![]() Note: All the operations are circular that is adding 1 to ‘z’ will give ‘a’ and subtracting 1. If the frequency of current character is odd, then decrement current character by x. Initialize the current lowest indices of policeman in pol and thief in thi variable as -1.Ģ Find the lowest index of policeman and thief.ģ If lowest index of either policeman or thief remain -1 then return 0.Ĥ If |pol – thi| <=k then make an allotment and find the next policeman and thief.ĥ Else increment the min(pol, thi) to the next policeman or thief found.Ħ Repeat the above two steps until we can find the next policeman and thief.īelow is the implementation of the above algorithm. Given a string s, the task is to encrypt the string in the following way: If the frequency of current character is even, then increment current character by x.
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