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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