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Tuesday, January 31, 2023

Maximum length of sequence formed from cost N


Given  N coins, the sequence of numbers consists of {1, 2, 3, 4, ……..}. The cost for choosing a number in a sequence is the number of digits it contains. (For example cost of choosing 2 is 1 and for 999 is 3), the task is to print the Maximum number of elements a sequence can contain.

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

Examples: 

Input: N = 11
Output: 10
Explanation: For N = 11 -> selecting 1 with cost 1,  2 with cost 1,  3 with cost 1,  4 with cost 1,  5 with cost 1,  6 with cost 1,  7 with cost 1,  8 with cost 1,  9 with cost 1, 10 with cost 2.
totalCost = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 2  = 11.

Input: N = 189
Output: 99

Naive approach: The basic way to solve the problem is as follows:

Iterate i from 1 to infinity and calculate the cost for current i if the cost for i is more than the number of coins which is N then i – 1 will be the answer.

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

Efficient Approach: The above approach can be optimized based on the following idea:

This Problem can be solved using Binary Search. A number of digits with given cost is a monotonic function of type T T T T T F F F F. Last time the function was true will generate an answer for the Maximum length of the sequence. 

Follow the steps below to solve the problem:

  • If the cost required for digits from 1 to mid is less than equal to N update low with mid.
  • Else high with mid – 1 by ignoring the right part of the search space.
  • For printing answers after binary search check whether the number of digits from 1 to high is less than or equal to N if this is true print high
  • Then check whether the number of digits from 1 to low is less than or equal to N if this is true print low.
  • Finally, if nothing gets printed from above print 0 since the length of the sequence will be 0.

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>

using 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, high = 1e9;

  

    while (high - low > 1) {

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

  

        

        

        if (totalDigits(mid) <= N) {

  

            

            low = mid;

        }

        else {

  

            

            high = mid - 1;

        }

    }

  

    

    if (totalDigits(high) <= N)

        cout << high << endl;

  

    

    else if (totalDigits(low) <= N)

        cout << low << endl;

  

    

    else

        cout << 0 << endl;

}

  

int main()

{

  

    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 finding the number of digits from 1 to N)
Auxiliary Space: O(1)

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