Count of integers in a range which have even number of odd digits and odd number of even digits

Difficulty Level : Expert
Last Updated : 13 Feb, 2020

Given a range [L, R], the task is to count the numbers which have even number of odd digits and odd number of even digits. For example,

8 has 1 even digit and 0 odd digit – Satisfies the condition since 1 is odd and 0 is even.
545 has 1 even digit and 2 odd digits – Satisfies the condition since 1 is odd and 2 is even.
4834 has 3 even digits and 1 odd digit – Does not satisfy the condition since there are odd numbers(i.e 1) of odd digits.

Examples:

Input: L = 1, R = 9
Output: 4
2, 4, 6 and 8 are the only integers from the
given range that satisfy the given conditions.

Input: L = 1, R = 19
Output: 4

Input: L = 123, R = 984
Output: 431

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

Case 1
There is a pattern in which these numbers occur between and (where 1<=k<=18).
Number of occurrences from
- 1 – 10 and 1 – 100 are 4
- 1 – 1000 and 1 – 10000 are 454
- 1 – 10000 and 1 – 100000 are 45454
and so on…
Case 2
- If the number of digits in a number is even then it cannot satisfy the given condition because we need an odd number(of digits) and an even number(of digits) to satisfy our condition and odd number + even number is always odd
- So if the number of digits for a given number(say n) is even then its number of occurrences from 1 is equal to the number of occurrences from $1$ to largest $10^{k}$ (1<=k<=18) which is less than n
- Example:
  Let n = 19, number of digits in 19 are 2
  Therefore number of occurrences from 1 – 19 = number of occurrences from 1 – 10 (since 10 the largest $10^{k}$ less than 19)
Case 3
If number of digits for a given number(say n) are odd then number of occurrences between $10^{k}+1$ and n is equal to

$\LARGE {\color{DarkGreen} occurrences = \begin{cases} &\left \lfloor ( n - ( 10^{i} + 1 ) )/2 \right \rfloor + 1; \text{ if } (10^{i}+1)\text{ or } {n} \text{ satisfy the condition} \\ &\left \lfloor ( n - ( 10^{i} + 1 ) )/2 \right \rfloor + 0; \text{ otherwise } \end{cases}}$

where $10^{i}$ is the largest $10^{k}$ less than n.

Implementation: Now we now how to calculate the number of occurrences from 1 to given n. Therefore,
Number of occurrences from L to R = NumberOfOccurrencesUpto(R) – NumberOfOccurrencesUpto(L – 1) where L is not equal to 1.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach 
#include <bits/stdc++.h> 
using namespace std; 
  
#define ll long long 
  
// Pattern table from Case 1 
map<ll, ll> values{{1, 0}, 
                   {10, 4}, 
                   {100, 4}, 
                   {1000, 454}, 
                   {10000, 454}, 
                   {100000, 45454}, 
                   {1000000, 45454}, 
                   {10000000, 4545454}, 
                   {100000000, 4545454}, 
                   {1000000000, 454545454}, 
                   {10000000000, 454545454}, 
                   {100000000000, 45454545454}, 
                   {1000000000000, 45454545454}, 
                   {10000000000000, 4545454545454}, 
                   {100000000000000, 4545454545454}, 
                   {1000000000000000, 454545454545454}, 
                   {10000000000000000, 454545454545454}, 
                   {100000000000000000, 45454545454545454}, 
                   {1000000000000000000, 45454545454545454}}; 
  
// Function that returns the number of 
// even and odd digits in val 
pair<ll, ll> count_even_odd(ll val)  
{ 
    ll even = 0, odd = 0; 
    while (val)  
    { 
        ll num = val % 10; 
        if (num % 2 == 0) 
            even++; 
        else
            odd++; 
        val /= 10; 
    } 
    return make_pair(even, odd); 
} 
  
