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++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count total number
// of digits from numbers 1 to N
int totalDigits(int N)
{

    int cnt = 0LL;
    for (int i = 1; i <= N; i *= 10)
        cnt += (N - i + 1);

    return cnt;
}

// Function to find Maximum length of
// Sequence that can be formed from cost
// N
void findMaximumLength(int N)
{

    int low = 1, high = 1e9;

    while (high - low > 1) {
        int mid = low + (high - low) / 2;

        // Check if cost for number of digits
        // from 1 to N is less than equal to N
        if (totalDigits(mid) <= N) {

            // atleast mid will be the answer
            low = mid;
        }
        else {

            // igonre right search space
            high = mid - 1;
        }
    }

    // Check if high can be the answer
    if (totalDigits(high) <= N)
        cout << high << endl;

    // else low can be the answer
    else if (totalDigits(low) <= N)
        cout << low << endl;

    // else answer will be zero.
    else
        cout << 0 << endl;
}

// Driver Code
int main()
{

    int N = 11;

    // Function Call
    findMaximumLength(N);

    int N1 = 189;

    // Function call
    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)

Related Articles: 

• Binary Search