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++
// 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)
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