, we can do an operation such that after the operation, \mathrm{maxf}_1 \le \lceil n/2\rceil.
Proof:
This means that if we have \mathrm{maxf}_1 \le \lceil n/2\rceil and \mathrm{maxf}_1 < n, we can do an operation and keep the first condition satisfied. Thus, we can do operations while \lceil n/2\rceil < n\ \Leftrightarrow\ (n+1)/2 < n\ \Leftrightarrow\ n+1 < 2n\ \Leftrightarrow\ 1 < 2n-n\ \Leftrightarrow\ n > 1. Thus we can always reach the state with n = 1. \square