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Most size of sequence fashioned from value N

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Most size of sequence fashioned from value N
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Given  N cash, the sequence of numbers consists of {1, 2, 3, 4, ……..}. The associated fee for selecting a quantity in a sequence is the variety of digits it comprises. (For instance value of selecting 2 is 1 and for 999 is 3), the duty is to print the Most variety of components a sequence can include.

Any aspect from {1, 2, 3, 4, ……..}. can be utilized at most 1 time. 

Examples: 

Enter: N = 11
Output: 10
Clarification: For N = 11 -> deciding on 1 with value 1,  2 with value 1,  3 with value 1,  4 with value 1,  5 with value 1,  6 with value 1,  7 with value 1,  8 with value 1,  9 with value 1, 10 with value 2.
totalCost = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 2  = 11.

Enter: N = 189
Output: 99

Naive strategy: The essential method to resolve the issue is as follows:

Iterate i from 1 to infinity and calculate the fee for present i if the fee for i is greater than the variety of cash which is N then i – 1 would be the reply.

Time Complexity: O(N * logN)
Auxiliary House: O(1)

Environment friendly Strategy: The above strategy might be optimized primarily based on the next concept:

This Downside might be solved utilizing Binary Search. Quite a lot of digits with given value is a monotonic operate of kind T T T T T F F F F. Final time the operate was true will generate a solution for the Most size of the sequence. 

Observe the steps beneath to resolve the issue:

  • If the fee required for digits from 1 to mid is lower than equal to N replace low with mid.
  • Else excessive with mid – 1 by ignoring the fitting a part of the search area.
  • For printing solutions after binary search test whether or not the variety of digits from 1 to excessive is lower than or equal to N if that is true print excessive
  • Then test whether or not the variety of digits from 1 to low is lower than or equal to N if that is true print low.
  • Lastly, if nothing will get printed from above print 0 for the reason that size of the sequence will probably be 0.

Beneath is the implementation of the above strategy:

C++

#embrace <bits/stdc++.h>

utilizing namespace std;

  

int totalDigits(int N)

{

  

    int cnt = 0LL;

    for (int i = 1; i <= N; i *= 10)

        cnt += (N - i + 1);

  

    return cnt;

}

  

void findMaximumLength(int N)

{

  

    int low = 1, excessive = 1e9;

  

    whereas (excessive - low > 1) {

        int mid = low + (excessive - low) / 2;

  

        

        

        if (totalDigits(mid) <= N) {

  

            

            low = mid;

        }

        else {

  

            

            excessive = mid - 1;

        }

    }

  

    

    if (totalDigits(excessive) <= N)

        cout << excessive << endl;

  

    

    else if (totalDigits(low) <= N)

        cout << low << endl;

  

    

    else

        cout << 0 << endl;

}

  

int important()

{

  

    int N = 11;

  

    

    findMaximumLength(N);

  

    int N1 = 189;

  

    

    findMaximumLength(N1);

  

    return 0;

}

Time Complexity: O(logN2)  (first logN is for logN operations of binary search, the second logN is for locating the variety of digits from 1 to N)
Auxiliary House: O(1)

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