// Function that returns True if num 
// satisfies the given condition 
bool satisfies_condition(ll num)  
{ 
    pair<ll, ll> answer = count_even_odd(num); 
    ll even = answer.first; 
    ll odd = answer.second; 
  
    if (even % 2 == 1 and  
         odd % 2 == 0) return true; 
    return false; 
} 
  
// Function to return the count of  
// numbers from 1 to val that  
// satisfies the given condition 
ll count_upto(ll val) 
{ 
    // If the value is already present 
    // in the values dict 
    if (values.find(val) != values.end())  
        return values[val]; 
  
    ll index = 1; 
    for (int i = 0;  
             i < to_string(val).length() - 1;  
             i++)  
         index *= 10; 
  
    // If the value is even 
    // Case 2 
    if (to_string(val).length() % 2 == 0)  
        return values[index]; 
  
    ll val_len = to_string(val).length(); 
    ll cnt = values[index]; 
  
    // Now the problem is to count the desired 
    // numbers from 10**(val_len-1) + 1 to val 
    ll left_end = index + 1; 
  
    // Case 3 
    // Eliminating all the even numbers 
    cnt += (val - left_end) / 2; 
    if (satisfies_condition(val) or  
        satisfies_condition(left_end))  
        cnt++; 
  
    return cnt; 
} 
  
// Driver Code 
int main()  
{ 
    // Input l and r 
    ll l = 123, r = 984; 
    ll right = count_upto(r); 
    ll left = 0; 
  
    if (l == '1') 
        left = 0; 
    else
        left = count_upto(l - 1); 
  
    cout << right - left << endl; 
    return 0; 
} 
  
// This code is contributed by 
// sanjeev2552 

Python3

# Python3 implementation of the approach 
  
# Pattern table from Case 1 
values = { 
    1: 0, 
    10: 4, 
    100: 4, 
    1000: 454, 
    10000: 454, 
    100000: 45454, 
    1000000: 45454, 
    10000000: 4545454, 
    100000000: 4545454, 
    1000000000: 454545454, 
    10000000000: 454545454, 
    100000000000: 45454545454, 
    1000000000000: 45454545454, 
    10000000000000: 4545454545454, 
    100000000000000: 4545454545454, 
    1000000000000000: 454545454545454, 
    10000000000000000: 454545454545454, 
    100000000000000000: 45454545454545454, 
    1000000000000000000: 45454545454545454, 
} 
  
# Function that returns the number of  
# even and odd digits in val 
def count_even_odd(val): 
    even = odd = 0
    while val > 0: 
        num = val % 10
        if num % 2 == 0: 
            even += 1
        else: 
            odd += 1
        val //= 10
  
    return even, odd 
  
# Function that returns True if num  
# satisfies the given condition 
def satisfies_condition(num): 
    even, odd = count_even_odd(num) 
    if even % 2 == 1 and odd % 2 == 0: 
        return True
    return False
  
  
# Function to return the count of numbers  
# from 1 to val that satisfies the given condition 
def count_upto(val): 
  
    # If the value is already present in the 
    # values dict 
    if int(val) in values: 
        return values[int(val)] 
  
    # If the value is even 
    # Case 2 
    if len(val) % 2 == 0: 
        return values[int('1' + '0' * (len(val) - 1))] 
  
    val_len = len(val) 
    count = values[int('1' + '0' * (val_len - 1))] 
  
    # Now the problem is to count the desired 
    # numbers from 10**(val_len-1) + 1 to val 
    left_end = int('1' + '0' * (val_len - 1)) + 1
  
    # Case 3 
    # Eliminating all the even numbers 
    count += (int(val) - left_end) // 2
  
    if satisfies_condition(int(val)) or satisfies_condition(left_end): 
        count += 1
    return count 
  
  
if __name__ == '__main__': 
  
    # Input L and R ( as a string ) 
    l, r = '123', '984'
  
    right = count_upto(r) 
  
    left = 0
    if(l == '1'): 
        left = 0
    else: 
        left = count_upto(str(int(l)-1)) 
  
    print(right - left) 

Output:

Time Complexity: O(logn)

My Personal Notes

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Python3