Approach 1
This is the brute force approach. The algorithm is as follows:
- Iterate from i = L to i = R.
- Initialize the answer as 0.
- Calculate the sum of digits of ‘i’.
- If the sum of digits is equal to ‘Y’, increment the answer.
- Return answer.
Now let us implement the above approach in C++ programming language.
Program
#include <iostream>
using namespace std;
// Function to check if the sum of digits of X is equal to Y.
bool checkSumDigit(int x, int y)
{
int sum = 0;
// Calculating sum of digits of 'X'.
while (x > 0)
{
sum += x % 10;
x /= 10;
}
// Checking if the sum of digits is equal to 'Y'.
return (sum == y);
}
// Function that returns the count of numbers from the range [L, R] whose sum of Digits is Y.
int countNumbersY(int L, int R, int Y)
{
int answer = 0;
// Iterating from 'L' to 'R'.
for (int i = L; i <= R; i++)
{
// Checking if the sum of digits of 'i' is equal to 'Y'.
if (checkSumDigit(i, Y))
{
// If yes, increment the answer.
answer++;
}
}
return answer;
}
int main()
{
int L, R, Y;
cout << " Enter the Value of L, R, Y: ";
cin >> L >> R >> Y;
cout << "Count of numbers from the range [L, R] whose sum of digits is Y: " << countNumbersY(L, R, Y) << endl;
return 0;
}
You can also try this code with Online C++ Compiler
Input
Enter the value of L, R, Y: 10 100 7
Output
Count of numbers from the range [L, R] whose sum of digits is Y: 7
Time Complexity
O((log(R!) / (L-1)!), where L and R are the integer ranges.
The time complexity checkSumDigit Function is O(log X), and Since checkSumDigit Function is called for every value of between ‘L’ to ‘R’, So the Time Complexity of checkNumberY is.
T(L, R) = log(L) + log(L + 1) + log(L + 2) + . . . + log(R - 1) + log(R)
T(L, R) = [log(1) + log(2) + log(3) + . . . + log(R)] - [log(1) + log (2) + log(3) + . . . + log(L - 1)]
T(L, R) = log(1 * 2 * 3 * . . . * R) - log(1 * 2 * 3 * . . . * (L - 1))
T(L, R) = log(R!) - log((L - 1)!)
T(L, R) = log(R! / (L - 1)!)
Space Complexity
O(1)
We are only declaring a fixed number of variables. So it takes constant space.
Approach 2
Here we will use the concept of digit DP. Digit DP can be used to solve the problems where we need to query over a range. This problem falls into the category of digit DP. Please have a look at the Digit DP blog. We first try to write a recursive solution to the problem and then try to apply digit DP to optimize it. We can choose from 0 to 9 for all three digits. That is, we will choose a digit for each position and then recurse for the other digits to build the sum equal to Y. Given below is the function definition.
helper(i, sum, tight) = i=09 helper(i+1, (sum - i), tight(i == end))
where sum denotes the total number of digits.
tight: Determine whether the total of the digits exceeds Y.
end: Stores the maximum potential value of a number's ith digit.
helper(N, Y, tight): Returns the number of numbers in the range [0, X] with a digit sum of Y.
Use DP to solve the problem by following the steps below.
- To compute and store the values of all subproblems of the aforementioned recurrence relation, create a 3D array called dp[N][Y][tight].
- Finally, dp[N][sum][tight] should be returned.
Program
#include <iostream>
#include <cstring>
using namespace std;
#define N 1000
// Function that counts the numbers in range 0 to STR whose sum of digits is SUM.
int helper(string str, int i, int sum, int tight, int dp[N][N][2])
{
// Check if count of digits in a number is
// greater than count of digits in str
if (i >= str.size() || sum < 0)
{
// If sum of digits of a
// number is equal to Y
if (sum == 0)
{
return 1;
}
return 0;
}
// Check if current subproblem has
// already been computed
if (dp[sum][i][tight] != -1)
{
return dp[sum][i][tight];
}
// Stores count of numbers whose
// sum of digits is Y
int ans = 0;
// Check if the number
// exceeds Y or not
int end = tight ? str[i] - '0' : 9;
// Iterate over all possible
// values of i-th digits
for (int j = 0; j <= end; j++)
{
// Update ans
ans += helper(str, i + 1, sum - j,
(tight & (j == end)), dp);
}
// Return ans
return dp[sum][i][tight] = ans;
}
// Function that returns the count of numbers from the range [L, R] whose sum of Digits is Y.
int countNumbersY(int L, int R, int Y)
{
if (R == 0 && Y == 0)
{
return 1;
}
// Stores numbers in the form
// of its equivalent string
string str = to_string(R);
// Stores overlapping subproblems
int dp[N][N][2];
// Initialize dp[][][]
memset(dp, -1, sizeof(dp));
// Stores count of numbers
// in the range [0, R]
int rightCount = helper(str, 0, Y,
true, dp);
// Update str
str = to_string(L - 1);
// Initialize dp[][][]
memset(dp, -1, sizeof(dp));
// Stores count of numbers in
// the range [0, L - 1]
int leftCount = helper(str, 0, Y,
true, dp);
return (rightCount - leftCount);
}
int main()
{
int L, R, Y;
cout << " Enter the Value of L, R, Y: ";
cin >> L >> R >> Y;
cout << "Count of numbers from the range [L, R] whose sum of digits is Y: " << countNumbersY(L, R, Y) << endl;
return 0;
}
You can also try this code with Online C++ Compiler
Input
Enter the value of L, R, Y: 20 70 10
Output
Count of numbers from the range [L, R] whose sum of digits is Y: 5
Time Complexity
O(Y * log(R)), where ‘Y’ is the sum and ‘R’ is the Upper Bound.
Space Complexity
O(Y * log(R)), where ‘Y’ is the sum and ‘R’ is the Upper Bound.
Check out this problem - Pair Sum In Array.
Also Read - Strong number in c
Key takeaways
In this blog, we learned about a new problem namely the Count of numbers from the range [L, R] whose sum of digits is Y. We solved it using two different methods. The first method is Brute force and gives a time complexity of O((log(R!) / (L-1)!). The second approach used dynamic programming and gave a time complexity of O(Y * log(R)). Stay tuned for more such problems.
240+ registered 
