Number of n-digits non-decreasing integers
Given an integer n > 0, which denotes the number of digits, the task to find the total number of n-digit positive integers which are non-decreasing in nature.
A non-decreasing integer is one in which all the digits from left to right are in non-decreasing form. ex: 1234, 1135, ..etc.
Note: Leading zeros also count in non-decreasing integers such as 0000, 0001, 0023, etc are also non-decreasing integers of 4-digits.
Input : n = 1 Output : 10 Numbers are 0, 1, 2, ...9. Input : n = 2 Output : 55 Input : n = 4 Output : 715
Naive Approach: We generate all possible n-digit numbers and then for each number we check whether it is non-decreasing or not.
Time Complexity : (n*10^n), where 10^n is for generating all possible n-digits numbers and n is for checking whether a particular number is non-decreasing or not.
If we fill digits one by one from left to right, the following conditions hold.
- If current last digit is 9, we can fill only 9s in remaining places. So only one solution is possible if current last digit is 9.
- If current last digit is less than 9, then we can recursively compute count using following formula.
a[i][j] = a[i-1][j] + a[i][j + 1] For every digit j smaller than 9. We consider previous length count and count to be increased by all greater digits.
We build a matrix a where a[i][j] = count of all valid i-digit non-decreasing integers with j or greater than j as the leading digit. The solution is based on below observations. We fill this matrix column-wise, first calculating a then using this value to compute a and so on.
At any instant if we wish to calculate a[i][j] means number of i-digits non-decreasing integers with leading digit as j or digit greater than j, we should add up a[i-1][j] (number of i-1 digit integers which should start from j or greater digit, because in this case if we place j as its left most digit then our number will be i-digit non-decreasing number) and a[i][j+1] (number of i-digit integers which should start with digit equals to greater than j+1). So, a[i][j] = a[i-1][j] + a[i][j+1].
Non-decreasing digits = 55
Time Complexity : O(10*n) equivalent to O(n).
If we observe, we can see that 0 has to be placed before 1-9, 1 has to be placed before 2-9 and so on. As we are asked to find non-decreasing integers, 111223 is a valid non-decreasing integer which means same digit can occur conscuetively.
Example 1: When N=2, we have 11C9, which is equal to 55.
Example 2: When N=5, we have 14C9, which is equal to 2002.
Number of Non-Decreasing digits: 24310
Time Complexity : O(n).
Space Complexity: O(n) .
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